9 Hypothesis Testing

Section 9.4 Exercises

CLARIFYING THE CONCEPTS

Question 9.197

1. What assumption is required for performing the test that is not required for the test? (p. 525)

9.4.1

The population standard deviation is known.

Question 9.198

2. What do we use to estimate the unknown population standard deviation ? (p. 525)

PRACTICING THE TECHNIQUES

CHECK IT OUT!

To do	Check out	Topic
Exercises 3–8	Example 18	One-tailed test for : critical value method
Exercises 9–14	Example 19	Two-tailed test for : critical value method
Exercises 15–20	Example 20	One-tailed test for : -value value method
Exercises 21–26	Example 21	Two-tailed test for : -value method
Exercises 27–30	Example 22	Estimating the -value for a test for
Exercises 31–36	Example 23	Equivalence between confidence intervals and two-tailed tests for
Exercises 37–40	Example 24	Interpreting software output

For Exercises 3–14, do the following:

State the hypotheses.
Calculate the critical value and state the rejection rule. Also, sketch the critical region.
Find the test statistic .
State the conclusion and the interpretation.

539

Question 9.199

3. vs. , , , ,

9.4.3

(a) (b) . Reject if .

(c) (d) Since is , the conclusion is reject . There is evidence at the 0.05 level of significance that the population mean is less than 22.

Question 9.200

4. vs. , , , ,

Question 9.201

5. vs. , , , , , population is normal

9.4.5

(a) (b) . Reject if .

(c) (d) Since is not , the conclusion is do not reject . There is insufficient evidence at the 0.01 level of significance that the population mean is greater than 11.

Question 9.202

6. vs. , , , , , population is normal

Question 9.203

7. A random sample of size 25 from a normal population yields and . Researchers are interested in finding whether the population mean exceeds 100, using level of significance .

9.4.7

(a) (b) . Reject if 2.492.

(c) (d) Since (d) is not , the conclusion is do not reject . There is insufficient evidence at the 0.01 level of significance that the population mean is greater than 100.

Question 9.204

8. A random sample of size 100 from a population with an unknown distribution yields a sample mean of −5 and a sample standard deviation of 5. Researchers are interested in finding whether the population mean is less than −4, using level of significance .

Question 9.205

9. vs. , ,

9.4.9

(a) (b) . Reject if or if .

(c) (d) Since is , the conclusion is reject . There is evidence at the 0.05 level of significance that the population mean differs from 102.

Question 9.206

10. vs. , , , ,

Question 9.207

11. vs. , , , , , population is normal

9.4.11

(a) (b) . Reject if or if .

(c) (d) Since is not and not , the conclusion is do not reject . There is insufficient evidence at the 0.10 level of significance that the population mean differs from 1000.

Question 9.208

12. vs. , , , , , population is normal

Question 9.209

13. A random sample of size 36 from a population with an unknown distribution yields and . Researchers are interested in finding whether the population mean differs from 9, using level of significance .

9.4.13

(a) (b) . Reject if or if .

(c) (d) Since is , the conclusion is reject . There is evidence at the 0.10 level of significance that the population mean differs from 9.

Question 9.210

14. A random sample of size 16 from a normal population yields and . Researchers are interested in finding whether the population mean differs from 1000, using level of significance .

For Exercises 15–26, do the following:

State the hypotheses and the rejection rule using the -value method.
Calculate the test statistic .
Find the -value. (Use technology or estimate the -value.)
State the conclusion and the interpretation.

Question 9.211

15. vs. , , , ,

9.4.15

(a) . Reject if the (b) (c) 0 (d) Since the is , the conclusion is reject . There is evidence at the 0.01 level of significance that the population mean is less than 10.

Question 9.212

16. vs. , , , ,

Question 9.213

17. vs. , , , , , population is normal

9.4.17

(a) . Reject if the (b) (c) 0.0285 (d) Since the is , the conclusion is reject . There is evidence at the 0.10 level of significance that the population mean is greater than 100.

