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Chapter 8 The Comparison of Two Populations

True/False Questions

1. The only assumption required for the paired-difference test is that the population ofdifferences is normally distributed.

Answer: True Type: Concept Difficulty: Easy

2. You always state H0

first.


3. When testing for the difference between two population means taken fromindependent random samples, the two samples must be the same size.

Answer: False Type: Concept Difficulty: Easy

4. The appropriate distribution when testing for the difference in two population means,assuming equal variances and small sample sizes is the t distribution.


5. When using the t distribution to test for a difference between two population meanstaken from independent samples the appropriate value for degrees of freedom isalways (n1 + n2 -2).

Answer: False Type: Concept Difficulty: Easy

6. H0

and H1

combined will always include all possible outcomes.


7. The F distribution is a symmetric distribution.

Answer: False Type: Concept Difficulty: Medium

8. Sample sizes are considered large if at least one sample is at least 30.


Aczel, Complete Business Statistics, Sixth Edition 1

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9. The two assumptions necessary to test for a difference in two population proportionsare that the sample sizes are large and that the samples are independent randomsamples from the two populations.

Answer: True Type: Concept Difficulty: Medium

10. The degrees of freedom for the paired-observation t-test are n - 2.


11. If the difference between two sample means is significant, then this is evidence thatthe two samples come from populations with equal means.


12. When the null hypothesis is not rejected, it is inferred that any difference in the twosample means, is due to sampling error, rather than a difference in the two populations.


13. When the sample evidence leads us to reject the null hypothesis that two populationproportions are equal, then we can conclude that the two population proportions aresignificantly different.


14. In testing for a significant difference in two population proportions, at = .01, if the

test statistic is 4.20, then the null hypothesis will not be rejected.


15. Paired observations are taken from two independent populations.


2 Aczel, Complete Business Statistics, Sixth Edition

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Multiple Choice Questions

16. When testing for the equality of two population means, using = 0.05 with n1

= 12

and n2

= 10, from independent samples, so long as the population variances can be

assumed equal, the critical values for t are:A) +2.704 and -2.704B) +1.96 and -1.96C) +2.086 and -2.086D) +1.645 and -1.645E) none of the above

Answer: C Type: Computation Difficulty: Medium

17. To test whether or not two population variances are equal, the appropriate distributionis:A) z distributionB) chi-square distributionC) F distributionD) t distribution with n

1+ n

2- 2 degrees of freedom

E) none of the above

Answer: C Type: Concept Difficulty: Medium

Use the following to answer question 18-22:

A company made a major change in its advertising theme this year and is interested inknowing whether there is any significant increase in sales over last year. The following data isthe sales in thousands for different stores over the country, and has been adjusted for inflation.Take the difference as (current year's sales - last year's sales).

S t o r e L a s t Y e a r ' s S a l e s C u r r e n t Y e a r ' s S a l e s

1 1 8 3 2 0 6

2 4 0 6 5 2 83 3 8 8 6 7 8

4 6 9 4 6 0 1

5 2 7 4 2 5 86 1 3 7 1 7 0

7 3 3 3 18 1 4 2 3 1 4 6 8


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18. State the null and alternative hypotheses to test the hypothesis that the change inadvertising has increased sales.

A) H0:

D> 0, H

1:

D 0

B) H0:

D 0, H

1:

D> 0

C) H0:

D 0, H

1:

D< 0

D) H0:

D= 0, H

1:

D> 0


Answer: B Type: Computation Difficulty: Medium

19. Find the critical value to test the hypothesis that the change in advertising has

increased sales, using = 0.05.

A) z > 1.645B) |z| > 1.96C) |t| > 2.365D) t > 1.895E) none of the above

Answer: D Type: Computation Difficulty: Medium

20. Compute the test statistic for this test.A) 1.116B) 1.244

C) -1.116D) -1.244E) none of the above


21. If the decision for this test is to fail to reject the null hypothesis, what is theappropriate conclusion?A) the change in advertising did increase sales significantlyB) sales have not changed at all from last year to this current yearC) there is not sufficient evidence to indicate that the change in advertising increased

salesD) sales have decreased from last year to this current yearE) none of the above



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22. Construct a 95% confidence interval for the average change in sales.A) 50.25 (2.365) (40.385)B) 50.25 (1.96) (40.385)

C) 50.25 (1.895) (40.385)D) 50.25 (2.306) (40.385)E) none of the above

Answer: A Type: Computation Difficulty: Medium

Use the following to answer questions 23-25:

An auto racing driver is given training and his timings on five different tracks are thenmeasured. The times before and after training are given below. The differences are taken asbefore after.

