chap 9

Fundamentals of HypothesisTesting: One-sample Tests

USING STATISTICS @ Oxford Cereals, Part II

ESIS.TESTING M ETHODOLOGY 9.4Null and Alternative HypothesesCritical Value of the Test Statistic

of Rejection and Nonrejectionin Decision Making Using Hypothesis-

Testing Methodology

TEST OF HYPOTHESISTHE MEAN (o KNOWN)

Critical Value Approach to Hypothesis Testingp-Value Approach to Hypothesis Testing

Connection Between Confidence IntervalEstimation and Hypothesis Testing

.TAIL TESTSCritical Value Approachp-Value Approach

9.5

9.6

t TEST OF HYPOTHESISFOR THE MEAN (o UNKNOWN)The Critical Value ApproachThep-Value ApproachChecking Assumptions

ZTEST OF HYPOTHESISFOR THE PROPORTIONThe Critical Value ApproachThep-Value Approach

POTENTIAL HYPOTH ESIS.TESTI NGPITFALLS AND ETHICAL IssUES

9.7 C' (CD.ROM TOPIq THE POWER OF A TEST

EXCEL COMPANION TO CHAPTER 9E9.l Using the ZTest for the Mean (o Known)E9.2 Using the I Test for the Mean (o Unknown)E9.3 Using the ZTest for the Proportion

In this chapter, you learn:r The basic principles of hypothesis testingI How to use hypothesis testing to test a mean or proportionr The assumptions of each hypothesistesting procedure, how to evaluate them,

and the consequences ifthey are seriously violatedr How to avoid the pitfalls involved in hypothesis testingr Ethical issues involved in hypothesis testing

328 CHAPTER NINE Fundamentals of Hypothesis Testing: One-Sample Tests

Using Statistics @ Oxford Cereals, Part l l

As in Chapter 7, you again find yourself as plant operationsOxford Cereals. You are responsible for monitoring the amount incereal box filled. Company specifications require a mean weight ofgrams per box. It is your responsibility to adjust the process whenmean fill weight in the population of boxes deviates from 368 grams.can vou rationallv make the decision whether or not to adiust thewhen it is impossible to weigh every single box as it is being filled?begin by selecting and weighing a random sample of 25 cerealAfter computing a sample mean, how do you proceed?

]n Chapter 7. you learned methods to determine whether a sample mean is consistentIknown population mean. In this Oxford Cereals scenario, you seek to use a sample nvalidate a claim about the population mean, a somewhat different problem. For thisproblem, you use an inferential method called hypothesis testing. Hypothesis testingthat you state a claim unambiguously. In this scenario, the claim is that the population368 grams. You examine a sample statistic to see if it better supports the stated claim,null hypothesis, or the mutually exclusive alternative (for this scenario, that themean is not 368 grams.)

In this chapter, you will learn several applications of hypothesis testing. You will learnto make inferences about a population parameter by analyzing dffirences between theobserved, the sample statistic, and the results you would expect to get if somehypothesis were actually true. For the Oxford Cereals scenario, hypothesis testing wouldyou to infer one of the following:

r The mean weight of the cereal boxes in the sample is a value consistent withwould expect if the mean of the entire population of cereal boxes is 368 grams.

r The population mean is not equal to 368 grams because the sample mean is sidifferent from 368 srams.

9.'I HYPOTHESIS.TESTING METHODOLOGYHypothesis testing typically begins with some theory, claim, or assertion about a particularmeter of a population. For example, your initial hypothesis about the cereal example isprocess is working properly, so the mean fill is 368 grams, and no corrective action is

The Null and Alternative HypothesesThe hypothesis that the population parameter is equal to the company specification isto as the null hypothesis. A null hypothesis is always one of status quo and is identifiedsymbol Ho. Here the null hypothesis is that the filling process is working properly, andfore the mean fill is the 368-gram specification. This is stated as

/10 : p :368

PtE 9 .1

9.1: Hypothesis-GstingMethodology 329

Even though information is available only from the sample, the null hypothesis is written interms of the population. Remember, your focus is on the population of all cereal boxes. Thesample statistic is used to make inferences about the entire filling process. One inference maybe that the results observed from the sample data indicate that the null hypothesis is false. If thenull hypothesis is considered false, something else must be true.

Whenever a null hypothesis is specified, an alternative hypothesis is also specified, and itmust be true if the null hypothesis is false. The alternative hypothesis,Ilr, is the opposite ofthe null hypothesis, /10. This is stated in the cereal example as

Hr :1 : " *368

The alternative hypothesis represents the conclusion reached by rejecting the null hypothesis. Thenull hypothesis is rejected when there is sufficient evidence from the sample information that thenull hypothesis is false. In the cereal example, if the weights of the sampled boxes are sufficientlyabove or below the expected 368-gram mean specified by the company, you reject the null hypoth-esis in favor of the alternative hypothesis that the mean fill is different from 368 grams. You stopproduction and take whatever action is necessary to correct the problem. If the null hypothesis isnot rejected, you should continue to believe in the status quo, that the process is working correctlyand therefore no corrective action is necessary. In this second circumstance, you have not proventhat the process is working correctly. Ratheq you have failed to prove that it is working incorrectly,and therefore you continue your belief(although unproven) in the null hypothesis.

In the hypothesis-testing methodology, the null hypothesis is rejected when the sample evi-dence suggests that it is far more likely that the alternative hypothesis is true. However, failureto reject the null hypothesis is not proofthat it is true. You can never prove that the null hypoth-esis is correct because the decision is based only on the sample information, not on the entirepopulation. Therefore, ifyou fail to reject the null hypothesis, you can only conclude that thereis insufficient evidence to warrant its rejection. The following key points summarize the nulland alternative hypotheses:

. The null hypothesis, i1s, represents the status quo or the current beliefin a situation.' The alternative hypothesis, F/,, is the opposite ofthe null hypothesis and represents a

research claim or specific inference you would like to prove., Ifyou reject the null hypothesis, you have statistical proofthat the alternative hypothesis is

correct.r Ifyou do not reject the null hypothesis, you have failed to prove the alternative hypothesis.

The failure to prove the alternative hypothesis, however, does not mean that you haveproven the null hypothesis.

r The null hypothesis, 116, always refers to a specified value of the population parameter(such as p), not a sample statistic (such as X).

r The statement of the null hypothesis always contains an equal sign regarding the specifiedvalue of the population parameter (for example, Ho: $:368 grams).

. The statement of the alternative hypothesis never contains an equal sign regarding thespecified value of the population parameter (for example, Hi V * 368 grams).

THE NULL AND ALTERNATIVE HYPOTHESES

You are the manager of a fast-food restaurant. You want to determine whether the waiting timeto place an order has changed in the past month from its previous population mean value of 4.5minutes. State the null and alternative hypotheses.

SOLUTION The null hypothesis is that the population mean has not changed from its previ-ous value of 4.5 minutes. This is stated as

Ho: trt:4.5


The alternative hypothesis is the opposite of the null hypothesis. Because the nullthat the population mean is 4.5 minutes, the alternative hypothesis is that the populationis not 4.5 minutes. This is stated as

Hr: 1t+ 4.5

The Critical Value of the Test StatisticThe logic behind the hypothesis-testing methodology is to determine how likely thehypothesis is to be true by considering the information gathered in a sample. In theCereal Company scenario, the null hypothesis is that the mean amount of cereal per boxthe entire filling process is 368 grams (that is, the population parameter specified bycompany). You select a sample of boxes from the filling process, weigh each box, andpute the sample mean. This statistic is an estimate of the corresponding parameter (theulation mean, p). Even if the null hypothesis is true, the statistic (the sample mean,likely to differ from the value of the parameter (the population mean, p) because of varidue to sampling. However, you expect the sample statistic to be close to the populationmeter if the null hypothesis is true. If the sample statistic is close to the populationter, you have insufficient evidence to reject the null hypothesis. For example, if themean is 367.9, you conclude that the population mean has not changed (that is, p:because a sample mean of 367.9 is very close to the hypothesized value of 368. Intuiyou think that it is likely that you could get a sample mean of 367.9 from a populationmean is 368.

However, if there is a large difference between the value of the statistic and thesized value of the population parameter, you conclude that the null hypothesis is false.example, if the sample mean is 320, you conclude that the population mean is not 368 (thatp I 368), because the sample mean is very far from the hypothesized value of 368. In suchcase, you conclude that it is very unlikely to get a sample mean of 320 if the populationis really 368. Therefore, it is more logical to conclude that the population mean is not equal368. Here you reject the null hypothesis.

Unfortunately, the decision-making process is not always so clear-cut. Determiningis "very close" and what is "very different" is arbitrary without clear definitions. Htesting methodology provides clear definitions for evaluating differences. Furthermore,enables you to quantify the decision-making process by computing the probability of gettinggiven sample result if the null hypothesis is true. You calculate this probability by determinithe sampling distribution for the sample statistic of interest (for example, the sampleand then computing the particular test statistic based on the given sample result. Becausesampling distribution for the test statistic often follows a well-known statistical distributisuch as the standardized normal distribution or I distribution. vou can use these distributito help determine whether the null hypothesis is true.

Regions of Rejection and NonrejectionThe sampling distribution of the test statistic is divided into two regions, a region of re(sometimes called the critical region) and a region of nonrejection (see Figure 9.1).

Ifthe test statistic falls into the region ofnonrejection, you do not reject the null hsis. In the Oxford Cereals scenario. vou conclude that there is insufficient evidence thatpopulation mean fill is different from 368 grams. If the test statistic falls into the rejregion, you reject the null hypothesis. In this case, you conclude that the population mean is368 erams.

9.1ions of rejectionnonrejection in

is testing

9.1: Hypothesis-Testing Methodology 33 I

CriticalValue

Region ofRejection

Region ofRejection

Crit icalVa lue

\ t r

Region ofNonrejection

The region of rejection consists of the values of the test statistic that are unlikely to occurif the null hypothesis is true. These values are more likely to occur if the null hypothesis isfalse. Therefore, ifa value ofthe test statistic falls into this rejection region, you reject the nullhypothesis because that value is unlikely if the null hypothesis is true.

To make a decision concerning the null hypothesis, you first determine the critical valueofthe test statistic. The critical value divides the nonrejection region from the rejection region.Determining this critical value depends on the size of the rejection region. The size of the rejec-tion region is directly related to the risks involved in using only sample evidence to make deci-sions about a population parameter.

Risks in Decision Making Using Hypothesis-Testing MethodologyWhen using a sample statistic to make decisions about a population parameter, there is a riskthat you will reach an incorrect conclusion. You can make two different types of errors whenapplying hypothesis-testing methodology, Type I and Type II errors.

A Type I error occurs ifyou reject the null hypothesis, I1o, when it is true and should notbe rejected. The probability of a Type I error occurring is c.A Type II error occurs ifyou do not reject the null hypothesis, F16, when it is false andshould be rejected. The probability ofa Type II error occurring is p.

In the Oxford Cereals scenario, you make a Type I error if you conclude that the populationmean fill is not 368 when it ls 368. This error causes you to adjust the filling process eventhough the process is working properly. You make a Type II error if you conclude that the pop-ulation mean fill ls 368 when itis not 368. Here, you would allow the process to continue with-out adjustment even though adjustments are needed.

The Level of Signiftcance (a) The probability of committing a Type I error, denoted bya (the lowercase Greek letter alpha), is referred to as the level of significance of the statisticaltest. Traditionally, you control the Type I error by deciding the risk level, o, that you are willingto have in rejecting the null hypothesis when it is true. Because you speci$ the level of signif-icance before the hypothesis test is performed, the risk of committing a Type I error, cr, isdirectly under your control. Traditionally, you select levels of 0.01 , 0.05, or 0. 10. The choice ofa particular risk level for making a Type I error depends on the cost of making a Type I error.After you specify the value for ct, you can then determine the critical values that divide therejection and nonrejection regions. You know the size ofthe rejection region because c is theprobability of rejection when the null hypothesis is true. From this, you can then determinethe critical value or values that divide the reiection and nonreiection resions.


The ConJidence Cofficienl The complement of the probability of aType I error, (1is called the confidence coefficient. When multiplied by l00o/o, the confidence coefficientthe confidence level that was studied when constructins confidence intervals (see Section 8.

The confidence coefficient, (1 - cr), is the probability that you will not reject the nullhypothesis, F1o, when it is true and should not be rejected. The confidence level of ahypothesis test is (l - cr) x 100%.

In terms of hypothesis-testing methodology. the confidence coefficient representsprobability of concluding that the value of the parameter as specified in the nullplausible when it is true. In the Oxford Cereals scenario, the confidence coefficientthe probability of concluding that the population mean fill is 368 grams when it is368 grams.

The $ Risk The probability of committing a Type II error is denoted by B (theGreek letter beta). Unllke a Type I error, which you control by the selection of cr, theity of making a Type II error depends on the difference between the hypothesized andvalues of the population parameter. Because large differences are easier to find thanones, ifthe difference between the hypothesized and actual values ofthe populationis large, B is small. For example, if the population mean is 330 grams, there is a small(P) that you will conclude that the mean has not changed from 368. However, if the dibetween the hypothesized and actual values of the parameter is small, B is large. Forif the population mean is actually 367 grams, there is alarge chance (B) that you willthat the mean is still 368 srams.

The Power of a Test The complement of the probability of a Type II error, (l - p), isthe power of a statistical test.

The power of a statistical test, (l - B), is the probability that you will reject the nullhypothesis when it is false and should be rejected.

In the Oxford Cereals scenario, the power of the test is the probability that you willrectly conclude that the mean fill amount is not 368 grams when it actually is not 368For a detailed discussion of the power of the test, see Section 9.7 on the Student CD-ROM.

Risfts in Decision Making: A Delicste Balance Table 9.1 illustrates thethe two possible decisions (do not reject /10 or reject H) that you can make in anytest. You can make a correct decision or make one of two types of errors.

T A B L E 9 . 1Hypothesis Testingand Decision Making

Actual Situation

Statistical I)ecision /In True I1n False

Do not reject Ho Correct decisionConfidence: (l - cr)

Type II errorP (Type II error) = p

Reject Ho Type I errorP (Type I error) : o

Correct decisionPower: (l - p)

One way to reduce the probability of making a Type II error is by increasing thesize. Large samples generally permit you to detect even very small differences betweenhypothesized values and the population parameters. For a given level of o,, increasing theple size decreases B and therefore increases the power ofthe test to detect that the null

9.1: Hypothesis-Testing Methodology 333

esiS, 116, is false. However, there is always a limit to your resources, and this affects the decisionas to how large a sample you can take. Thus, for a given sample size, you must consider thetrade-offs between the two possible types of errors. Because you can directly control the risk ofType I error, you can reduce this risk by selecting a smaller value for o. For example, if the neg-ative consequences associated with making a Type I error are substantial, you could select a :

0.01 instead of 0.05. However, when you decrease cx, you increase B, so reducing the risk of aType I error results in an increased risk of a Type II error. However, if you wish to reduce F, youcould select a larger value for cr. Therefore, if it is important to try to avoid a Type II error, youcan select o of 0.05 or 0.10 instead of 0.01.

