10- 1 chapter ten mcgraw-hill/irwin © 2005 the mcgraw-hill companies, inc., all rights reserved
TRANSCRIPT
10- 1
Chapter
Ten
McGraw-Hill/Irwin
© 2005 The McGraw-Hill Companies, Inc., All Rights Reserved.
10- 2
Chapter TenOne-Sample Tests of One-Sample Tests of HypothesisHypothesis
GOALSWhen you have completed this chapter, you will be able to:ONEDefine a hypothesis and hypothesis testing.
TWO Describe the five step hypothesis testing procedure.
THREEDistinguish between a one-tailed and a two-tailed test of hypothesis.
FOURConduct a test of hypothesis about a population mean.
10- 3
Chapter Ten continued
GOALSWhen you have completed this chapter, you will be able to:
FIVE Conduct a test of hypothesis about a population proportion.
SIXDefine Type I and Type II errors.
SEVENCompute the probability of a Type II error.
One-Sample Tests of HypothesisOne-Sample Tests of Hypothesis
10- 4
What is a Hypothesis?What is a Hypothesis?
Twenty percent of all customers at Bovine’s Chop House return for another meal within a month.
What is a What is a Hypothesis?Hypothesis?
A statement about the value of a population parameter developed for the purpose of testing.
The mean monthly income for systems analysts is $6,325.
10- 5
What is Hypothesis Testing?What is Hypothesis Testing?
Hypothesis testingHypothesis testing
Based on sample
evidence and probability
theory
Used to determine whether the hypothesis is a reasonable statement
and should not be rejected, or is
unreasonable and should be rejected
10- 6
Hypothesis TestingHypothesis Testing
D o n o t re jec t n u ll R e jec t n u ll an d accep t a lte rn a te
S tep 5 : Take a sam p le , a rrive a t a d ec is ion
S tep 4 : F orm u la te a d ec is ion ru le
S tep 3 : Id en tify th e tes t s ta tis t ic
S tep 2 : S e lec t a leve l o f s ig n ifican ce
S tep 1 : S ta te n u ll an d a lte rn a te h yp o th eses
10- 7
Alternative Hypothesis H1:
A statement that is accepted if the sample data provide evidence that the
null hypothesis is false
Null Hypothesis H0
A statement about the value of a population
parameter
Step One: State the null and alternate hypothesesStep One: State the null and alternate hypotheses
10- 8
Three possibilities regarding
means
H0: = 0H1: = 0
H0: < 0H1: > 0
H0: > 0H1: < 0
Step One: State the null and alternate Step One: State the null and alternate hypotheseshypotheses
The null hypothesis
always contains equality.
3 hypotheses about means
10- 9
Step Two: Select a Level of Step Two: Select a Level of Significance.Significance.
The probability of rejecting the null
hypothesis when it is actually true; the level of
risk in so doing.
Rejecting the null hypothesis when it is actually true
Accepting the null hypothesis when it is actually false
Level of SignificanceLevel of Significance
Type I ErrorType I Error
Type II ErrorType II Error
10- 10
Step Two: Select a Level of Significance.Step Two: Select a Level of Significance.
Researcher
Null Accepts Rejects
Hypothesis Ho Ho
Ho is true
Ho is false
Correct
decision
Type I error
Type II
Error
Correct
decisionRisk Risk tabletable
10- 11
Step Three: Select the test statistic.Step Three: Select the test statistic.
A value, determined from sample information, used to determine whether or not to reject the null hypothesis.
Examples: z, t, F, 2
Test statistic Test statistic zz Distribution as a Distribution as a test statistictest statistic
n/
X
z
The z value is based on the sampling distribution of X, which is normally distributed when the sample is reasonably large (recall Central Limit Theorem).
10- 12
Step Four: Formulate the decision rule.Step Four: Formulate the decision rule.
Critical value: The dividing point between the region where the null hypothesis is rejected and the region where it is not rejected.