T r a c k B e f o r e A f t e r1 2 2 6 1 9 8

2 7 0 6 7 0 1

3 5 5 9 5 8 9

4 9 7 5 8 9 2

5 2 8 0 2 6 4

23. Compute the average difference, taking each difference as before after.A) 20.4B) 102

C) 25.5D) 32.4E) none of the above

Answer: A Type: Computation Difficulty: Easy

24. State the null hypothesis to test that the training was beneficial to this driver.

A) H0:

D= 0

B) H0:

D 0

C) H0

: D

> 0

D) H0:

D 0


Answer: B Type: Concept Difficulty: Easy


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25. Was the training beneficial to this driver, at = 0.01?

A) YesB) No



A major educational testing service is evaluating the performance of its new automated essaygrading system. Student essays are scanned into a computer-recognizable format and thenevaluated by the computer program. The software will not be moved into second-level testingunless there is compelling evidence that it finds more grammatical errors than human graders.

A random sample of 100 essays evaluated by the software was compared to a different sample

of 100 essays evaluated by humans. Results were as follows:

Evaluator Sample Average Sample Std.Dev.Software 7.2 1.738Humans 5.2 1.627

It is assumed that both software grading (population 1) and human grading (population 2) arenormally distributed, with equal population variances.

26. State the null and alternative hypotheses to test whether this software does pick outmore errors.

A) H0: 1 - 2 0, H1: 1 -2 > 0

B) H0:

1-

2= 0, H

1:

1-

2 0

C) H0:

1-

2 0, H

1:

1-

2< 0

D) H0:

1-

2> 0, H

1:

1-

2 0


Answer: C Type: Concept Difficulty: Easy

27. Find the critical points to test whether this software does find more errors, at = 0.05.

A) +1.96B) +1.645C) +1.282D) +2.575E) +2.33

Answer: B Type: Computation Difficulty: Easy


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28. Compute the p-value for the test of this hypothesis.A) cannot compute from the information givenB) about one-half

C) very close to oneD) very close to zeroE) none of the above


29. Construct a 90% confidence interval for the average difference between the number oferrors picked by the software and those found by human graders.A) 2 (1.96) (0.9552)B) 2 (2.33) (0.9552)C) 2 (1.645) (0.9552)

D) 2 (1.28) (0.9552)E) none of the above


30. When testing for the equality of two population proportions, the F distribution is:A) sometimes appropriateB) never appropriateC) only appropriate if both sample sizes are less than 30D) only appropriate if at least one sample is at least 30E) used when the two variances are not equal

Answer: B Type: Concept Difficulty: Medium


The proportion of cars stolen that do not have a burglar alarm installed is 12% in a sample of100 cars. The proportion of cars stolen that do have a burglar alarm installed is 8%, also in asample of 100 cars. Assume cars with an alarm are population 1 and those without alarms arepopulation 2.


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31. State the null hypothesis to test the hypothesis that there is a difference in proportionof cars stolen with and without alarms.A) H

0: p

1- p

2= 0

B) H0: p

1- p

2 0

C) H0: p

1- p

2 0

D) H0: p

1- p

2 0

E) H0: p

1- p

2> 0

Answer: A Type: Concept Difficulty: Medium

32. Compute the p-value for this test of the hypothesis that there is a difference inproportion of cars stolen, with and without burglar alarms.

A) 0.6554B) 0.6528C) 0.3472D) 0.3264E) 0.1736

Answer: C Type: Computation Difficulty: Hard

33. Find the critical value or values for the test of the hypothesis that there is a difference

in the two proportions, at = 0.05.