In the Oxford Cereals scenario, the risk of a Type I error involves concluding that the meanfill amount has changed from the hypothesized 368 grams when it actually has not changed.The risk of a Type II error involves concluding that the mean fill amount has not changed fromthe hypothesized 368 grams when it actually has changed. The choice of reasonable values fora and p depends on the costs inherent in each type of error. For example, if it is very costly tochange the cereal-fill process, you would want to be very confident that a change is neededbefore making any changes. In this case, the risk of a Type I error is more important, and youwould choose a small cr. However, if you want to be very certain of detecting changes from amean of 368 grams, the risk of a Type II error is more important, and you would choose ahigher level of cr.

Applying the Conceptsthe Basics

9.1 You use the symbol l'1n for which hy-pothesis?

9.2 You use the symbol H, for which hy-pothesis?

9.3 What symbol do you use for the chance ofcommitting a Type I error?

9.4 What symbol do you use for the chance ofcommitting a Type II error?

9.5 What does I - cr represent?

What is the relationship of o to a Type I error?

What is the relationship of B to a Type II error?

How is power related to the probability of making aII error?

9.9 Why is it possible to reject the null hy-pothesis when it is true?

9.10 Why is it possible to not reject the nullhypothesis when it is false?

11 For a given sample size, if cr is reduced from 0.05 tol, what happens to p?

12 ForHr : p :100, Hr : t r t * l00 ,andforasampleo f s izewhy is p larger ifthe actual value ofp is 90 than ifthe

9.13 In the U.S. legal system, a defendant ispresumed innocent until proven guilty. Considera null hypothesis, fls, that the defendant is inno-

cent, and an alternative hypothesis, Hr,that the defendantis guilty. A jury has two possible decisions: Convict thedefendant (that is, reject the null hypothesis) or do not con-vict the defendant (that is, do not reject the null hypothe-sis). Explain the meaning of the risks of committing eithera Type I or Type II error in this example.

9.14 Suppose the defendant in Problem 9.13 is pre-sumed guilty until proven innocent, as in some other judi-cial systems. How do the null and alternative hypothesesdiffer from those in Problem 9.13? What are the meaningsof the risks of committing either a Type I or Type II errorhere?

9.15 The U.S. Food and Drug Administration (FDA) isresponsible for approving new drugs. Many consumergroups feel that the approval process is too easy and" there-fore, too many drugs are approved that are later found to beunsafe. On the other hand, a number of industry lobbyistsare pushing for a more lenient approval process so thatpharmaceutical companies can get new drugs approvedmore easily and quickly (R. Sharpe, "FDA Tries to FindRight Balance on Drug Approvals," The Wall StreetJournal,April20, 1999,p. A24). Consider a null hypothe-sis that a new, unapproved drug is unsafe and an alternativehypothesis that a neq unapproved drug is safe.value of u is 75?

a.

b.

334 CHAPTER NINE Fundamentals of Hvoothesis Testinc: One-Samplc Tcsts

Explain the risks of committing a Type I or Type II error.Which type of error are the consumer groups trying toavoid? Explain.

c. Which type of error are the industry lobbyists tryrng toavoid'? Explain.

d. How would it be possible to lower the chances of bothType I and Type Il errors'?

9.16 As a resul t o f cornpla ints f rom both stu-dents and faculty about lateness, the registrar at alarge university wants to adjust the scheduledclass times to allow for adequate travel t ime

between classes and is ready to undertake a study. Untilnoq the registrar has believed that there should be 20 min-utes between scheduled classes. State the null hvpothesis.Hn, and the alternative hypothesis, H,.

9.17 The operations manager at a clothing factoryto detennine whether a new machine is producing aular tvne of cloth accordins to the manufacturer'scations, which indicate that the cloth should have abreaking strength of 70 pounds. State the null and altive hvootheses.

9.18 The manager of a paint supply store wants t0mine whether the amount of oaint contained in 1cans purchased from a nationally known manufiactually averages I gallon. State the null and alhypotheses.

9.19 The aualitv control manaser at a lisht bulbneeds to determine whether the mean life of a largement of l ight bulbs is equal to the specified value ofhours. State the null and alternative hvootheses. !

ffiT/-sEAffi

9.2 ZTEST OF HYPOTHESIS FOR THE MEAN (o KNOWN)Now that you have been introduced to hypothesis-testing methodology, recall that in the UskStatistics scenario on page 328, Oxford Cereals wants to determine whether the cereal-frprocess is working properly (that is, whether the mean fill throughout the entire packagitprocess remains at the specified 368 grams, and no corrective action is needed). To evaluatetl368-gram requirement, you take a random sample of 25 boxes, weigh each box, and then eviuate the difference between the sample statistic and the hypothesized population parameterl

comparing the mean weight (in grams) from the sample to the expected mean of 368 grar

specified by the cornpany. The null and alternative hypotheses are

H6: P = 363

Hr: 'p, + 368

When the standard deviation, o, is known, you use the Z test if the population is normally dtributed. If the population is not normally distributed, you can still use the Z test if the samlsize is large enough for the Central Limit Theorem to take effect (see Section 7.4). Equati(9.1) defines the Z-test statistic for determining the difference between the sample mean,and the population mean, p, when the standard deviation, o, is known.

ZTEST OF HYPOTHESIS FOR THE MEAN (o KNOWN)

X-pZ - (e.1)

o

G

ln Equation (9.1)the numerator measures how far (in an absolute sense) the obserrsample mean, X, is fiom the hypothesized mean,lu. The denorninator is the standard errotthe mean, so Z represents the difference between X and p in standard error units.

The firitical Value Appnoach to Hypothesis Testing

The observed value of the Z test statistic, Equation (9.1), is compared to crit ical values.'

critica) values are expressed as standard\zed Z va\ues \t\at rs, rn standard der,ratron um\s)

example, if you use a level of significance of 0.05, the size of the rejection region is (

Because the rejection region is divided into the two tails of the distribution (this is called a47'22.'t2 - / .'// J) :J.'z,z:.' 27.4.2 i:)z:zt szzzt. ?11/-/,1.2-z;t.t .2//)2,2-?;z'.zzy',z . ) ,z'72'zf;),:':;,,23:-,.).:.2:t22.1

,dS

ic-fi-)anla-

.'r-on'er

ve

ryp-75

9.2: ZTest of Hypothesis for the Mean (o Known) 335

each tail of the normal distribution results in a cumulative area of 0.025 below the lower criti-cal value and a cumulative area of 0.975 below the upper critical value. According to the cumu-lative standardized normal distribution table (Table E.2), the crit ical values that divide therejection and nonrejection regions are -1.96 and +1.96. Figure 9.2 i l lustrates that if the mean isactually 368 grams, as Hn claims, the values of the test statistic Zhave a standardized normaldistribution centered at Z :0 (which corresponds to an X value of 368 grams). Values of Zgreater than +1.96 or less than -1.96 indicate that X is so far from the hypothesized p:368that it is unlikely that such a value would occur if F1n were true.

FIGURE 9.2Testing a hypothesisabout the mean(o known) at the 0.05level of s igni f icance

FIGURE 9.3Testing a hypothesisabout the mean cerealweight (o known) at 0.05level of s igni f icance

0I

IReg ion o f

NonrejectionI

i

Reg ion o fRejection

rgillrglerl-,yIS

Cr i t i ca lVa lue

Therefore, the decision rule is

Re jec t Ho l f Z >+1 .96

o r i f Z < - 1 . 9 6 ;otherwise, do not reject Hn.

Suppose that the sample of 25 cereal boxes indicates a sample mean, X, of 312.5 grams, andthe population standard deviation, o, is assumed to be l5 grams. Using Equation (9.1) on page 334,

- x -u 372 .5 -368. = - = -----------:--:_- = Tl.Jv

o l )r Jzs\l n

Because the test s tat is t ic Z - +1.50 is between -1.96 and +1.96, you do not re ject Hn (seeFigure 9.3). You continue to believe that the mean fi l l amount is 368 grams. To take intoaccount the possibility ofa Type II error, you state the conclusion as "there is insufficient evi-dence that the mean fill is different from 368 grams."

t -

en

i -1 .96

Region ofRejection

0 + '1 .50 +1.96 I ZReg ion o f Reg ion o f

Nonrejection Rejection


Exhibit 9.1 provides a summary of the critical value approach to hypothesis testing.

EXHIBIT 9.1 THE SIX-STEP METHOD OF HYPOTHESIS TESTINGl. State the null hypothesis, Hr, and the alternative hypothesis,11,.2. Choose the level of significance, o, and the sample size, n. The level of signifi

based on the relative importance of the risks of committing Type I and Type IIthe problem.Determine the appropriate test statistic and sampling distribution.Determine the critical values that divide the rejection and nonrejection regions.Collect the sample data and compute the value of the test statistic.Make the statistical decision and state the manaserial conclusion. If the testfalls into the nonrejection region, you do not reject the null hypothesis,110. Iftkstatistic falls into the rejection region, you reject the null hypothesis. Theconclusion is written in the context of the real-world oroblem.

APPLYING THE SIX-STEP METHOD OF HYPOTHESIS TESTINGAT OXFORD CEREALS

State the six-step method of hypothesis testing at Oxford Cereals.

SOLUTION

Step I State the null and alternative hypotheses. The null hypothesis, lln, is alwaysstatistical terms, using population parameters. In testing whether the mean fill isgrams, the null hypothesis states that p equals 368. The alternative hypothesis,llalso stated in statistical terms, using population parameters. Therefore, thehypothesis states that p is not equal to 368 grams.

Step 2 Choose the level of significance and the sample size. You choose the level of sicance, o, according to the relative importance of the risks of committing Type IType II errors in the problem. The smaller the value of c, the less risk there is ofing a Type I error. In this example, a Type I error is to conclude that themean is not 368 grams when it is 368 grams. Thus, you will take corrective actionthe filling process even though the process is working properly. Here, c : 0.05selected. The sample, n, is 25.

Step 3 Select the appropriate test statistic. Because o is known from information aboutfilling process, you use the normal distribution and the Z test statistic.

Step 4 Determine the rejection region. Critical values for the appropriate test statisticselected so that the rejection region contains atotal arca of cr when I1n is true andnonrejection region contains alotal area of I - cr when 11n is true. Because c =

in the cereal examDle. the critical values of the Z-test statistic are -1 .96 and +1.The rejection region is therefore Z < -1.96 or Z > +1.96. The nonrejection region- 1 . 9 6 < Z < + 1 . 9 6 .

Step 5 Collect the sample data and compute the value of the test statistic. In theexample. X :372.5. and the value of the test s tat is t ic is Z: +1.50.

Step 6 State the statistical decision and the managerial conclusion. First, determiwhether the test statistic has fallen into the rejection region or the nonrejection,region. For the cereal example, Z: +1.50 is in the region of nonrejection because-1.96 < Z-- +1.50 < +1.96. Because the test statistic falls into the nonrejection region,the statistical decision is to not reject the null hypothesis,llo. The managerial conclu-sion is that insufficient evidence exists to prove that the mean fill is different from 368grams. No corrective action on the filling process is needed.

J .

4.5 .6 .

EXAMPLE 9 .2

LE 9 .3


REJECTING A NULL HYPOTHESIS

You are the manager of a fast-food restaurant. You want to determine whether the populationmean waiting time to place an order has changed in the past month from its previous populationmean value of 4.5 minutes. From past experience, you can assume that the population is nor-mally distributed with a population standard deviation of 1.2 minutes. You select a sample of 25orders during a one-hour period. The sample mean is 5.1 minutes. Use the six-step approachlisted in Exhibit 9.I to determine whether there is evidence at the 0.05 level of significance thatthe population mean waiting time to place an order has changed in the past month from its pre-vious population mean value of 4.5 minutes.

SOLUTION

Step 1 The null hypothesis is that the population mean has not changed from its previousvalue of 4.5 minutes:

Ho: 1t:4.5

The alternative hypothesis is the opposite of the null hypothesis. Because the nullhypothesis is that the population mean is 4.5 minutes, the alternative hypothesis is thatthe population mean is not 4.5 minutes:

Hr : S t+ 4 .5

Step 2 You have selected a sample of n : 25. The level of significance is 0.05 (that is,c r : 0 . 0 5 ) .

Step 3 Because o is known, you use the normal distribution and the Z test statistic.

Step4 Because c:0.05, the cr i t ical values of the Ztest stat ist ic are -1.96 and +1.96.The reject ion region is Z < -1.96 or Z > +1.96. The nonreject ion region is- 1 . 9 6 < Z < + 1 . 9 6 .

Step 5 You collect the sample data and compute X = S.t. Using Equation (9.1) on page 334,you compute the test statistic:

x -1 tZ_

Step 6 Because Z: 2.50 > 1.96, you reject the null hypothesis. You conclude that there is evr-dence that the population mean waiting time to place an order has changed from itsprevious value of 4.5 minutes. The mean waiting time for customers is longer nowthan it was last month.

The pValue Approach to Hypothesis TestingMost modern software, including Microsoft Excel, computes the p-value when performing atest ofhypothesis.

Thep-value is the probability of getting a test statistic equal to or more extreme than thesample result, given that the null hypothesis, F16, is true.

Thep-value, often referred to as the observed level of significance, is the smallest level atwhich 110 can be rejected.

The decision rules for rejecting 110 in thep-value approach are

r If thep-value is greater than or equal to cx, do not reject the null hypothesis.I If thep-value is less than c, reject the null hypothesis.

o

G=#=2.50

E

3 3 8 CHAPTER NINE Fundamentals of Hypothesis Testing: One-Sample Tests

Many people confuse these rules, mistakenly believing that a highp-value is grounds fortion. You can avoid this confusion by remembering the following mantra:

lf thep-value is low, then 110 must go.

To understand the p-value approach, consider the Oxford Cereals scenario. Youwhether the mean fill was equal to 368 grams. The test statistic resulted in a Z va\te of *

and you did not reject the null hypothesis because +1.50 was less than the upper criticalof +1.96 and more than the lower critical value of -1.96.

To use the p-value approach for the two-tail test, yov find the probability of gettingstatistic Z that is equal to or more extreme than l.50 standard deviation units from thea standardized normal distribution. In other words, you need to compute the probabilityvalue greater than +1.50, along with the probability of a Z value less than -1.50. Tableshows that the probability of a Z value below -1.50 is 0.0668. The probability of a value+1.50 is 0.9332, and the probability of a value above +1.50 is 1 - 0.9332:0.0668.thep-value for this two-tail test is 0.0668 + 0.0668:0.1336 (see Figure 9.4). Thus, thebility of a result equal to or more extreme than the one observed is 0.1336. Because 0.1greater than cr : 0.05, you do not reject the null hypothesis.