0 1.65
D o not
re ject
[P robability = .95]
R egion of
re jection
[P robability= .05]
C ritica l va lue
Sampling DistributionSampling DistributionOf the Statistic Of the Statistic zz, a, aRight-Tailed Test, .05Right-Tailed Test, .05Level of SignificanceLevel of Significance
10- 13
Reject the null hypothesis and accept the alternate hypothesis if
Computed -z < Critical -z
or
Computed z > Critical z
Decision Rule
Decision Rule
10- 14
Using the p-Value in Hypothesis Testing
If the p-Value is larger than or equal to the significance level, , H0 is not rejected.
pp-Value-ValueThe probability, assuming that the null hypothesis is true, of finding a value of the test statistic at least as extreme as the computed value for the test
Calculated from the probability distribution function or by computer
Decision Rule
If the p-Value is smaller than the significance level, , H0 is rejected.
10- 15
> .0 5 .1 0p
> .0 1 .0 5p
Interpreting p-valuesInterpreting p-values
SOME evidence Ho is not true
> .0 0 1 .0 1p
STRONG evidence Ho is not true
VERY STRONG evidence Ho is not true
10- 16
Step Five: Make a decision.Step Five: Make a decision.
MovieMovie
10- 17
One-Tailed Tests of Significance
One-Tailed Tests of SignificanceOne-Tailed Tests of Significance
The alternate
hypothesis, H1, states a direction
H1: The mean yearly commissions earned by
full-time realtors is more than $35,000. (µ>$35,000)
H1: The mean speed of trucks traveling on I-95 in Georgia is less than 60 miles per hour. (µ<60)
H1: Less than 20 percent of the customers pay cash for their gasoline purchase. 20)
10- 18
One-Tailed Test of Significance
.
0 1.65
D o not
re ject
[P robability = .95]
R egion of
re jection
[P robability= .05]
C ritica l va lue
Sampling DistributionOf the Statistic z, aRight-Tailed Test, .05Level of Significance
10- 19
Two-Tailed Tests of Significance
H1: The mean price for a gallon of
gasoline is not equal to $1.54.
(µ ne $1.54).
No direction is specified in the alternate hypothesis H1.
H1: The mean amount spent by customers at the
Wal-mart in Georgetown is
not equal to $25.
(µ ne $25).
Two-Tailed Tests of SignificanceTwo-Tailed Tests of Significance
10- 20
Two-Tailed Tests of SignificanceTwo-Tailed Tests of Significance
Regions of Nonrejection and Rejection for a Two-Tailed Test, .05 Level of Significance
0 1.96
D o not
re ject
[P robability = .95]
R egion of
re jection
[P robability= .025]
C ritica l va lue-1.96
R egion of
re jection
[P robability= .025]
C ritica l va lue
10- 21
Testing for the Population Mean: Large Sample, Population Standard Deviation
Known
n/
X
z
Test for the population Test for the population mean from a large sample mean from a large sample with population standard with population standard
deviation knowndeviation known
10- 22
Example 1
The processors of Fries’ Catsup indicate on the label that the bottle contains 16 ounces of catsup. The standard deviation of the process is 0.5 ounces. A sample of 36 bottles from last hour’s production revealed a mean weight of 16.12 ounces per bottle. At the .05 significance level is the process out of control? That is, can we conclude that the mean amount per bottle is different from 16 ounces?
10- 23
EXAMPLE 1
Step 1 State the null and the alternative hypotheses
H0: = 16H1: 16
Step 3Identify the test statistic. Because we know the population standard
deviation, the test statistic is z.
Step 2 Select the significance level. The significance level is .05.
Step 4 State the decision rule. Reject H0 if z > 1.96
or z < -1.96 or if p < .05.
Step 5Make a decision and interpret the results.
10- 24
Example 1
44.1365.0
00.1612.16
n
Xz
oComputed z of 1.44
< Critical z of 1.96,
op of .1499 > of .05,
Do not reject the null hypothesis.
The p(z > 1.44) is .1499 for a two-tailed test.
Step 5: Make a decision and interpret the results.
We cannot conclude the
mean is different from 16 ounces.
10- 25
Testing for the Population Mean: Large Sample, Population Standard Deviation Unknown
zX
s n
/
Testing for the Testing for the Population Mean: Population Mean:
Large Sample, Large Sample, Population Standard Population Standard Deviation UnknownDeviation Unknown
Here is unknown, so we estimate it with the sample
standard deviation s.