A) |z| > 1.96

B) |z| > 1.645C) |z| > 2.33D) z > 1.96E) z < -1.645


34. Construct a 90% confidence interval for the difference between the proportion ofstolen cars with alarms and the proportion of stolen cars without alarms.A) 0.04 (1.28)*(0.0422)B) 0.04 (1.96)*(0.0422)

C) 0.04 (2.575)*(0.0422)D) 0.04 (1.645)*(0.0422)E) 0.04 (1.96)*(0.0422)



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35. If the decision on this hypothesis test is to reject the null hypothesis that the two

proportions are equal, at = 0.10, the correct interpretation is:

A) there is no evidence that there is a significant difference in the two proportions

B) there is a significant difference in the two proportionsC) cars without alarms are stolen twice as often as cars with alarmsD) an alarm on a car will reduce the likelihood that it is stolen by 10%E) none of the above

Answer: B Type: Concept Difficulty: Medium

36. Calculate the pooled variance for the following sample data.

S a m p l e m e a n S a m p l e V a r i a n c e S a m p l e S i z e4 0 1 0 1 2

3 0 1 2 1 5

A) 3.33B) 124.64C) 11.12D) 34.4E) none of the above



A tire manufacturer claims that his tires last at least 1,500 miles more than his competitor's.Two independent samples of 10 tires each are subjected to wear and the average life of thecompany's tires is found to be 16,700 miles, with a standard deviation of 1,700 miles. Thecompetitor's tires last a mean of 15,100 miles, with a standard deviation of 1,350 miles.

37. State the null hypothesis to test whether the manufacturer's claim is true. Assume thatthe manufacturer's tires are population 1 and the competitor's tires are population 2.

A) H0:

1-

2 1,500

B) H0:

1-

2 0

C) H0: 1 - 2 1,500D) H

0:

1-

2 0

E) H0:

1-

2= 1,500



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38. Find the critical value or values to test the claim that the manufacturer's tires last on

the average at least 1,500 more miles that the competitor's tires. Use = 0.05.

A) 1.330

B) 1.734C) 1.645D) 2.101E) 1.96


39. Compute the test statistic for this test.A) 2.33B) 9.16C) 1.53

D) 0.146E) none of the above

Answer: D Type: Computation Difficulty: Hard

40. The average stock price of a particular company is $34 before a dividend isannounced, for the last nine months, with a standard deviation of $2. After thedividend is announced, the average stock price is $37, with a standard deviation of$2.50 for the next nine months. State the null and alternative hypotheses to testwhether this data provides evidence that a dividend release raises the stock price.Assume that population I is before the dividend release and population 2 is after.

A) H0: 1 - 2 > 0, H1: 1 - 2 0

B) H0:

1-

2= 0, H

1:

1-

2 0

C) H0:

1-

2< 0, H

1:

1-

2 0

D) H0:

1-

2 0, H

1:

1-

2< 0

E) H0:

1-

2 0, H

1:

1-

2> 0


41. Using two independent samples, two population means are compared to determine if a

difference exists. The number in the first sample is 15 and the number in the secondsample is 12. How many degrees of freedom are associated with the critical t-value?A) 27B) 26C) 25D) None of the above.



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42. If a researcher is testing the difference in population mean before and after someexperimental intervention, the use of the ___________ test will reduce the___________ error rate without a corresponding increase in the risk of an

inappropriately rejected null hypothesis.A) Paired-observation t; Type IB) Paired-observation t; Type IIC) Independent-samples z; Type ID) Independent-samples z; Type IIE) None of the above

Answer: B Type: Concept Difficulty: Hard


The rate of consumption of disposable tooling is a significant cost driver at a smallmanufacturing firm which employs 12 workers. Recently, on a trial basis the firm switchedsuppliers of its tooling and observed the following changes in rates of consumption across itsemployees:

Employee

Monthly Consumption

Old New

Charles 72 59Caroline 59 57Emma 44 37Andrew 51 51

Roy 91 66Colleen 73 63Edgar 55 60Gregory 80 72Janet 72 67Brenda 57 48Susan 62 53Michael 65 67

The firm will permanently switch suppliers only if it has substantial evidence ( = 0.05) that

consumption is lowerunder the new supplier.


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43. What are the most appropriate null and alternate hypotheses for this scenario?

A) H0:

1

20; H

1:

1

2> 0

B) H0

: 1

2

= 0; H1

: 1

2

0

C) H0:

D0; H

1:

D< 0

D) H0:

D0; H

1:

D< 0

E) H0:

D= 0; H

1:

D0

Answer: D Type: Concept Difficulty: Medium

44. What is an appropriate decision rule to use for hypothesis testing in this situation?A) Reject H

0if the observed t is less than -1.796

B) Reject H0 if the absolute value of the observed t is greater than 2.074C) Reject H