FIGURE 9.4Finding a p-valuefor a two-tail test

-1 .50 0 +1 .50 Z

In this example, the observed sample mean is 372.5 grams,4.5 grams above thesized value, and thep-value is 0.1336. Thus, if the population mean is 368 grams, there is13.36% chance that the sample mean differs from 368 grams by more than 4.5 grams (thatiis2372.5 grams or < 363.5 grams). Therefore, while 312.5 is above the hypothesized value368, a result as extreme as or more extreme than 372.5 is not highly unlikely when thetion mean is 368.

Unless you are dealing with a test statistic that follows the normal distribution, computithe p-value can be very difficult. However, Microsoft Excel routinely computes the p-valueits hypothesis-testing procedures. Figure 9.5 displays a Microsoft Excel worksheet forcereal-filling example discussed in this section.

FIGURE 9.5Microsoft Excel Z-testresults for the cereal-fil lexamote

-s/s0RT(87)-(88 - 8tyBt1

-lloRXSlllv(85/21-HORISII{V(I - Blilzl-2' (1 - xoRxsolsr(ABspr4))-lF(817 < 85, 'ReJoct ihe null hypotheCr',

"0o not .c.rccl lic null hlrpoihcria')

.8684III

i

ftrI

Fr'Ia

9.2: ZTestof Hypothesis for the Mean (o Known) 339

Exhibit 9.2 provides a summary of thep-value approach for hypothesis testing.

EXHIBfT 9.2 THE FIVE-STEP p-VALUE APPROACH TO HYPOTHESIS TESTINGl. State the null hypothesis, Ho, and the alternative hypothesis, ,F1,.2. Choose the level of significance, d, and the sample size, n. The level of significance is

based on the relative importance of the risks of committingType I andType II errors inthe problem.

3. Determine the appropriate test statistic and sampling distribution.4. Collect the sample data, compute the value of the test statistic, and compute thep-value.5. Make the statistical decision and state the managerial conclusion. If thep-value is

greater than or equal to s, you do not reject the null hypothesis, .F10. If thep-value isless than o(, you reject the null hypothesis. Remember the mantra: If thep-value is loqthen.I/o must go. The managerial conclusion is written in the context of the real-worldproblem.

REJECTING A NULL HYPOTHESIS, USING THE p-VALUE APPROACH

You are the manager of a fast-food restaurant. You want to determine whether the populationmean waiting time to place an order has changed in the past month from its previous value of4.5 minutes. From past experience, you can assume that the population standard deviation is1.2 minutes. You select a sample of 25 orders during a one-hour period. The sample mean is 5.1minutes. Use the five-step approach of Exhibit 9.2 above to determine whether there is evi-dence that the population mean waiting time to place an order has changed in the past monthfrom its previous population mean value of 4.5 minutes.

SOLUTION

Step 1 The null hypothesis is that the population mean has not changed from its previousvalue of 4.5 minutes:

H6 trt: 4.5

The alternative hypothesis is the opposite of the null hypothesis. Because the nullhypothesis is that the population mean is 4.5 minutes, the alternative hypothesis is thatthe population mean is not 4.5 minutes:

H r : 1 t + 4 . 5

Step 2 You have selected a sample of n : 25. You choose a 0.05 level of significance (that is,a = 0.05 ).

Step 3 Select the appropriate test statistic. Because o is known, you use the normal distribu-tion and the Z test statistic.

Step4 You col lect the data and compute X: S. l . Using Equat ion (9.1) on page334,youcompute the test statistic as follows:

MPLE 9 .4

_ X - tt - - -

or

1 n

5 . 1 - 4 . 5 _ ) 5 01 .2

]ETo find the probability of getting a test statisti c Z that is equal to or more extreme than2.50 standard deviation units from the center of a standardized normal distribution,you compute the probability of a Z value greater than 2.50 along with the probability


of a Z value less than -2.50. From Table 8..2, the probability of a Z value belowis 0.0062. The probability of a value below +2.50 is 0.9938. Therefore, theof a value above +2.50 is 1 - 0.9938 : 0.0062. Thus. the p-value for thisis 0.0062 + 0.0062 :0.0124.

Step 5 Because the p-value : 0.0124 < cx : 0.05, you reject the null hypothesis. Youthat there is evidence that the population mean waiting time to place anchanged from its previous population mean value of 4.5 minutes. The meantime for customers is longer now than it was last month.

A Connection Between Confidence Interval Estimationand Hypothesis Testing

This chapter and Chapter 8 discuss the two major components of statistical inference:dence interval estimation and hypothesis testing. Although they are based on the sameconcepts, they are used for different purposes. In Chapter 8, confidence intervals wereestimate parameters. In this chapter, hypothesis testing is used for making decisionsspecified values of population parameters. Hypothesis tests are used when trying toa parameter is less than, more than, or not equal to a specified value. Proper interpretationconfidence interval, however, can also indicate whether a parameter is less than, morenot equal to a specified value. For example, in this section, you tested whether themean fill amount was different from 368 grams by using Equation (9.1) on page334:

Z _

lnstead of testing the null hypothesis that p: 368 grams, you can reach the same conclusiconstructing a confidence interval estimate of p. If the hypothesized value of p:368 istained within the interval, you do not reject the null hypothesis because 368 would not besidered an unusual value. However, if the hypothesized value does not fall into the interval,reject the null hypothesis because "p : 368 grams" is then considered an unusual value,Equation (8.1) on page 287 and the following data:

-n = 25. X = 372.5grams. o = 15 grams

for a confidence level of 95oh (corresponding to a 0.05 level of significance-that

xtz

372.s+(1.e6)+12s

372.5+ 5 .88

so that

3 6 6 . 6 2 ( u ( 3 7 8 . 3 8

Because the interval includes the hypothesized value of 368 grams, you do not reject the nuhypothesis. There is insufficient evidence that the mean fill amount over the entire fillinprocess is not 368 grams. You reached the same decision by using two-tail hypothesis testing.

X -p,o_T

N N

o-T\1 n

the Basics

9.20 If you use a 0.05 level of significance in a(two-tail) hypothesis test, what will you decide ifthe computed value of the test statistic Z is +2.21?

9.21 If you use a 0.l0level of significance in a(two-tail) hypothesis test, what is your decisionrule for rejecting a null hypothesis that the popu-

mean is 500 if you use the Z test?

9,22 lf you use a 0.01 level of significance in a(two{ail) hypothesis test, what is your decisionrule for rejecting 110: yt : 12.5 if you use the Z test?

9.23 What is your decision in Problem 9.22 ifthe computed value of the test statistic Z is1 .6 t?

9.24 Suppose that in a two-tail hypothesis test,you compute the value of the test statistic Z as+2.00. What is thep-value?

9.25 ln Problem 9.24, what is your statisticaldecision ifyou test the null hypothesis at the 0. l0level ofsignificance?

9.26 Suppose that in a two-tail hypothesis test,you compute the value of the test statistic Z as-1.38. What is thep-value?

9.27 In Problem 9.26, what is your statisticaldecision if you test the null hypothesis at the 0.01level of significance?

the Concepts

9.28 The operations manager at a clothing fac-tory needs to determine whether a new machineis producing a particular type ofcloth according

manufacturer's specifications, which indicate thatcloth should have a mean breaking strength of 70

and a standard deviation of 3.5 pounds. A samplepieces of cloth reveals a sample mean breaking

of69.1 pounds.there evidence that the machine is not meetingmanufacturer's specifications for mean breaking

h? (Use a 0.05 level of significance.)thep-value and interpret its meaning.

is your answer in (a) if the standard deviation is1.75 pounds?What is your answer in (a) if the sample mean is 69

and the standard deviation is 3.5 pounds?

9.29 The manager of a paint supply store wantsto determine whether the mean amount of paint


contained in l-gallon cans purchased from a nationallyknown manufacturer is actually 1 gallon. You know fromthe manufacturer's specifications that the standard devia-tion of the amount of paint is 0.02 gallon. You select a ran-dom sample of 50 cans, and the mean amount of paint perl-gallon can is 0.995 gallon.a. Is there evidence that the population mean amount is

different from 1.0 gallon (use o: 0.01)?b. Compute thep-value and interpret its meaning.c. Construct a 99oh confidence interval estimate of the

population mean amount of paint.d. Compare the results of (a) and (c). What conclusions do

you reach?

9.30 The quality control manager at a light bulb factoryneeds to determine whether the mean life of a large ship-ment of light bulbs is equal to 375 hours. The populationstandard deviation is 100 hours. A random sample of 64light bulbs indicates a sample mean life of 350 hours.a. At the 0.05 level of significance, is there evidence that

the population mean life is different from 375 hours?b. Compute thep-value and interpret its meaning.c. Construct a 95o/o confidence interval estimate of the

population mean life of the light bulbs.d. Compare the results of (a) and (c). What conclusions do

you reach?

9.31 The inspection division of the Lee County Weightsand Measures Department is interested in determiningwhether the proper amount of soft drink has been placed in2-liter bottles at the local bottling plant of a large nationallyknown soft-drink company. The bottling plant hasinformed the inspection division that the standard deviationfor 2-liter bottles is 0.05 liter. A random sample of 1002-liter bottles selected from this bottling plant indicates asample mean of 1.99 liters.a. At the 0.05 level of significance, is there evidence that

the population mean amount in the bottles is differentfrom 2.0 liters?

b. Compute thep-value and interpret its meaning.c. Construct a 95oh confidence interval estimate of the

population mean amount in the bottles.d. Compare the results of (a) and (c). What conclusions do

you reach?

9.32 A manufacturer of salad dressings uses machines todispense liquid ingredients into bottles that move along afilling line. The machine that dispenses dressings is work-ing properly when the mean amount dispensed is 8 ounces.The population standard deviation of the amount dispensedis 0.15 ounce. A sample of 50 bottles is selected periodi-cally, and the filling line is stopped if there is evidence that


the mean amount dispensed is different from 8 ounces.Suppose that the mean amount dispensed in a particularsample of 50 bottles is 7.983 ounces.a. Is there evidence that the population mean amount is dif-

ferent from 8 ounces? (Use a 0.05 level of significance.)b. Compute thep-value and interpret its meaning.c. What is your answer in (a) if the standard deviation is

0.05 ounce?d. What is your answer in (a) if the sample mean is 7.952

ounces and the standard deviation is 0.15 ounce?

9.33 ATMs must be stocked with enough cash to satisfycustomers making withdrawals over an entire weekend. Butif too much cash is unnecessarily kept in the ATMs, thebank is forgoing the opportunity of investing the money

and earning interest. Suppose that at a particularthe population mean amount of money withdrawnAIMs per customer transaction over the weekend iswith a population standard deviation of $30.a. If a random sample of 36 customer transactions

cates that the sample mean withdrawal amount isis there evidence to believe that the populationwithdrawal amount is no longer $160? (Use a 0.05of significance.)

b. Compute thep-value and interpret its meaning.c. What is your answer in (b) if you use a 0.01

significance?d. What is your answer in (b) if the standard devi

$24 (use cr: 0.05)?

9.3 ONE.TAIL TESTSIn Section 9.2, hypothesis-testing methodology has been used to examine the questionwhether the population mean amount of cereal filled is 368 grams. The alternative(Ht: p * 368) contains two possibilities: Either the mean is less than 368 grams, or themore than 368 grams. For this reason, the rejection region is divided into the two tails ofsampling distribution of the mean.

In many situations, however, the alternative hypothesis focuses on a particularone such situation occurs in the following application: A company that makescheese is interested in determining whether some suppliers that provide milk for the processioperation are adding water to their milk to increase the amount supplied to the processingation. It is known that excess water reduces the freezing point of the milk. The freezing pointnatural milk is normally distributed, with a mean of -0.545' Celsius (C). The standardtion of the freezing temperature of natural milk is known to be 0.008"C. Because the ccompany is only interested in determining whether the freezing point of the milk is lesswhat would be expected from natural milk, the entire rejection region is located in the lowerof the distribution.

The Critical Value ApproachSuppose you wish to determine whether the mean freezing point of milk is less than -0.5450.To perform this one-tail hypothesis test, you use the six-step method listed in Exhibit 9.1 onpage 336.

Step 1 Ho: It>--0.545"C.

Hi trt< -0.545"C.

The alternative hypothesis contains the statement you are trying to prove. If the con-clusion of the test is "reject Ho," there is statistical proof that the mean freezing pointof the milk is less than the natural freezing point of -0.545'C. If the conclusion of tnetest is "do not reject lln," then there is insufficient evidence to prove that the meanfreezing point is less than the natural freezingpoint of -0.545"C.

Step 2 You have selected a sample size of n: 25. You decide to use o, : 0.05.

Step 3 Because o is known, you use the normal distribution and the Z-test statistic.

Step 4 The rejection region is entirely contained in the lower tail of the sampling distributionof the mean because you want to reject 110 only when the sample mean is significantly

L -

)

n)l

9.3: One-Tai lTests 343

less than -0.545'C. When the entire rejection region is contained in one tail of the sam-pling distribution of the test statistic, the test is called a one-tail or directional test.When the alternative hypothesis includes the /e.r.i than sign. the crit ical value of Z mustbe negative. As shown in Table 9.2 and Figure 9.6, because thc entire rejcction regiolr isin the lower tail of the standardized normal distribution and contains an area of 0.05. thecr i t ica l va luc of the Ztest s tat is t ic is -1.645, hal lway between -1.64 and -1.65.

T l re dec i s i on ru le i s

Rejec t Ho i f Z <-1 .645;otherwise, do not reject H,,.

.02 .03 [b4 ^ot .06 .07 .08 .09TABLE 9 .2

Finding the Cr i t ica lValue of the Z TestStatistic from theStandardized NormalDistribution for a One-TailTest with o( = 0.05

FIGURE 9.6Onetail test ofhypothesis for a mean(o known) at the 0.05level of s igni f icance

:8 .03597 .0446

:n ? < |

.0436

:.0314.0392.0485

:.0301.037s.0465

03

rf, S

.S

e

St ' r r r tc : f . t t r t t , t t l l fut r t t T,r l r l r 'E.)

' i ' -1.645 0

Reg ion o f Reg ion o fRejection Nonre ject ion

Step 5 You select a sample of 25 containers of rrrilk and find that the sarnple rrean freezingpo in t equa ls -0 .550"C. Us ing r r :25 . X : -0 .550"C. o - 0 .008 'C, and Equat ion (9 .1 )on page 334,

t .

doD

:-rft -

eni l

- F -poT

1 t l

-0.ss0 - ( -0.s4s)- _ 1 t ? {0.008

,E

t

I

Step 6 Because Z : -3.125 < - I .645. you rc'ject the null hypothesis (see Figure 9.6). You con-clude that the rnean freezing point of the rri lk provided is below -0.545'C. The com-pany should pursue an irrvestigation of thc milk supplier because the mean freezingpoint is significantly below what is expected to occur by chance.