As long as the sample size n > 30, z can be approximated
using
10- 26
Example 2
Roder’s Discount Store chain issues its own credit card. Lisa, the credit manager, wants to find out if the mean monthly unpaid balance is more than $400. The level of significance is set at .05. A random check of 172 unpaid balances revealed the sample mean to be $407 and the sample standard deviation to be $38.
Should Lisa conclude that the population mean is greater than $400, or is it reasonable to assume that the difference of $7 ($407-$400) is due to chance?
10- 27
Example 2 Example 2
Step 1
H0: µ < $400
H1: µ > $400
Step 2The significance
level is .05.
Step 3 Because the sample is large
we can use the z distribution as the test
statistic.
Step 4H0 is rejected if
z > 1.65 or if p < .05.
Step 5Make a decision and interpret the
results.
10- 28
42.217238$
400$407$
ns
Xz
The p(z > 2.42) is .0078 for a one-
tailed test.
oComputed z of 2.42
> Critical z of 1.65,
op of .0078 < of .05.
Reject H0.
Step 5Make a decision and interpret the
results.
Lisa can conclude that the mean unpaid balance
is greater than $400.
10- 29
Testing for a Population Mean: Small Sample, Population Standard Deviation Unknown
ns
Xt
/
The critical value of t is determined by its degrees of
freedom equal to n-1.
Testing for a Testing for a Population Mean: Population Mean:
Small Sample, Small Sample, Population Population
Standard Deviation Standard Deviation UnknownUnknown
The test statistic is the t
distribution.
10- 30
Example 3
The current rate for producing 5 amp fuses at Neary Electric Co. is 250 per hour. A new machine has been purchased and installed that, according to the supplier, will increase the production rate. The production hours are normally distributed. A sample of 10 randomly selected hours from last month revealed that the mean hourly production on the new machine was 256 units, with a sample standard deviation of 6 per hour.
At the .05 significance level can Neary conclude that the new machine is faster?
10- 31
Step 4 State the decision rule.There are 10 – 1 = 9 degrees of freedom.
Step 1
State the null and alternate hypotheses.
H0: µ < 250
H1: µ > 250
Step 2 Select the level of
significance. It is .05.
Step 3 Find a test statistic. Use the t distribution since is not known and n < 30.
The null hypothesis is rejected if t > 1.833 or, using the p-value, the null hypothesis is rejected if p < .05.
10- 32
Example 3Example 3
162.3106
250256
ns
Xt
oComputed t of 3.162 >Critical t of 1.833 op of .0058 < a of .05
Reject Ho
The p(t >3.162) is .0058 for a one-
tailed test.
Step 5 Make a decision and interpret the
results.
The mean number of amps produced is more than 250 per
hour.
10- 33
n
pz
)1(
The sample proportion is p and is the population proportion.
The fraction or percentage that indicates the part of the population or sample having a particular trait of interest.
sampledNumber
sample in the successes ofNumber p
ProportionProportion
Test Statistic for Testing a Single Population Proportion
10- 34
Example 4Example 4
In the past, 15% of the mail order solicitations for a certain charity resulted in a financial contribution. A new solicitation letter that has been drafted is sent to a sample of 200 people and 45 responded with a contribution. At the .05 significance level can it be concluded that the new letter is more effective?
10- 35
Example 4Example 4
Step 1State the null and the alternate hypothesis.
H0: p < .15 H1: p > .15
Step 2Select the level of
significance. It is .05.
Step 3Find a test statistic. The z distribution is the test statistic.
Step 4State the decision rule.The null hypothesis is rejected if z is greater than 1.65 or if p < .05.
Step 5Make a decision and interpret the results.
10- 36
Example 4
97.2
200
)15.1(15.
15.200
45
)1(
n
pz
Because the computed z of 2.97 > critical z of 1.65, the p of .0015 < of .05, the null hypothesis is rejected. More than 15 percent responding with a pledge. The new letter is more effective.
p( z > 2.97) = .0015.
Step 5: Make a decision and interpret the results.