0if the observed t is less than -2.200

D) Reject H0

if the absolute value of the observed z is greater than 1.96

E) Reject H0

if the absolute value of the observed z is less than -1.645


45. Compute the test statistic for this test:A) t = -2.97B) t = -1.42

C) t = -0.38D) z = -0.96E) z = -1.37

Answer: A Type: Computation Difficulty: Hard


A manufacturing firm uses two different processes to evaluate the quality of incomingcomponents of a particular class. In a recent test of the processes' effectiveness, a randomsample was taken consisting of 5,000 components, 2,500 that had been evaluated by process A

and 2,500 that had been evaluated by process B. Of the 2,500 that had been evaluated byprocess A, 87 had been misclassified. Process B generated 65 misclassifications in its sample.Assume that this evaluation was undertaken to determine if there is sufficient evidence to

conclude that process B leads to a lower miscalculation rate. Assume also that is set at 0.05.


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46. Which of the following is the appropriate null hypothesis to test in this instance?A) H

0: p

A- p

B0

B) H0

: pA

- pB

0

C) H0: p

A- p

B= 0

D) H0: p

A p

B0

E) None of the above

Answer: A Type: Concept Difficulty: Medium

47. What is the appropriate critical value for the test statistic in this instance?A) Z < - 1.96B) Z < - 1.645C) Z > 2.33D) Z > 1.96E) Z > 1.645

Answer: E Type: Computation Difficulty: Medium

48. What is the p-value for the observed test statistic in this instance?A) 0.0351B) 0.0708C) 0.1401D) 0.3085E) 0.2266

Answer: A Type: Computation Difficulty: Hard

49. Before a researcher can proceed with hypothesis tests involving a potential differencein population means, she hopes to first establish that the two population variances areequal. In a random sample of 18 observations from each of populations 1 and 2, she

observes sample variances of, respectively, 33 and 47. Given = 0.10, the critical

lower and upper values for F are, respectively:A) F

LOWER= 0.32, F

UPPER= 3.10

B) FLOWER

= 0.37, FUPPER

= 2.67

C) FLOWER

= 0.39, FUPPER

= 2.60

D) FLOWER

= 0.44, FUPPER

= 2.27

E) FLOWER

= 0.45, FUPPER

= 2.22



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Short Answer Questions

50. Before selecting a method for studying whether two populations have different means,

a researcher first assesses whether or not their variances are different. Suppose herandomly samples from each population and observes the following:

S a m p l e S i z e ( n ) S a m p l e S t d . D e v . ( s )

P o p u l a t i o n 1 1 0 7 . 9 4

P o p u l a t i o n 2 1 6 4 . 7 9

State:A) Null and alternate hypotheses

B) Critical values for the test statistic at = 0.10

C) The observed test statistic and a corresponding decision

Answer:

A) H0:

2

1=

2

2, H

1:

2

1

2

2

B) FLOWER

= 0.33, FUPPER

= 2.59

C) FOBSERVED

= 2.748; Reject H0

Type: Computation Difficulty: Hard

51. Guests of a popular resort destination are given the opportunity to purchasememberships in a resort company's international condominium club. Managementbelieves that guests who stay in oceanfront rooms are more inclined to purchase

memberships than those who stay in marshview rooms. Suppose 100 oceanfrontguests were randomly sampled and 36 of them purchased memberships. Out of 100marshview guests, only 29 purchased memberships. Is this sufficient evidence to

conclude (at = 0.05) that oceanfront guests are more likely to purchase

memberships?

Answer: No, given Zcrit

= 1.645 and Zobserved

= 1.06, this is not sufficient evidence to

conclude that the two populations differ in their rate of membership purchasesType: Computation Difficulty: Medium

52. Monthly travel and entertainment expenditures for a random sample of 16 top-levelbusiness executives were observed before and after an IRS ruling on the deductibilityof certain business expenses. The average difference was $-123.20, with a standard

deviation of $115.43. Does it appear the IRS ruling significantly (at = 0.05) affected

these executives behavior?

Answer: Yes, given |tcrit

| = 2.13 and an observed t of -4.27, we would reject a null

hypothesis of no difference and conclude there has been a change in behavior.


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Type: Computation Difficulty: Medium


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53. A survey was conducted to see if the proportion of men and women liking this brandof jeans differed. In a sample of 100 men and 90 women, 62 of the men liked thejeans, and 66 of the women liked the jeans. Conduct a test of the hypothesis that there

is no difference in the proportions of men and women liking this brand of jeans.Assume = 0.05.