The p-Value l\pproach

Use the five steps l isted in Exhibit 9.2 on page 339 to i l lustrate the test from the preceding page

using the 7r-value approach.

Steps l-3 Thcse steps are the sarne as in the crit ical value approach.

Step 4 Z - -3.125 (see step 5 ofthe crit ical value approach). Because the alternative hypoth-csis indicates a re.jectiorr region entirely in the lollet'tatl of the san-rpling distribLrtion of

FIGURE 9.7Determining thep-value for aone-tail test

FIGURE 9.8Microsoft ExcelZtestresults for the milkproduction example

See Section E9.1 to createthis.

EXAMPLE 9 .5

344 cHAPTER NrNEFundamentals of Hypothesis Testing: One-Sample Tests

the Z test statistic, to compute the p-value, you need to find the probability thatvalue will be /ess thanthe test statistic of -3.125. From Table E.2,the probabilitythe Z value will be less than -3.125 is 0.0009 (see Figures 9.7 and 9.8).

.86/5ARTF4-(88 _ B4yB11

-iloRrslw(B$-ll0RllS0lST@12)-lF{816 < Bli, ?eJea tha null hypothrdr',

"llo Dot roJ.c{ the null hnothcclrJ

Step 5 Thep-value of 0.0009 is less than cr: 0.05. You reject.Flo. You conclude that thefreezing point of the milk provided is less than -0.545oC. The company shouldan investigation of the milk supplier because the mean freezing point is signifiless than what is expected to occur by chance.

A ONE-TAIL TEST FOR THE MEAN

A company that manufactures chocolate bars is particularly concerned that the mean weighta chocolate bar not be greater than 6.03 ounces. Past experience allows you to assume thatstandard deviation is 0.02 ounces. A sample of 50 chocolate bars is selected, and themean is 6.034 ounces. Using the o : 0.01 level of significance, is there evidence that theulation mean weight of the chocolate bars is greater than 6.03 ounces?

SOLUTION Using the critical value approach,

Step I l1o: p ( 6.03

11,: trr > 6.03

Step 2 You have selected a sample size of n: 50. You decide to use o: 0.01.

Step 3 Because o is known, you use the normal distribution and the Z test statistic.

Step 4 The rejection region is entirely contained in the upper tail of the sampling distribuof the mean because you want to reject lln only when the sample mean is signifigreater than 6.03 ounces. Because the entire rejection region is in the upper tail ofstandardized normal distribution and contains anarea of 0.01. the critical value ofZtest statistic is 2.33.

-3.125

9.3: One-Tail Tests 345

The decision rule is

Reject Hoif Z> 2.33;otherwise, do not reject.Flo.

Step 5 You select a sample of 50 chocolate bars, and the sample mean weight is 6.034 ounces.Using z : 50, X : 6.034, o : 0.02, and Equation (9.1) on page 334,

=w#&=r'4r4So

X -V

o

1 n

Z_

Step 6 Because Z:1.414 <2.33, you do not reject the null hypothesis. There is insufficientevidence to conclude that the population mean weight is greater than 6.03 ounces.

To perform one-tail tests of hypotheses, you must properly formulate HoandFl,. A sum-mary of the null and alternative hypotheses for one-tail tests is as follows:

1. The null hypothesis, I1s, represents the status quo or the current belief in a situation.2. The alternative hypothesis, F11, is the opposite of the null hypothesis and represents a

research claim or specific inference you would like to prove.3. Ifyourejectthe null hypothesis, youhave statistical proofthatthe alternative hypothesis is

correct.4. Ifyou do not reject the null hypothesis, then you have failed to prove the alternative

hypothesis. The failure to prove the alternative hypothesis, however, does not mean thatyou have proven the null hypothesis.

5. The null hypothesis (11e) always refers to a specified value of the population parameter(such as p), not to a sample statistic (such as X ).

6. The statement of the null hypothesis always contains an equal sign regarding the specifiedvalue of the parameter (for example, Ho: IL > -0.545'C).

7. The statement of the alternative hypothesis never contains an equal sign regarding thespecified value of the parameter (for example, Hri F < -0.545"C).

9.38 Suppose that in a one-tail hypothesis testwhere you reject Ho only in the upper tail, youcompute the value of the test statistic Z tobe +2.00.

icance?

9.37 ln Problem 9.36, what is your statisticaldecision if the computed value of the Ztest statis-t ic is -1.15?

What is thep-value?

9.39 In Problem 9.38, what is your statisticaldecision if you test the null hypothesis at the 0.05level of significance?

9.40 Suppose that in a one-tail hypothesis test where youreject Ho only in the lower tail, you compute the value ofthe test statistic Z as -1.38. What is thep-value?

9.41 In Problem 9.40, what is your statistical decision ifyou test the null hypothesis at the 0.0 I level of significance?

9.42 In a one-tail hypothesis test where you reject flo onlyin the lower tail, you compute the value of the test statisticZ as +1.38. What is the p-value?

the Basics

9.34 In a one-tail hypothesis test where youreject I1o only in the upper tail, what is the criti-cal value of the Z test statistic at the 0.01 level of

,|

9.35 In Problem 9.34,what is your statisticaldecision if the computed value of the Ztest statis-tic is 2.39?

9.36 In a one-tail hypothesis test where youreject Ho only in the lower tail, what is the criti-cal value of the Ztest statistic at the 0.01 level of


9-43 t r r Prol r lenr 9.42' what is the stat is t ical dccis ion i i you

tcst the nul l hypothcsis ar t thc 0.Ot level of s igni f icance' /

Applying the Concepts

9.44 The Glen Valley Steel Company manufac-tures steel bars. Ifthe production process is work-ing properly, it turns out steel bars that are nor-

mally distributed with mean length of at least 2.8 feet, witha standard deviation of 0.20 foot (as determined from engi-neering specifications on the production equipmentinvolved). Longer steel bars can be used or altered, butshorter bars must be scrapped. You select a sample of 25bars, and the mean length is 2.73 feet. Do you need to adjustthe production equipment?a. If you test the null hypothesis at the 0.05 level of signif-

icance, what decision do you make using the criticalvalue approach to hypothesis testing?

b. Ifyou test the null hypothesis at the 0.05 level ofsignif-icance, what decision do you make using the p-valueapproach to hypothesis testing?

c. Interpret the meaning of the p-value in this problem.d. Compare your conclusions in (a) and (b).

9.45 You are the manager of a restaurant that deliverspizza to college dormitory rooms. You have just changedyour delivery process in an effort to reduce the mean timebetween the order and completion of delivery from the cur-rent25 minutes. From past experience, you can assume thatthe population standard deviation is 6 minutes. A sample of36 orders using the new delivery process yields a samplemean of 22.4 minutes.a. Using the six-step critical value approach, at the 0.05

level of significance, is there evidence that the popula-tion mean delivery time has been reduced below the pre-vious population mean value of 25 minutes?

b. At the 0.05 level of significance, use the five-stepp-value approach.

c. Interpret the meaning of thep-value in (b).d. Compare your conclusions in (a) and (b).

9.46 Children in the United States account directly for $36billion in sales annually. When their indirect influence overproduct decisions from stereos to vacations is considered, the

total econornic spending aff-ected by children in the

States is $290 billion. It is estimated that by age 10, amakes an average of tnore than five trips a weck to a(M. E . Go ldberg , G. J . Gorn , L . A . Peracch io ,

G. Bamossy, "Understanding Materialism AmongJoumal of Consumer Psychology,2003, l3(3) pp.2Suppose that you want to prove that children in youraverage more than five trips a week to a store. Let psent the population mean number of times children incity make trips to a store.a. State the null and alternative hypotheses.b. Explain the meaning of the Type I and Type II

the context ofthis scenario.c. Suppose that you carry out a similar study in the city

which you live. Based on past studies, you assumethe standard deviation of the number of trios to theis 1.6. You take a sample of 100 children and findthe mean number of trips to the store is 5.47. At0.01 level of significance, is there evidence thatpopulation mean number of trips to the store isthan 5 per week?

d. Interpret the meaning of the p-value in (c).

9.47 The waiting time to place an order at a branchfast-food chain durine the lunch hour has had amean of 3.55 minutes, with a population standardof l.l minutes. Recently. in an effort to reduce the waititime, the branch has experimented with a system in whithere is a single waiting line. A sample of 100during a recent lunch hour was selected, and theirwaiting time to place an order was 3.18 minutes. Athat the population standard deviation of the waiting tihas not chansed from 1.1 minutes.a. At the 0.05 level of significance, using the critical

approach to hypothesis testing, is there evidence thatpopulation mean waiting time to place an order isthan 3.55 minutes?

b. At the 0.05 level of significance, using the p-approach to hypothesis testing, is there evidence thatpopulation mean waiting time to place an order isthan 3.55 minutes?

c. Interpret the meaning of the p-value in this problem.d. Compare your conclusions in (a) and (b).

9.4 t TEST OF HYPOTHESIS FOR THE MEAN (o UNKNOWN)In most hypothesis-testing situations concerning the population mean F, you do not knowpopulation standard deviation, o. Instead, you use the sample standard deviation, S. If youassume that the population is normally distributed" the sampling distribution of the mean fol-lows a r distribution with r - I degrees of freedom. If the population is not normally distrib-uted" you can still use the / test if the sample size is large enough for the Central Limit Theoremto take effect (see Section 7.4). Equation (9 .2) defines the test statistic / for determining the dif-ference between the sample mean. X, and the population mean, p, when the sample standarddeviation.,S. is used.

)dldre1d

lr"

8) .ityre-)ur

i i n

/ i n;hat.orethatthethe

ater

o f attionltioniting'hich

mersneaniumetime

,alue

t theless

aluet theless

z theyoufol-

trib-)remdif-

dard

9.4: tTest of Hypothesis fbr thc Mean (o Unknown) 347

t TEST OF HYPOTHESIS FOR THE MEAN (o UNKNOWN)Y - r t

(e.2)

where the test statistic t follows a I distribution having n - I degrees of freedom.

To il lustrate the use of this 1test, return to the Using Statistics sccnario concerning theSaxon Home lmprovement Company on page 284. Over the past f ive years, the mean amountper sales invoice is S120. As an accountant for the company, you need to inform the financedepartment if this amount changes. In other words. the hypothesis test is r-rsed to try to provethat the mean amount per sales invoice is increasing or decreasing.

.t

G

The Crit ical Value Approach

To perform this two-tail hypothesis test, youpage 336.

Step I

use the six-step rnethod listed in Exhibit 9.1 on

H , , : p - $ 1 2 0H , : p * $ 1 2 0

The alternative hypothesis contains the stater.nent you are trying to prove. If the nullhypothesis is rejected then there is statistical proof that the population mean amountper sales invoice is no longer S120. If the statistical conclusion is "do not reject H,,."then you wi l l conclude that there is insuf f ic ient ev idence to prove that the meanamoLlnt differs from the long-term rnean of $ 120.

Step 2 You have selected a sample of n - 12. You decide to use o - 0.05.

Step 3 Because o is unknown, you use the t distribution and the I test statistic for this exam-ple. You rnust assume tlrat the population of sales invoices is nonnally distributed. Thrsassurnption is discussed on page 346.

S tep4 Fo rag i vensamp les i ze .n . t he tes t s ta t i s t i c l f o l l owsa ld i s t r i bu t i onw i thn - l deg reesof l 'rcedom. The crit ical values of the r distribution with l2 - I : l l degrees of free-dom are found in Table E.3, as i l lustrated in Table 9.3 and Figure 9.9. The alternativehypothesis . Hr .V+ $120 is two-ta i led. Thus, the area in the re ject ion region of theI d is t r ibut ion 's le f t ( lower) ta i l is 0.025, and the area in the re ject ion region of theI d is t r ibut ion 's r ight (upper) ta i l is a lso 0.025.

Upper-Tail Areas

Degrees of Freedom ) < .05.10

TABLE 9 .3Determining theCr i t ica lValue f rom thetTable for an Area of0.025 in Each Tai l wi th'1 1 Degrees of Freedom

228 1

I23A

5b

789

1 0

r .00000 . 8 1 6 50.76490.14070.72670 . 1 r 7 60 . 7 1 I I0.70640.70270.6998

3.07771 . 8 8 5 61 . 6 3 7 71.s3321 .47 59r .43981.41491 . 3 9 6 81 . 3 8 3 01 . 3 7 2 2

6 . 3 1 3 82.92002.35342 . l 3 l 82 . 0 1 s 0r.94321.89461.85951 . 8 3 3 r1 . 8 1 2 5

31.82076.96464.54073.74693.36493 . t 4 2 12.99802.89652.82142.76382.7181

63.6s149.92485.84094.60414.03223.70743.49953.3s543.24983. t6933 . 1 0 5 8

062027824164106469

.025

Sour< e. E.ttruttt,tl li ont ThhlL: 1,. -l

FIGURE 9.10Microsoft Excel resultsfor the one-samolet test of sales invoices

318 CHAPTER NINE Fundamenrals of Hypothesis Testing: one-Sa'.rpre Tests

FIGURE 9.9Test ing.a hypothesisabout the mean(o unknown) at the0.05 level of s igni f icancewith 11 degrees offreedom

| -2.2010

t tRegion o fRejection

Crit icalVa lue

' 0I

IRegion of

NonrejectionI

+2.2010 | t . ,

Reg ion o fRejection

From the r table as given in Table 8.3, a portion of which is shown in Table 9.3,critical values are +2.2010. The decision rule is

Reject Hoif t . - t rr : _2.2010o r i f l > t r r - + 2 . 2 0 1 0 ;

otherwise, do not reject Hn.

Step 5 The data in the file FilFlllttltl are the amounts (in dollars) in a random sample ofsales invoices.

108.98 152.22 I I I .45 1 10.59 t2t .46 107 .2o93.32 9 t .97 I I L56 7s .7 | 128.58 l3s . I I

Using Equations (3.1) and (3.10) on pages 97 and 107, or Microsoft Excel, as shownin Figure 9. 10,

S"L " I

X = i = t = $ 1 1 2 . 8 5n

From Equation (9.2) on page 347 ,

1 1 2 . 8 5 - 1 2 0

Crit icalVa lue

20.80

E= - 1 . 1 9 0 8

-B8rS0RT{861-86 -1-6n , ar;/Brt

-{nilvps,8.t2)}-nilv{85,8121-TDlsT(A8s{Bt3}, Bl2, 2t-lF(818 < &i, -Reject rte n{lt hirpothesb-,

'Do not roJecf $. null hypoth€b"l

x - t t

.str

1 n


A B

a large sample size

available, S estimates o

enough that there

little diff e ren ce betwe en

t andZ distributions.

9.11Excel

statistics forinvoice data

9.4: tTestofHypothesis forthe Mean (oUnknown) 349

Step 6 Because -2.2010 < l: -1.1908 < 2.2010, you do not reject F1o. You have insufficientevidence to conclude that the mean amount per sales invoice differs from $120. Youshould inform the finance department that the audit suggests that the mean amount perinvoice has not chansed.

The pValue Approach

Steps 1-3 These steps are the same as in the critical value approach.

Step 4 From the Microsoft Excel worksheet of Figure 9.10, t: -1.1908 and the p-value :

0.2588 (see step 5 ofthe critical value approach).

Step 5 The Microsoft Excel results in Figure 9.10 give the p-value for this two-tail test as0.2588. Because the p-value of 0.2588 is greater than cr : 0.05, you do not reject 110.The data provide insufficient evidence to conclude that the mean amount per salesinvoice differs from $120. You should inform the finance department that the auditsuggests that the mean amount per invoice has not changed. Thep-value indicates thatif the null hypothesis is true, the probability that a sample of 12 invoices could have amonthly mean that differs by $7.15 or more from the stated $120 is 0.2588. In otherwords, if the mean amount per sales invoice is truly $120, then there is a25.88%ochance of observing a sample mean below $112.85 or above $127.15.

In the preceding example, it is incorrect to state that there is a 25.88% chance that the nullhypothesis is true. This misinterpretation of the p-value is sometimes used by those not prop-erly trained in statistics. Remember that the p-value is a conditional probabiliry calculated byassuming that the null hypothesis is true. In general, it is proper to state the following:

If the null hypothesis is true, there is a (p-value)*1000/o chance of observing a sampleresult at least as contradictory to the null hypothesis as the result observed.

Checking AssumptionsYou use the one-sample r test when the population standard deviation, o, is not known and is esti-mated using the sample standard deviation,l ,S. To use the one-sample / test, the data are assumedto represent a random sample from a population that is normally distributed. In practice, as longas the sample size is not very small and the population is not very skewed, the I distribution pro-vides a good approximation to the sampling distribution of the mean when o is unknown.

There are several ways to evaluate the normality assumption necessary for using the / test.You can observe how closely the sample statistics match the normal distribution's theoreticalproperties. You can also use a histogram, stem-and-leaf display, box-and-whisker plot, or nor-mal probability plot. For details on evaluating normaliry see Section 6.3 on page 234.

Figure 9.11 presents descriptive statistics generated by Microsoft Excel. Figure 9.12 is aMicrosoft Excel box-and-whiskerplot. Figure 9.13 is a Microsoft Excel normal probability plot.

fodeStandard Dcvladon

VarlancsXurtoglrSkormoerRangetlnlmomXnxlmumSumCountLargcctfl|Smallodfl)

fl2.qrfi6.txls:f

l r lm#vA

20.73799f32.t5650.1727|a0.1rnn

70-5175.71

$7J,.1354.21

12tna,T5.f I

Section E3.1 to create


'1

FIGURE 9.12Microsoft Excel box-and-whisker plot for thesales invoice data


Box.and4llhisker Plot of Invoice Amount

FIGURE 9.13Microsoft Excel normalprobabi l i ty plot for thesales invoice data

120


E too

5 ro

: 5 0

Because the mean is very close to the median, the points on the normal probabilit)

appear to be increasing approximately in a straight line, and the box-and-whisker plot ap

approximately symmetrical. You can assume that the population of sales invoices is ap1

mately normally distributed. The normality assumption is valid, and therefore the aud

results are valid.The / test is a robust test. It does not lose power if the shape of the population de

somewhat from a normal distribution, particularly when the sample size is large enou

enable the test statistic I to be influenced by the Central Limit Theorem (see Sectior

However, you can make erroneous conclusions and can lose statistical power if you ut

/ test incorrectly. If the sample size, n, is small (that is, less than 30) and you cannot (

make the assumption that the underlying population is at least approximately norl

distributed, other nonparametric testing procedures are more appropriate (see referr

1and2).

Normat Probability Plot of lnvoice Amount

the Basics

9.48 If, in a sample of n : 16 selected from anormal population, X : 56 and .9: 12, what is

value ofthe t-test statistic ifyou are testingis I1o: p: 50?

9.48, how many degrees of freedom areone-sample I test?

9.48 and 9.49. what are the critical val-the level ofsignificance, cr, is 0.05 and the alter-

s,11r, is

Problems 9.48,9,49, and 9.50, what is your statis-ion if the alternative hypothesis, F/,, is

in a sample of n: 16 selected from a left-skewedX = 65 and ̂ S: 21, would you use the / test tohypothesis Hoi lr:60? Discuss.

in a sample of n: 160 selected from a left-skewedX = 65 and ̂ S: 21, would you use the I test tohypothesis Ho: lt: 60? Discuss.

the Concepts

payment of medical claims can add to the costcare. An article (M. Freudenheim. "The Check Is

Mail," The New York Times, May 25,2006, pp.reported that the mean time from the date of ser-

date of payment for one insurance company wasduring a recent period. Suppose that a sample of

claims is selected durins the latest timeThe sample mean time from the date of service toofpayment was 39.6 days, and the sample standardwas 7.4 days.the 0.05 level of significance, is there evidence

re population mean has changed from 41.4 days?is your answer in (a) if you use the 0.01 level of

is your answer in (a) if the samplemean is 38.210.7 days?and the sample standard deviation is

article (N. Hellmich, "'Supermarket Guru'Has aMantra;' USA Today, June 1 9, 2002, p. 70) claimedtypical supermarket trip takes a mean of 22 min-

that in an effort to test this claim. vou selectof 50 shoppers at a local supermarket. The meantime for the sample of 50 shoppers is 25.36 min-

9.4; tTestof Hypothesis for the Mean (o Unknown) 351

0.10 level of significance, is there evidence that the meanshopping time at the local supermarket is different from theclaimed value of 22 minutes?

9.56 You are the manager of a restaurant for afast-food franchise. Last month, the mean wait-ing time at the drive-through window, as mea-

sured from the time a customer places an order until thetime the customer receives the order, was 3.7 minutes. Thefranchise helped you institute a new process intended toreduce waiting time. You select a random sample of 64orders. The sample mean waiting time is 3.57 minutes,with a sample standard deviation of 0.8 minute. At the 0.05level of significance, is there evidence that the populationmean waiting time is now less than 3.7 minutes?

9.57 A manufacturer of chocolate candies uses machinesto package candies as they move along a filling line.Although the packages are labeled as 8 ounces, the com-pany wants the packages to contain 8.17 ounces so that vir-tually none of the packages contain less than 8 ounces. Asample of 50 packages is selected periodically, and thepackaging process is stopped ifthere is evidence that themean amount packaged is different from 8.17 ounces.Suppose that in a particular sample of 50 packages, themean amount dispensed is 8.159 ounces, with a samplestandard deviation of0.05l ounce.a. Is there evidence that the population mean amount

is different from 8.17 ounces? (Use a 0.05 level ofsignificance.)

b. Determine thep-value and interpret its meaning.

9.58 The data in the file@@contains prices (indollars) for two tickets, with online service charges, largepopcorn, and two medium soft drinks at a sample of sixtheatre chains:

36.15 31.00 35.05 40.25 33.75 43.00

Source: Extractedfrom K. Kelly, "The Multiplex Under Siege," TheWall Street Journal, December 24-25, 2005, pp. Pl, P5.

a. At the 0.05 level of significance, is there evidence thatthe mean price for two tickets, with online servicecharges, large popcorn, and two medium soft drinks, isdifferent from $35?

b. Determine thep-value in (a) and interpret its meaning.c. What assumption about the population distribution is

needed in (a) and (b)?d. Do you think that the assumption stated in (c) is seri-

ously violated?

9.59 In NewYork State, savings banks are permitted to sella form oflife insurance called savings bank life insurance(SBLI). The approval process consists of underwriting,a standard deviation of 7.24 minutes. Usins the


which includes a review of the application, a medical infor-mation bureau check, possible requests for additional med-ical information and medical exams, and a policy compila-tion stage in which the policy pages are generated and sentto the bank for delivery. The ability to deliver approved poli-cies to customers in a timely manner is critical to the prof-itability of this service. During a period of one month,aran-dom sample of 27 approved policies is selected, and thetotal processing time, in days, is recorded (as stored in the

@$file):

73 19 16 64 28 28 31 90 60 56 31 56 22 18

4s 48 l7 17 r7 9 t 92 63 50 51 69 16 t7

a. In the past, the mean processing time was 45 days. Atthe 0.05 level of significance, is there evidence that themean processing time has changed from 45 days?

b. What assumption about the population distribution isneeded in (a)?

c. Do you think that the assumption stated in (b) is seri-ously violated? Explain.

9.60 The following data (see the !@ file) representthe amount of soft-drink filled in a sample of 50 consecu-tive 2-liter bottles. The results, listed horizontally in theorder of being filled, were:

2.109 2.086 2.066 2.075 2.065

2.031 2.029 2.025 2.029 2.023

2.0t2 2.012 2.012 2.010 2.005

1.994 1.986 1.984 1.98r 1.973

t .963 1.957 t .951 1.9s1 1.947

2.057 2.052 2.044 2.036 2.038

2.020 2.015 2.014 2.013 2.0t4

2.003 1,999 1.996 1.997 1.992

1.975 t .971 t .969 r .966 1.967

r.941 1.941 1.938 1.908 1.894

a. At the 0.05 level of significance, is there evidence thatthe mean amount of soft drink filled is different from2.0 liters?

b. Determine thep-value in (a) and interpret its meaning.c. Evaluate the population distribution assumption you

made in (a) graphically. Are the results of (a) valid? Why?d. Examine the values of the 50 bottles in their sequential

order, as given in the problem. Is there a pattern to theresults? If so, what impact might this pattern have on thevalidity of the results in (a)?

9.61 One of the major measures of the quality of serviceprovided by any organization is the speed with which itresponds to customer complaints. A large family-helddepartment store selling furniture and flooring, includingcarpet, had undergone a major expansion in the past severalyears. In particular, the flooring department had expandedfrom 2 installation crews to an installation supervisor, ameasurer, and l5 installation crews. Last year there were50 complaints concerning carpet installation. The follow-ing data (stored in the @$ file) represent the num-ber of days between the receipt of a complaint and the res-olution of the complaint:

54 5 35 137 31 27 t52 2

l l t 9 t 2 6 l l 0 l l 0 2 9 6 1 3 5

12 4 165 32 29 28 29 26

1 3 1 0 5 2 7 4 5 2 3 0 2 2

33 68

123 8l 74

94 31 26

2 5 l 1 4

36 26 20

a. The installation supervisor claims that the meanof days between the receipt of a complaint and thetion of the complaint is 20 days or less. At the 0.05ofsignificance, is there evidence that the claim is not(that is, that the mean number of days is greater than

b. What assumption about the population distributionyou make in (a)?

c. Do you think that the assumption made in (b) isously violated? Explain.

d. What effect might your conclusion in (c) have onvalidity of the results in (a)?

9.62 A manufacturing company produces steelfor electrical equipment. The main component part ofhousine is a steel troush that is made out of a Isteel coil. It is produced using a 250-ton progressivepress with a wipe-down operation that puts twoforms in the flat steel to make the troush. The difrom one side ofthe form to the other is critical becauseweatherproofing in outdoor applications. Therequires that the width of the trough be between 8.31 iand 8.61 inches. The data file@@[!contains the wiof the troughs, in inches, for a sample of n:49.

8.312 8.343 8.317 8.383 8.348 8.410 8.351 8.373 8.481 8.4n

8.476 8.382 8.484 8.403 8.414 8.419 8.385 8.465 8.498 8.447

8.436 8.413 8.489 8.414 8.481 8.4r5 8.479 8.429 8.458 8.462

8.460 8.444 8.429 8.460 8.4t2 8.420 8.410 8.405 8.323 8.420

8.396 8.44'7 8.405 8.439 8.411 8.427 8.420 8.498 8.409

At the 0.05 level of significance, is there evidencethe mean widths of the trouehs is different from 8.inches?

b. What assumption about the population distributionyou need to make in (a)?

c. Do you think that the assumption made in (a) hasseriously violated? Explain.

9.63 One operation of a steel mill is to cut pieces ofinto parts that are used in the frame for front seats inautomobile. The steel is cut with a diamond sawrequires the resulting parts to be within +0.005 inch oflength specified by the automobile company. The data ithe file @!lf!! come from a sample of 100 steel parts. Themeasurement reported is the difference, in inches, betweenthe actual length of the steel part, as measured by ameasurement device, and the specified length of the steelpart. For example, a value of -0.002 represents a steel partthat is 0.002 inch shorter than the specified leneth.

7

5

J

J

)er.u-/elueD?rst

ri-

he

z

7

2

0

ratt6

Jo

3n

a. At the 0.05 level of significance, is there evidence thatthe mean difference is not equal to 0.0 inches?

b. Determine the p-value in (a) and interpret its meaning.c. What assumption about the differences between the

actual length ofthe steel part and the specified length ofthe steel part must you make in (a)'i

d. Evaluate the assumption in (c) graphically. Are theresults of (a) valid? Why/

9.64 In Problem 3.63 on page 138, you were introduced toa tea-bag-filling operation. An important quality character-istic of interest for this process is the weight of the tea inthe individual bags. The data in the file [ll[!![! is anordered array of the weight, in grams, of a sample of 50 teabags produced during an eight-hour shift.a. Is there evidence that the mean amount of tea per bag is

different from 5.5 grams (use a : 0.0l)?b. Construct a99%o confidence interval estimate of the popu-

lation mean amount of tea per bag. Interpret this interval.c. Compare the conclusions reached in (a) and (b).

9.5: ZTest of Hypothesis for the proportion 353

9.65 Although many people think they can put a meal onthe table in a short period of time, a recent article reportedthat they end up spending about 40 minutes doing so(N. Hellmich, 'Americans

Go for the euick Fix forDinner," USA Today, February 14,2006). Suppose anotherstudy is conducted to test the validity of this statement. Asample of 25 people is selected, and the length of t ime toprepare and cook dinner (in minutes) is recorded, with thefollowing results (stored in the file E[lf[E)

44.0 51.9 49.7 40.0 55.5 33.0 43.4 4t.3 45.2 40.7 4t.t 4g.t 30.945.2 55.3 52.1 55.1 38.8 43.1 39.2 58.6 49.8 43.2 47.9 46.6

a. Is there evidence that the population mean ume ro pre-pare and cook dinner is greater than 40 minutes? Use alevel of significance of 0.05.

b. What assumptions are made to perform the test in (a) /c. Do you think the assumption needed in (a) is seriously

violated? Explain.d. Determine the p-value and interpret its meaning.

gshegech€eceofnyLCS

hs

9.5 Z TEST OF HYPOTHESIS FOR THE PROPORTIONIn some situations, you want to test a hypothesis about the proportion of successes in the popu-lation, n' rather than testing the population mean. To begin, you select a random sample andcompute the sample proportion, p: Xln. You then compare the value of this statistic to thehypothesized value of the parameter, n, in order to decide whether to reject the null hypothesis.