Answer: z = -1.66, do not reject the null Type: Computation Difficulty: Hard

54. A soap manufacturer changes the color of the wrapper of the soap from red to green inthe hope of increasing sales. The mean difference in sales over 30 stores is found to be$150 per month, with a standard deviation of $120. Find the p-value for this test of thehypothesis that the change in color of the wrapper has increased sales.

Answer: p approximately 0 Type: Computation Difficulty: Medium

55. Two samples of 64 students each from two colleges in a city spend an average of $540and $590, with standard deviations of $25 and $29 per month on their living expenses,respectively. State the null and alternative hypotheses to test the claim that the livingexpenses of the students are the same.

Answer: H0:

1-

2= 0, H

1:

1-

2 0 Type: Concept Difficulty: Medium

56. A commercial bank wants to lower the fluctuations in its deposit holdings andannounces a minimum amount required to be held in its checking accounts. Thevariance in 16 of its accounts before the minimum limit was 2,500 dollars squared,

and after the change it is 2025 dollars squared. Has the change in policy decreased the

variance of account holdings? Assume = 0.05.

Answer: F = 1.23, do not reject the null; no. Type: Computation Difficulty: Hard

57. Compute the p-value for a two-tailed test of the difference in two means, with bothsample sizes at least 30, if the test statistic is z = 2.50.

Answer: 0.0124 Type: Computation Difficulty: Medium

58. A survey was conducted to see if the proportion of men and women liking this brandof jeans differed. In a sample of 100 men and 90 women, 62 of the men liked thejeans, and 66 of the women liked the jeans. Construct a 95% confidence interval forthe difference in the proportion of men and women liking these jeans.

Answer: [-0.0186, +0.2452 Type: Computation Difficulty: Medium


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59. Two samples of 64 students each from two colleges in a city spend an average of $540and $555, with standard deviations of $25 and $29 per month on their living expenses,respectively. Compute the p-value for this test of the hypothesis that the living

expenses of these students are the same.

Answer: 0.0018 Type: Computation Difficulty: Medium

60. A sample of seven students who were given a GMAT test before and after they tookreview classes had the following scores:

S t u d e n t B e f o r e A f t e r

1 5 0 0 5 1 02 6 1 0 6 5 0

3 4 3 0 4 8 04 4 8 0 4 6 05 4 4 0 4 1 06 5 5 0 6 0 0

7 6 9 0 7 5 0

At an 0.01 level of significance, can we conclude that the instructional classes werehelpful?

Answer: t = 1.66, do not reject the null, no. Type: Computation Difficulty: Hard

61. Two control groups of 25 members each were given cigarettes with and without aspecial filter. Their physical condition was examined after one month by testing thetime it took them to climb a certain flight of stairs. The mean time for the smokerswith the filter was 25 seconds, with a standard deviation of 3 seconds, and the time forthe smokers without the filter was 32 seconds, with a standard deviation of 2.5seconds. Construct a 99% confidence interval for the mean difference in the time ittakes the two groups to climb the stairs.

Answer: [4.90, 9.09] Type: Computation Difficulty: Medium

62. In fifty different localities, the cable company gives free access to all cable channelsfor a weekend, as a promotional gesture. The mean proportion of customers who hadthe premium channels before the promotion was 20%. The mean proportion ofcustomers who had the premium channels after the promotion was 26%. Is the

increase significant at = 0.05?

Answer: z = 0.71, do not reject the null, no. Type: Computation Difficulty: Hard


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63. Your company is interested in a new method of advertising. They test the new methodin eleven markets with the following sales in thousands of dollars: $124; $157; $98;$190; $103; $135; $149; $176; $200; $180; and $256. Using the old method of

advertising, these same markets had the following sales in thousands of dollars: $142;$158; $97; $212; $116; $135; $150; $184; $209; $180; and $269. Is the difference

significant at = 0.05?

Answer: t = 3.13529, P = .0106, reject the null hypothesis; the new procedure is betterthan the old. Type: Computation Difficulty: Medium

64. Your company is interested in a new method of advertising. In order to test the newmethod they have selected thirty-one control and thirty one experimental markets.Mean sales in the control markets were 134,630 with a standard deviation of 5,290.Mean sales in the control markets were 138,780 with a standard deviation of 5,730. At

the 95% confidence level can you conclude that the new method is better than the old?

Answer: Yes, Z = 2.95 Type: Computation Difficulty: Medium

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