If the number of successes (X) and the number of failures (n - X) are each at least five, thesampling distribution of a proportion approximately follows a normal distribution. you use theZ test for the proportion given in Equation (9.3) to perform the hypothesis test for the differ-ence between the sample proportion, p, and the hypothesized population proportion, n.

ONE SAMPLE Z TEST FOR THE PROPORTIONn - f lz-- : (e.3)

whereNumber of successes in the sample

Sample rirr-

p : sample proportion of successes

n: hypothesized proportion ofsuccesses in the population

and the test statistic Z approximately follows a standardized normal distribution.

Alternatively, by multiplying the numerator and denominator by n, you can write the Z teslstatistic in terms of the number of successes, x, as shown in Equation (9.4).

Z TEST FOR THE PROPORTION IN TERMS OF THE NUMBER OF SUCCESSES

V

n

reltnrdneinne,'nerelrrt

X - n n---

lnn( l - n)

n(1 - n )n

Z_ (e.4)


To illustrate the one-sample Z test for a proportion, consider a survey of independentcery owners. In the survey, the owners were asked to consider what they believed to bebiggest competitive threat. Some noted large-chain supermarkets, wholesale clubs, andindependent grocers. The most common response, given by 78 of the l5l owners, was thatviewed Wal-Mart as their biggest competitive threat (Extracted from J. TarnowskiD. Chanil, "To Serve and Connectl' The Progressive Grocer, January 1,2006,pp.60-65),this survey, the null and alternative hypotheses are stated as follows:

Hn: n:0.50 (hat is, half of all independent grocery owners view Wal-Mart as theirbiggest competitive threat)

Hr: n* 0.50 (that is, either less than half or more than half of all independentowners view Wal-Mart as their biggest competitive threat)

FIGURE 9.14Two-tail test ofhypothesis for theproportion at the 0.05level of s ioni f icance

The Critical Value ApproachBecause you are interested in whether the proportion of independent grocery ownersview Wal-Mart as their biggest competitive threat is 0.50, you use a two-tail test. If youthe a : 0.05 level of significance, the rejection and nonrejection regions are set up asFigure 9.14, and the decision rule is

Re jec t Ho i f Z<- l .96or i f Z> +1.96 ;otherwise, do not reject 110.

Region ofRejection

Because 78 ofthepetitive threat,

151 of independent grocery owners view Wal-Mart as their biggest

= 0 .5166

Using Equation (9.3),

Z_ 0.5166 - 0 .s0 o'0166 = 0.40690.0407

or, using Equation (9.4).

78p = -' l s l

X -nn

,,lnn(l - n)

78 - (151X0 .50 ) = 2 .5 =0 .4069./t s t1o.soy1o.so; 6.144r

ru(l - r)n

0.50(l - 0.s0)

Z _

1

3 .45

7 -

g1 0 .1 1 .l 1

't3' 1 4 '

1 5 :16

1 8 '

ro-reirnerreylnd

For

vholects i n

FIGURE 9.15Microsoft Excelresults for the surveyof independent groceryowners and Wal-Martas biggest competit ivethreat


9.5: ZTcst of Hypothesis for the Proport ion 355

Because -1.96 < Z:0.4069 < 1.96, you do not reject Ho. There is insufficient evidence that theproportion of independent grocery owners who view Wal-Mart as their biggest competit ivethreat is not 0.50. Figure 9.15 presents Microsoft Excel results for these data. In other words,the current belief that half of al1 independent grocers view Wal-Mart as their bigget competit ivethreat is not contradicted bv the Z test.

-S187-soRT(8411 - B{)r87}-{810 - 84}1811

-t{oRf,stilv(852}-llORf,SlNV(1 ,852)-2' (1 - ilORilSDrsT(ABS(Br?)))-lF{817 < 85, -Rejad rhe null hypothe€is",

To not rejecr lho ilull hypoth6ls'J

The p-Value Approach

As an alternative to the crit ical value approach, you can compute the p-value. For this two-tailtest in which the rejection region is located in the lower tail and the upper tail, you need to findthe area below a Z value of -0.4069 and above a Z value of +0.4069. Figure 9.15 reports ap-value of 0.684 L Because this value is greater than the selected level of significance (o : 0.05),vou do not reiect the null hvnothesis.

TESTING A HYPOTHESIS FOR A PROPORTION

A fast-food chain has developed a new process to ensure that orders at the drive-through arefilled correctly. The previous process filled orders correctly 85% of the time. Based on a sampleof 100 orders using the new process, 94 were fil led correctly. At the 0.01 level of significance,can you conclude that the new process has increased the proportion oforders fil led correctly?

SOLUTION The null and alternative hypotheses are

Hu: n3 0.85 (that is, the proportion of orders fil led correctly is less than or equal to 0.85)Hr: n> 0.85 (that is, the proportion oforders fil led correctly is greater than 0.85)

Using Equat ion (9.3) on page 353,

Y q 4p = 1 ) _ = 0 . 9 4

n 1 0 0

o o g- i < 1

0.0357

The 7r-value for Z > 2.52 is 0.0059.Using the critical value approach, you reject Ho if Z> 2.33. Using thep-value approach,

you re jec t t ln i f thep-va lue < 0 .01 . Because Z :2 .52>2.33 or thep-va lue :0 .0059 < 0 .01 ,you reject Hn. You have evidence that the new process has increased the proportion of correctorders above 0.85.

Z

EXAMPLE

- / |Jr( r -

n

0.8s(L q.8l)100

ffiffiffi


Learning the Basics

9.66 If, in a random sample of 400 items, 88 aredefective, what is the sample proportion of defec-tive items?

9.67 In Problem 9.66, if the null hypothesis isthat20o/o of the items in the population are defec-tive, what is the value of the Z-test statistic?

9.68 In Problems 9.66 and 9.67, suppose youare testing the null hypothesis Hn: n : 0.20against the two-tail alternative hypothesis Hr: n*

0.20 and you choose the level of significance cr : 0.05.What is your statistical decision?

Applying the Concepts

9,69 Late payment of medical claims can add to the costof health care. An article (M. Freudenheim, "The Check IsNot in the Mail," The New York Times, May 25,2006, pp.Cl, C6) reported that for one insurance company, 85.1% ofthe claims were paid in full when first submitted. Supposethat the insurance company developed a new payment sys-tem in an effort to increase this percentage. A sample of200 claims processed under this system revealed that 180of the claims were paid in full when first submitted.a. At the 0.05 level of significance, is there evidence that the

proportion of claims processed under this new system ishigher than the article reported for the previous system?

b. Compute the p-value and interpret its meaning.

9.70 The high cost of gasoline in the spring of2006 had many people reconsidering their sum-mer vacation plans. A survey of 464 Cincinnati-

area adults found that 255 said that they were planning onmodifying or canceling their summer travel plans becauseof high gas prices ("Will You Travel?" The CincinnatiEnquirer, May 3,2006, p. Al0.)a. Use the six-step critical value approach to hypothesis-

testing and a 0.05 level of significance to try to provethat the majority of Cincinnati-area adults were plan-ning on modifying or canceling their summer travelplans because ofhigh gas prices.

b. Use the five-stepp-value approach to hypothesis testingand a 0.05 level of significance to try to prove that themajority of Cincinnati-area adults were planning onmodifying or canceling their summer travel plansbecause ofhigh gas prices.

c. Compare the results of (a) and (b).

9.71 A Wall Street Journql article suggested that age biaswas a growing problem in the corporate world (C. Hymowitz,"Top Executives Chase Youthful Appearance, but Miss RealIssue," TheWall StreetJournal, February 17,2004,p. BI). In

2001, an estimated 78% of executives believed thatwas a serious problem. lna2004 study by ExecuNet,the executives surveyed considered age bias a seriouslem. The sample size for the 2004 study was not dtSuppose 50 executives were surveyed.a. At the 0.05 level of significance, use the six-step

value approach to hypothesis-testing to try tothe 2004 proportion of executives who believedbias was a serious problem was higher than thevalue of 0.78.

b. Use the five-step p-value approach. Inte4ing of thep-value.

c. Suppose that the sample size used was(a) and (b).

d. Discuss the effect that sample size had on the ouof this analysis an4 in general, the effect thatsize plays in hypothesis testing.

9.72 A Wall Street Journal poll ("What's NewsThe Wall Street Journal, March 30,2004, p. D7)respondents if they trusted energy-efficiency ratings onand appliances; 552 responded yes, and 531 respondeda. At the 0.05 level of significance, use the six-step cri

value approach to hypothesis-testing to try to provethe percentage of people who trust energy-efficiratings differs from 50%.

b. Use the five-step p-value approach. Interpret theing of thep-value.

9.73 One ofthe biggest issues facing e-retailers is theto reduce the proportion of customers who cancel theiractions after they have selected their products. It hasestimated that about half of prospective customerstheir transactions after they have selected therr(B. Tedeschi, "E-Commerce, a Cure forAbandonedCarts: A Web Checkout System That Eliminates the NeedMultiple Screens," The NevvYorkTimes, February 14,p. C3). Suppose that a company changed its Web site socustomers could use a single-page checkout processthan multiple pages. A sample of 500 customers whoselected their products were provided with the newsystem. Of these 500 customers, 2 I 0 cancelled theirtions after they have selected their products.a. At the 0.01 level of significance, is there evidence

the proportion of customers who selected productsthen cancelled their transaction was less than 0.50the new system?

b. Suppose that a sample of 100 customers whoselected their products were provided with thecheckout system and that 42 of those customers cancelled their transactions after they had selected theiproducts. At the 0.01 level of significance, is there evidence that the proportion of customers who selecter

and then cancelled their transaction was less0,50 with the new system?

the results of(a) and (b) and discuss the effectsample size has on the outcome, and, in general, in

More professional women than ever before are forgo-because of the time constraints of their

However, many women still manage to find time tothe corporate ladder and set time aside to have chil-

A survey of I 87 attendees at Fortune Magazine's MostWomen in Business summit in March 2002 found

133 had at least one child (C. Hymowitz, "Women

9.6 POTENTIAL HYPOTHESIS-TESTING PITFALLSAND ETHICAL ISSUESTo this point, you have studied the fundamental concepts of hypothesis testing. You have usedhypothesis testing to analyze differences between sample estimates (that is, statistics) andhypothesized population characteristics (that is, parameters) in order to make business deci-sions concerning the underlying population characteristics. You have also learned how to eval-uate the risks involved in making these decisions.

When planning to carry out a test ofthe hypothesis based on a survey, research study, ordesigned experiment, you must ask several questions to ensure that you use proper methodol-ogy. You need to raise and answer questions such as the following in the planning stage:

What is the goal of the survey, study, or experiment? How can you translate the goal into anull hypothesis and an alternative hypothesis?Is the hypothesis test a two-tail test or one-tail test?can you select a random sample from the underlying population of interest?What kinds of data will you collect from the sample? Are the variables numerical or cate-gorical?

5. At what significance level, or risk of commiuing a Type I error, should you conduct thehypothesis test?

6. Is the intended sample size large enough to achieve the desired power of the test for thelevel of significance chosen?

7. What statistical test procedure should you use and why?8. What conclusions and interpretations can you reach from the results of the hypothesis test?

Failing to consider these questions early in the planning process can lead to biased orincomplete results. Proper planning can help ensure that the statistical study will provide objec-tive information needed to make good business decisions.

Statistical SigniftcanceVersus Practical Signifrcance You need to make the distinc-tion between the existence of a statistically significant result and its practical significance in thecontext within a field of application. Sometimes, due to a very large sample size, you may get aresult that is statistically significant but has little practical significance. For example, suppose thatprior to a national marketing campaign focusing on a series of expensive television commercials,you believe that the proportion of people who recognize your brand is 0.30. At the completion ofthe campaign, a survey of 20,000 people indicates that 6,168 recognized your brand. A one-tailtest trying to prove that the proportion is now greater than 0.30 results in ap-value of 0.0047 andthe correct statistical conclusion is that the proportion of consumers recognizing your brand namehas now increased. Was the campaign successful? The result of the hypothesis test indicates a sta-tistically significant increase in brand awareness, but is this increase practically important? Thepopulationproportionisnowestimatedat6,16S120,000:0.3084, or3}.S4Vo.Thisincreaseislessthan l%o more than the hypothesized value of 30o/o. Did the large expenses associated with the

9.6: Potential Hypothesis-Testing Pitfalls and Ethical Issues 357

Plotting Mix of Work and Family Won't Find perfect plan,"The Wall Street Journal, June 11, 2002,p. Bl). Assume thatthe group of 187 women was a random sample from thepopulation of all successful women executives.a. What was the sample proportion of successful women

executives who had children?b. At the 0.05 level of significance, can you state that more

than half of all successful women executives had children?c. At the 0.05 level of significance, can you state that more

than two-thirds of all successful women executives hadchildren?

d. Do you think the random sample assumption is valid?Explain.

1.

2.3.4.

358 CHAPTER NINE Fundamentals of Hypothesis Testing: One-sample Tests

marketing campaign produce a result with a meaningful increase in brand awareness?the minimal real-world impact an increase of less than lo/o has on the overall marketinsand the huge expenses associated with the marketing campaign, you should conclude thatcampaign was not successful. On the other han4 if the campaign increased brand awareness20o/o, you could conclude that the campaign was successful.

Ethical Issues You also need to distinguish between poor research methodologyunethical behavior. Ethical considerations arise when the hypothesis-testing process is manilated. Some of the areas where ethical issues can arise include the use of human subiectsexperiments, data collection method, informed consent from human subjects beingthe type of test (one-tail or two-tail test), the choice of the level of significance, datathe cleansing and discarding ofdata, and the failure to report pertinent findings.

Informed Consent from Human Respondents Being "Treated" Ethicalerations require that any individual who is to be subjected to some "treatment" in an ement be made aware of the research endeavor and any potential behavioral or physicaleffects. The person should also provide informed consent with respect to participation.

Data Collection Method-Randomization To eliminate the possibility ofbiases in the results, you must use proper data collection methods. To draw meaningfulsions, the data must be the outcome of a random sample from a population or from anment in which a randomization process was used. Potential respondents should not beted to self-select for a study, nor should they be purposely selected. Aside from theethical issues that may arise, such a lack of randomization can result in serious coverageor selection biases that destroy the integrity ofthe study.

Type of Test-Two-Tail or One-Tsil Test If prior information is available that leadsyou to test the null hypothesis against a specifically directed alternative, then a one-tail test ismore powerful than a two-tail test. However, if you are interested only in diferences from thenull hypothesis, not in the direction of the difference, the two-tail test is the appropriate proce-dure to use. For example, if previous research and statistical testing have already establithe difference in a particular direction, or ifan established scientific theory states that it issible for results to occur in only one direction, then a one-tail test is appropriate. It is neverappropriate to change the direction ofa test after the data are collected.

Choice of Level of SigntJicance, a. In a well-designed study, you select the level ofsignificance, cr, before data collection occurs. You cannot alter the level of significance afterthe fact to achieve a specific result. You should always report thep-value, not just the conclu-sions ofthe hypothesis test.

Data Snooping Data snooping is never permissible. It is unethical to perform a hypothe-sis test on a set ofdata, look at the results, and then specify the level ofsignificance or decidewhether to use a one-tail or two-tail test. You must make these decisions before the data are col-lected in order for the conclusions to be valid. In addition, you cannot arbitrarily change or dis-card extreme or unusual values in order to alter the results ofthe hypothesis tests.

Cleansing and Discarding of Data In the data preparation stage of editing, coding,and transcribing, you have an opportunity to review the data for any extreme or unusual values,After reviewing the data, you should construct a stem-and-leaf display and/or a box-and-whisker plot to determine whether there are possible outliers to double-check against theoriginal data.

The process of data cleansing raises a major ethical question. Should you ever remove avalue from a study? The answer is a qualified yes. If you can determine that a measurement lsincomplete or grossly in error because of some equipment problem or unusual behavioral

Summary 359

occurrence unrelated to the study, you can discard the value. Sometimes you have no choice-an individual may decide to quit a particular study he or she has been participating in before afinal measurement can be made. In a well-designed experiment or study, you should decide, inadvance, on all rules regarding the possible discarding of data.

Reporting of Findings In conducting research, you should document both good and badresults. You should not just report the results of hypothesis tests that show statistical signifi-cance but omit those for which there is insufficient evidence in the findings. In instances inwhich there is insufficient evidence to reject lllo, you must make it clear that this does not provethat the null hypothesis is true. What the result does indicate is that with the sample size used,there is not enough information to disprove the null hypothesis.

Summary To summarize, in discussing ethical issues concerning hypothesis testing, thekey is intent.You must distinguish between poor data analysis and unethical practice. Unethicalpractice occurs when researchers intentionally create a selection bias in data collection, manip-ulate the treatment of human subjects without informed consent, use data snooping to select thetype oftest (two-tail or one-tail) and/or level ofsignificance, hide the facts by discarding val-ues that do not support a stated hypothesis, or fail to report pertinent findings.

o (CD-ROM Topicl THE POWER OF A TESTSection 9.1 defines Type I and Type II errors and the power of the test. In $@Ef[E[on theStudent CD-ROM, the power of the test is examined in greater depth.

chapter presented the foundation of hypothesis test- improve a cereal-fill process. The three chapters that followbuild on the foundation of hypothesis testing discussedhere.

Figure 9.16 presents a roadmap for selecting the cor-rect one-sample hypothesis test to use.

learned how to perform Z and t tests on the popu-mean and the Z test on the population proportion.

learned how an operations manager of a produc-ility can use hypothesis testing to monitor and

9.16for selecting

test

9.7

One-Sample

.- l@

I

L

Cateqorical tY9" Numerical

l - D;1"-

lrg; zr"*t to, tt e No --

6 -

LP.*"il; f .*n?*nlJ

lir

t$*

fTest ::*i.t

Tf-.i:".-


In deciding which test to use, you should ask the fol-lowing questions:

. Does the test involve a numerical variable or a categori-cal variable? If the test involves a categorical variable,use the Ztestfor the proportion.

. If the test involves a numerical variable. do youthe population standard deviation? Ifyou know theulation standard deviation. use the Z test for theyou do not know the population standard deviatithe I test for the mean.

Table 9.4 provides a list of topics covered in this

Tlpe of Data

Numerical Categorical

T A B L E 9 . 4Summary of Topicsin Chaoter 9

Tlpe ofAnalysis

Hypothesis test concerninga single parameter

Z test ofhypothesis for themean (Section 9.2)

I test ofhypothesis for themean (Section 9.4)

Z test ofhypothesisproportion (Section

ZTestof Hypothesis for the Mean (o Known)

Z _

/Test of Hypothesis for the Mean (o Unknown)

x -1 t

o (level ofsignificance) 331alternative hypothesis (111) 329

B risk 332confidence coefficient (l - cr) 332confidence level 332critical value 331data snooping 358directional test 343hypothesis testing 328

(e.1)

(e.2)

level of significance (a) 331null hypothesis (F1s) 328one-tail test 343p-value 337power of a statistical test 332randomization 358region ofnonrejection 330region ofrejection 330robust 350

One Sample ZTest for the Proportion

ZTest for the Proportion in Termsof the Number of Successes

Z _

X -p ,oT=

"tln

,sT,

X - n n---

lnn(r - n)

Checking Your Understanding9.75 What is the difference between a null hypothesis, F16,and an alternative hypothesis, 11,?

9.75 What is the difference between a Type I error and aType II error?

sample proportion 353I test of hypothesis for a mean 346test statistic 330two-tail test 334Type I error 331Type II error 331Z test of hypothesis for the mean 3Ztestfor the proportion 353Ztest statistic 334

9.77 What is meant by the power of a test?

9.78 What is the difference between a one-tail test andtwo-tail test?

9.79 What is meant by ap-value?

p -n

How can a confidence interval estimate for the pop-mean provide conclusions to the correspondingsis test for the population mean?

What is the six-step critical value approach tois testing?

What are some of the ethical issues involved withing a hypothesis test?

the ConceptsAn article in Marketing News (T. T. Semon,ider a Statistical Insignificance Test," MarketingFebruary l,1999) argued that the level of signifi-used when comparing two products is often too

is, sometimes you should be using an ct, valuethan 0.05. Specifically, the article recounted testing

ion of potential customers with a preference forI over product2.The null hypothesis was that theion proportion of potential customers preferringI was 0.50, and the alternative hypothesis was that

not equal to 0.50. The p-value for the test was 0.22.article suggested that in some cases, this should be

evidence to reject the null hypothesis.the null and alternative hypotheses for this exam-

ple in statistical terms.Explain the risks associated with Type I and Type II

in this case.would be the consequences ifyou rejected the nullhesis for a p-value of 0.22?do you think the article suggested raising theof s?would you do in this situation?is your answer in (e) if thep-value equals 0.12?if it equals 0.06?

La Quinta Motor Inns developed a computer model topredict the profitability of sites that are being consid-as locations for new hotels. If the computer model pre-large profits, La Quinta buys the proposed site anda new hotel. If the computer model predicts small or

profits, La Quinta chooses not to proceed with that(Extracted from S. E. Kimes and J. A. Fitzsimmons,

ing Profitable Hotel Sites at La Quinta Motorl' Interfaces, Vol. 20, March-April 1990, pp.12-20).decision-making procedure can be expressed in the

is-testing framework. The null hypothesis is that theis not a profitable location. The alternative hypothesis isthe site is a profitable location.

ain the risks associated with committing a Type Iin this case.in the risks associated with committing a Type II

error in this case.Which type of error do you think the executives at La

Chapter Review Problems 361

d. How do changes in the rejection criterion affect theprobabilities of committing Type I and Type II errors?

9.85 In 2006, Visa wanted to move away from its long-running television advertising theme of "Visa, it's every-where you want to be." During the Winter Olympics, Visafeatured Olympians in commercials with a broader message,including securiry check cards, and payment technologiessuch as contactless processing. One of the first commercialsfeatured snowboarder Lindsey Jacobellis being coached tocalm down before a big race by imagining that her VisaCheck Card got stolen. A key metric for the success of tele-vision advertisements is the proportion of viewers who "likethe ads a lot." Harris Ad Research Service conducted a studyof 903 adults who viewed the new Visa advertisement andreported that 54 indicated that they "like the ad a lot."According to Harris, the proportion of a typical televisionadvertisement receiving the "like the ad a lot" score is 0.21(Extracted from T. Howard "Visa to Change Strategies inUpcoming Ads," usatoday.com, January 23, 2006.).a. Use the six-step critical value approach to hypothesis

testing and a 0.05 level of significance to try to provethat the new Visa ad is less successful than a typical tele-vision advertisement.

b. Use the five-step p-value approach to hypothesis testingand a 0.05 level of significance to try to prove that thenew Visa ad is less successful than a typical televisionadvertisement.

c. Compare the results of (a) and (b).

9.86 The owner of a gasoline station wants to study gaso-line purchasing habits by motorists at his station. He selectsa random sample of 60 motorists during a certain week,with the following results:

. The amount purchased was .F : ll.3 gallons,^t : 3.1 gal lons.

. 1l motorists purchased premium-grade gasoline.a. At the 0.05 level of significance, is there evidence that

the mean purchase was different from l0 gallons?b. Determine thep-value in (a).c. At the 0.05 level of significance, is there evidence that

fewer than 20%o of all the motorists at the station pur-chased premium-grade gasoline?

d. What is your answer to (a) if the sample mean equals10.3 gallons?

e. What is your answer to (c) if 7 motorists purchased pre-mium-grade gasoline?

9.87 An auditor for a government agency is assigned thetask of evaluating reimbursement for office visits to physi-cians paid by Medicare. The audit was conducted on a sam-ple of 75 of the reimbursements, with the following results:

. ln 12 of the office visits, an incorrect amount ofreimbursement was provided.

. The amount of reimbursement was X : S93.70.^s: s34.55.inta Motor Inns are trying hard to avoid? Explain.

362 CHAPTER NINE Fundamentals of Hypothesis Testing: One-sample Tests

a. At the 0.05 level of sigaificance, is there evidence thatthe mean reimbursement was less than $I00?

b. At the 0.05 level of significance, is there evidence thatthe proportion of incorrect reimbursements in the popu-lation was greater than 0. 10?

c. Discuss the underlying assumptions of the test used in (a).d. What is your answer to (a) if the sample mean equals $90?e. What is your answer to (b) if 15 office visits had incor-

rect reimbursements?

9.88 A bank branch located in a commercial district of acity has developed an improved process for serving cus-tomers during the noon-to-1:00 p.m. lunch period. Thewaiting time (defined as the time the customer enters theline until he or she reaches the teller window) of all cus-tomers during this hour is recorded over a period of I week.A random sample of 15 customers (see the data file

!l@!) is selected, and the results are as follows:

4.21 5.55 3.02 5.13 4.77 2.34 3.54 3.204.50 6 .10 0 .38 5 .12 6 .46 6 . r9 3 .79

a. At the 0.05 level of significance, is there evidence thatthe mean waiting time is less than 5 minutes?

b. What assumption must hold in order to perform the testin (a)?

c. Evaluate the assumption in (b) through a graphicalapproach. Discuss.

d. As a customer walks into the branch office during thelunch hour, she asks the branch manager how long shecan expect to wait. The branch manager replies,'Almostcertainly not longer than 5 minutes." On the basis of theresults of (a), evaluate this statement.

9.89 A manufacturing company produces electrical insula-tors. If the insulators break when in use, a short circuit islikely to occur. To test the shength of the insulators, destruc-tive testing is carried out to determine how much force isrequired to break the insulators. Force is measured byobserving the number of pounds of force applied to theinsulator before it breaks. The following data (stored in the

[!@! file) are from 30 insulators subject to this testing:

Force (in the Number of Pounds Requiredto Break the Insulator)

t ,87 0 1,7 28 1,656 1,6 1 0 1,634 1,7 84 1,522 1,696 1,592 1,662

1,866 1,7 64 1,734 1,662 1,734 1,7 7 4 1,550 1,756 1,7 62 1,866

1,820 1,7 44 1,788 1,688 1,810 1,7 52 1,680 1,8 1 0 1,652 1,736

a. At the 0.05 level of significance, is there evidence thatthe mean force is greater than 1,500 pounds?

b. What assumption must hold in order to perform the testin (a)?

c. Evaluate the assumption in (b) through a graphicalapproach. Discuss.

d. Based on (a), what can you conclude about the strengthof the insulators?

9.9O An important quality characteistic used by theufacturer of Boston and Vermont asphalt shingles isamount of moisture the shineles contain when thevpackaged. Customers may feel that they haveproduct lacking in quality if they find moisture andshingles inside the packaging. In some cases,moisture can cause the sranules attached to the shinsletexture and coloring purposes to fall offthe shingle,ing in appearance problems. To monitor the amountmoisture present, the company conducts moistureshingle is weighed and then dr ied. The shingle isreweighed, and, based on the amount of moisture takenof the product, the pounds of moisture per 100 squareare calculated. The company would like to show thatmean moisture content is less than 0.35 pounds persquare feet. The data file @@fs includes 36ments (in pounds per 100 square feet) for Boston shiand 31 for Vermont shinsles.a. For the Boston shingles, is there evidence at the

level of significance that the mean moistureless than 0.35 pounds per 100 square feet?

b. Interpret the meaning of thep-value in (a).c. For the Vermont shingles, is there evidence at the

level of significance that the mean moisture contentless than 0.35 pounds per 100 square feet?

d. Interpret the meaning of thep-value in (c).e. What assumption must hold in order to perform the

in (a) and (c)?f. Evaluate the assumption in (e) for Boston shingles

Vermont shingles by using a graphical approach. Di

9.91 Studies conducted by the manufacturer ofand Vermont asphalt shingles have shown product weito be a major factor in the customer's perception ofMoreover, the weight represents the amount of rawals being used and is therefore very important to thepany from a cost standpoint. The last stage oftheline packages the shingles before the packages areon wooden pallets. Once a pallet is full (a pallet forbrands holds l6 squares of shingles), it is weighed andmeasurement is recorded. The data file [!fts@the weight (in pounds) from a sample of 368 palletsBoston shingles and 330 pallets of Vermont shingles.a. For the Boston shingles, is there evidence that the

weight is different from 3,150 pounds?b. Interpret the meaning of thep-value in (a).c. For the Vermont shingles, is there evidence that

mean weight is different from 3,700 pounds?d. Interpret the meaning of thep-value in (c).e. What assumption is needed for (a) and (c)?f. Evaluate this assumption through a graphical

Discuss.

9.92 The manufacturer of Boston and Vermontshingles provides its customers with a 20-year warrantymost of its products. To determine whether a shingle

long as the warranty period accelerated-life testing isat the manufacturing plant. Acceleratedlife test-the shingle to the stresses it would be subject to

ime of normal use in a laboratory setting via anthat takes only a few minutes to conduct. In this

shingle is repeatedly scraped with a brush for a short

References 363

lnterpret the meaning of thep-value in (a).For the Vermont shingles, is there evidence that themean granule loss is different from 0.50 grams?Interpret the meaning of thep-value in (c).What assumption is needed for (a) and (c)?Evaluate this assumption through a graphical approach.Discuss.

b.c.

d.e.f.

of time, and the shingle granules removed by theare weighed (in grams). Shingles that experience

amounts of granule loss are expected to last longer inuse than shingles that experience high amounts of

ule loss. The data file !@ contains a sample ofts made on the company's Boston shingles

140 measurements made on Vermont shineles.the Boston shingles, is there evidence that the mean

loss is different from 0.50 srams?

its monitoring of the blackness of the news-print, first described in the Chapter 6 Managing

'Springtille Herald case, the production department of

newspaper wants to ensure that the mean blackness of thefor all newspapers is at least 0.97 on a standard scaleich the target value is 1.0. A random sample of 50

has been selected, and the blackness of eachhas been measured (and stored in the !fi! file).

the sample statistics and determine whether there

your knowledge about hypothesis testing in this Webwhich continues the cereal-fill-packaging dispute

Casefrom Chapter 7.Concerned about statements of groups such as the

mers Concerned About Cereal Cheaters (CCACC)the Web Case for Chapter 7). Oxford Cerealstly conducted a public experiment concerningpackaging. The company claims that the results of

experiments refute the CCACC allegations that 3.

4.Cereals has been cheating consumers by packag-

cereals at less than labeled weiehts. Review theCereals' press release and supporting documents

describe the experiment at the company's Web site,

Bradley, J. Y, Distribution-Free Stqtistical Zesls (UpperSaddle River, NJ: Prentice Hall, 1968).Daniel, W., Applied Nonparametric Statistics, 2nd ed.(Boston: Houghton Mifflin, 1990).

Report Writing Exercises9.93 Referring to the results of Problems 9.90-9.92 con-cerning Boston and Vermont shingles, write a report thatevaluates the moisture level, weight, and granule loss of thetwo types of shingles.

is evidence that the mean blackness is less than 0.97. Writea memo to management that summarizes your conclusions.

Blackness of 50 Newspapers

0.854 1.023 1.005 1.030 1.2t9 0.977 r .044 0.778 t . t22 1. t141.091 1.086 1.141 0.931 0.723 0.934 1.060 1.047 0.800 0.8891.012 0.695 0.869 0.734 1.131 0.993 0.762 0.814 1.108 0.805r.223 r .024 0.884 0.799 0.870 0.898 0.621 0.818 l . l l3 1.2861.052 0.678 1.162 0.808 1.012 0.859 0.951 L l t2 1.003 0.972

www.prenhall.com/Sprin gvilte/OC_FullUp.htm, andthen answer the following:

1. Are the results of the independent testing valid? Why orwhy not? If you were conducting the experiment, is thereanything you would change?

Do the results support the claim that Oxford Cereals isnot cheating its customers?

Is the claim of the Oxford Cereals CEO that many cerealboxes contain morethan 368 grams surprising? Is it true?

Could there ever be a circumstance in which the resultsof the public experiment snd the CCACC's results areboth correct? Explain.

Microsoft Excel 2007 (Redmond, WA: Microsoft Corp.,2007\.

,

3 .

E9.1 USING THE Z TEST FORTHE MEAN (o KNOWN)

You conduct a Ztest for the mean (o known) by either select-

ing the PHStat2 ZTest for the Mean, sigma known proce-

dure or by making entries in the [f!1filworkbook'

Usinq PHStat2 Z Testfor the Mean, Sigma Known

Select PHStat ) One-Sample Tests ) ZTest for the

Mean, sigma known. In the Z Test for the Mean, sigma

known, dialog box (shown below), enter values for the

Null Hypothesis, the Level of Significance, and the

Population Standard Deviation. If you know the sample

size and sample mean of your sample, click Sample

364 EXCEL coMPANIoN to chapter 9

O€t6

l{I Ftypotha*s:

Lcvd of Sigr*icrnca:

Poptidkrr Stardard Dcviathn :

Satr$ staHdlcsOptims

A Empte**stks Knsnn

Sarple Sael

Ssrplc Mcan;

a sdndc Statistks

I

TaSWkns

rd T$n-TC Test

|* lSpcr-TailTart

f Uxner-td te*

0utfut Options

Urifig,rytl

r--*-;

Ttle; T---

Statistics Known and enter those values. Otherwise, cli

Sample Statistics Unknown and enter the cell range

your sample as the Sample Cell Range. Then click one

the test options, enter a title as the Title' and click OK'

Using Z Mean.xls

You open and use either the ZMean TT or ZMean-All

worksheets of the fif,!1fi! workbook to apply the Z test

for the mean (o known). These worksheets use the

NORMSINVe<n function to determine the lower and

upper critical values and use the NORMSDIST(Z value)

function to compute the p-values from the Z value calculated

in cell 812. The ZMean-TT worksheet (see Figure 9'5 on

page 338) applies the two-tail Z test to the Example 9'2

cereal-fill example. The ZMean-All worksheet (see Figure

E9.l) includes both the two-tail test and the upper- and

lower-tail tests on one worksheet. To adapt these work-

sheets to other problems, change (if necessary) the null

hypothesi s, level of si gnificance, population standard devi'

ation, sample size, and sample mean values in the tinted

cells 84 through 88.lf you want the ZMean-'\l\ worksheet to show only

one of the single-tail tests, first make a copy of that work'

sheet (see the Excel Companion to Chapter l)' For a lower'

tail-test-only worksheet, select and delete rows 25 through

28 and then select and delete rows 14 through 19. Foran

upper-tail-test-only worksheet, select and delete rows 14

throush 24.

About the lF Function

Both worksheets of the EEEEEIE *orkbook and all other

hypothesis-testing worksheets presented in the rest ofthe

book use the IF (comparison, what to do if comparison

holds, what to do if comparison fails) function to suggest

whether you should reject the null hypothesis. (You

should always verify the results of worksheets and not

blindly accept results reported by formulas containing the

IF function.)You need to supply three things when using an IF

function in a formula. The comparison is an algebraic or

logical comparison between two things. In Figure E9'1,

the p-values are compared to the level of significance'

E9.2: Using the I Test for the Mean (o Unknown) 365

FIGURE E9.1ZMean_All worksheet

1I3t5b7II

101 1

ZTest oflhe

lntermediale Calculalions

K

ff

15 Lowelt b

l7 o-Valrte1g'19

2n

nZJ

)123

x272S

What to do if comparison holds(alls are either text val-ues to display (as in Figure E9.1) or numeric values or for-mulas to use as the cell contents. In Figure E,9. l, "Do notreject the null hypothesis" is displayed in rows 18, 23, and28 because all three p-values (in cells B1l,822, and 827)are greater than the level of significance in cell 85. (By theway, text values in IF functions must be enclosed by a pairof quotation marks as they are in Figure E9.1.)

You should always enter formulas containing the IFfunction as a single, continuous line, even though such for-mulas are generally shown in this text on two physical lines(as in Figure E9.1). For the hypothesis-testing worksheets,the formulas containing the IF function have been enteredinto column A cells so that their text values disolav acrosscolumns A and B.

E9.2 USING THE t TEST FORTHE MEAN (o unknown)

You conduct a / test for the mean (o unknown) by either

selecting the PHStat2 t Test for the Mean, sigma unknownprocedure orby making entries in the@l$frworkbook.

Using PHStat2 t Testfor the Mean, Sigma Unknown

Select PHStat ) One-Sample Tests ) t Test for theMean, sigma unknown. In the t Test for the Mean, sigmaunknown dialog box (shown at right), enter values for theNull Hypothesis and the Level of Significance. lf you

know the sample size, sample mean, and sample standarddeviation, click Sample Statistics Known and enter those

-86isQRT{87}*(88 - B{lrB11

=l,lORMSll{V{85/21=l{ORlrlSll{V{t - 8512}=2 - {1 - }|ORh|SO|ST{ABS{812D}*lF{Bt7 < 85, "Re.lect the null hypothesis",

'llo rot reJecl lhe null hypothesis'J

*HORHSINV{B5l-NORIiSDTST{812}*lF(822 < 85, "Rej6cr tho nrll hypotfteris",

"Do ool rejgct ths null hypolheels']

-xoRtislNv{l - 85)-1 " ilOR['tSDtST{812}-lF{827 < 85, "Rojeclthe n$ll hypolhesl$",

"0o ttot reject lhc rull hyporhesic'J

values. Otherwise, click Sample Statistics Unknown andenter the cell range of your data as the Sample Cell Range.Then, click one of the test options, enter a title as the Title,

and click OK.

tutaNr.$ l-lypsthesisr

Level of $gr$ficancer

Sempb Stalntics OSionsr; ssnple Stsfristic$ Knsiln

Sanple 5izer

Sarple Meen:

SsfiSle Stondard Devi&ionr

{*' 5ample Ssistics Ur*q"rary*n

I

I

;

t*

t-* *** --_:r-

Test 0plionsr Two-TdTes*

r-' 1$Pr-TeilTest

,l".' Lower-Tnil Te*

Odp.rt Options

Titler :

- -t*"*-i ry

l i_..__a-K _. . j l csrcd j

I J

14

A BTd for thr linort-b of tlu Xrat

23 Oata

rnol|r.*E lwrl of Slonficarca 0.5 r Slzr

r larll llit.fiI r Silndanl D.Yldon m,9

lntarmsdiste Celculationel 1 ilandard Enor ofthg Mean 6.trlt(

armes of FmsdomTrd Staddc l.tllr

14Two.Tdl Tet

1E ow.r Ctl{crl Vilu. 2milt7 )oa1 ua 2.Z$11

r.Vr||r. 0.19 Do nol ral6cl $. null lwooth.drn21 Loser-Tall Ted2. Low.r Crl0c.l V!lu. -tn o-V.lu. 0.t29(

Ilo not rel.cl ft. null lmolh*x

UDO'I.T'TI Id

zl rDal crldcrl yrlua t.lgt!0r70aB o-V!lr|.

Do nol rcle cl th. null hwolhedr

---

366 ExcEL coMPANIoN to chapter 9

FIGURE E9.2TMean Allworksheet

Using T Mean.xlsYou open and use either the TMean_TT or TMean_Allworksheets of the ft!fi! workbook to apply the I testfor the mean (o unknown). These worksheets use theTlNY(l-conJidence level, degrees offreedom) function todetermine the lower and upper critical values and to com-pute the p-values. They also use the TDIST(ABS(I),degrees of freedom,fails) function in which ABS(r) is theabsolute value of the / test statistic and tuils is either I for aone-tail test or 2 for a two-tail test.

The TMean_TT worksheet (see Figure 9.10 on page348) applies the two-tail I test to the Section 9.4 salesinvoices example. The TMean_All worksheet (see FigureE9.2) includes both the two-tail test and the upper- andlower-tail tests on one worksheet. To adapt these work-sheets to other problems, change (if necessary) the nullhypothesis, level of significance, sample size, samplemean, and sample standard deviation values in the tintedcells 84 through 88. To better understand how a messagegets displayed in these worksheets, read the 'About the IFFunction"part ofSection E9.1 onpage 364.

If you want the TMean_All worksheet to show onlyone of the single-tail tests, first make a copy of that work-sheet (see the Excel Companion to Chapter l). For a lower-tail-test-only worksheet, select and delete rows 26 through29 and then select and delete cell range Al5:B20. For anupper-tail-test-only worksheet, select and delete cell rangeAl5:B25. These instructions ask you to select and deletecertain cell ranges in order to preserve the D20:E23 calcu-lations area cell range (which is not shown in Figure E9.2).

-B0xsORT{86}- 8 6 - 1-{87 -il}Alr

--(nl{vps,814)-nilvGt, Br2l-TI,IST(ABS(Bt3), Br2, 2)-lF{818 < Bli, 'Rsl.ct tln null lrypolbedr',

1)o not r.r..t ilr. null hypo{hrlbJ

-{n|wc.85, Bt2,,- lF(Bl3< 0,EA,ff i l-lFlBil3 < Hi, 'ReJcc{ th. nufl lrSpothe*','Do mt rcJ.c{$. null hypothrdr'}

-milvg. s,8r4)-lF(Bl3< O,E/l.,r7'l-lF(ElE < 8li, 'R.r.cl lfte null hpolhcrb',

'0o not rdea rhc mll lrypo&crb'|llel-6eu

Coll El2: -TIIIST0ABS{B|3}, 812, t}C-cllg}3: -1 -sn

E9.3 USING THE Z TEST FORTHE PROPORTION

You conduct a Ztest for the proportion by either sethe PHStat2 Z Test for the Proportion procedure or bying entries in the EIEEEE@ workbook.

Using PHStat2 Z Test for the PropoftionSelect PHStat ) One-Sample Tests ) Z Test forProportion. Inthe Z Test for the Proportion dialog(shown below), enter values for the Null Hypothesis,

i ildttypnthdr

i tarulofgfrftrmr

i frfr$crof 9ffir

i Sandcszol

l

ii

It trp-TdTastr $pr-f*rr*f uru-rdta*

, * r f

ot( Crt6l

ngk-

heoxhe

Level of Significance, the Number of Successes, and the

Sample Size. Click one of the test options, enter a title as

the Title, and click OK.

Using Z Proportion.xlsYou open and use either the ZProp-TT or ZProp-Allworksheets of the EEEEEEEEIE workbook Io apply the Ztest for the proportion. These worksheets use theNORMSINV(P<X) function to determine the lower andupper critical values and the NORMSDIST(Z value) func-tion to compute the p-values.

The ZProp_TT worksheet (see Figure 9.15 on page355)applies the two-tail Ztestro the Section 9.5 competrtivethreat example. The ZProp-All worksheet (see Figure E9.3)

FIGURE E9.3ZProp_All worksheet

E9.3: Usinc the ZTest fbr the Prooortion 367

includes both the two-tail test and the upper- and lower-tailtests on one worksheet. To adapt these worksheet to otherproblems, change (if necessary) the null hypothesis, level

of s igni f icance, number of successes, and sample s izevalues in the tinted cells 84 through 88. To better under-stand how messages get displayed in these worksheets,read the 'About the IF Function" part of Section E9.l onpage 364.

lf you want the ZProp-All worksheet to show only oneof the single-tail tests, first make a copy of that worksheet(see the Excel Companion to Chapter 1). For a lower-tail-test-only worksheet, select and delete rows 25 through 28

and then select and delete rows 14 through 19. For an

upper-tail-test-only worksheet, select and delete rows 14

throush 24.

-46187-soRT(8r11 - 8CFB4-{810 - B4yB1 I

-t{0RMSlt{V{85,2i=NORfttSlNV{l .85,?}*2' (1 . r'loRilSDlsT{ABs{812)}}*lF{817 < 85, 'Tsjecl lhe null hypotlresi*",

'?o not feject the '|ull hypotbesis'J

-ltoRMsrilv{85}-NoRllilSDlsT{812}*lF(822 < 85,'?ejedthe null hyporhasls",

"Do nol rajecl lhe null hypcthesis'J

-l{oRils|l{v{l - 85}*1 - t{oRfrlsDrsT(B12}*lF{827 < 85, "Rejesttbe null hypothesis".

"0o not rejecl the null hypothesis"]

chap 9

Documents

applications hypothesis

null hypothesis

oxford cereals scenario

sample chapter

usingstatistics oxford

excelcompanionto chapter

oxford cerealsscenario

value consistent of