Question 9.214

18. vs. , , , , , population is normal

Question 9.215

19. A random sample of size 400 from a population with an unknown distribution yields a sample mean of 230 and a sample standard deviation of 5. Researchers are interested in finding whether the population mean is greater than 200, using level of significance .

9.4.19

(a) . Reject if the (b) (c) 0 (d) Since the is , the conclusion is reject . There is evidence at the 0.05 level of significance that the population mean is greater than 200.

Question 9.216

20. A random sample of size 100 from a population with an unknown distribution yields and . Researchers are interested in finding whether the population mean is less than 28, using level of significance .

Question 9.217

21. vs. , , , ,

9.4.21

(a) . Reject if the (b) (c) 1 (d) Since the is not , the conclusion is do not reject . There is insufficient evidence at the 0.01 level of significance that the population mean differs from 25.

Question 9.218

22. vs. , , , ,

Question 9.219

23. vs. , , , , , population is normal

9.4.23

(a) . Reject if the (b) (c) 0.8616 (d) Since the is not , the conclusion is do not reject . There is insufficient evidence at the 0.10 level of significance that the population mean differs from 3.14.

Question 9.220

24. vs. , , , , , population is normal

Question 9.221

25. A random sample of size 9 from a normal population yields and . Researchers are interested in finding whether the population mean differs from 0, using level of significance .

9.4.25

(a) . Reject if the (b) (c) 0.0003 (d) Since the is , the conclusion is reject . There is evidence at the 0.05 level of significance that the population mean differs from 0.

Question 9.222

26. A random sample of size 16 from a normal population yields and . Researchers are interested in finding whether the population mean differs from 2.0, using level of significance .

For Exercises 27–30, use the table to estimate the -value for the hypothesis tests in the indicated exercises.

Question 9.223

27. Exercise 3

9.4.27

Question 9.224

28. Exercise 4

Question 9.225

29. Exercise 9

9.4.29

Question 9.226

30. Exercise 10

For Exercises 31–36, confidence interval is given. Use the confidence interval to test, using level of significance , whether differs from each of the indicated hypothesized values.

Question 9.227

31. A 95% confidence interval for is (1, 4).

Hypothesized values are

9.4.31

	Value of	Where lies in relation to 95% confidence interval (1, 4)	Conclusion of hypothesis test
(a)	0	Outside	Reject
(b)	2	Inside	Do not reject
(c)	5	Outside	Reject

Question 9.228

32. A 99% confidence interval for is (57, 58).

Hypothesized values are

55.5
59.5
57.5

Question 9.229

33. A 90% confidence interval for is (−20, −10).

Hypothesized values are

−21
−5
−12

9.4.33

	Value of	Where lies in relation to 90% confidence interval (–20, −10)	Conclusion of hypothesis test
(a)	–21	Outside	Reject
(b)	–5	Outside	Reject
(c)	–12	Inside	Do not reject

Question 9.230

34. A 95% confidence interval for is (2010, 2015).

Hypothesized values are

2012
2007
2014

Question 9.231

35. A 95% confidence interval for is (−1, 1).

Hypothesized values are

1.5
−1.5
0

9.4.35

	Value of	Where lies in relation to 95% confidence interval (−1, 1)	Conclusion of hypothesis test
(a)	1.5	Outside	Reject
(b)	–1.5	Outside	Reject
(c)	0	Inside	Do not reject

Question 9.232

36. A 95% confidence interval for is (19,570, 20,105).

Hypothesized values are

20,000
21,000
19,571

For Exercises 37–40, software output from a test for is provided. For each, examine the indicated software output, and provide the following steps:

Step 1 State the hypotheses and the rejection rule.
Step 2 Find .
Step 3 Find the -value.
Step 4 State the conclusion and interpretation.

540

Use level of significance for each hypothesis test.

Question 9.233

37. TI-83/84 output

9.4.37

Step 1

We will reject if the .

Step 2

Step 3

Step 4 The is . Therefore we reject . There is evidence at level of significance that the population mean is greater than 98.6.

Question 9.234

38. Minitab output

Question 9.235

39. SPSS output

9.4.39

Step 1

We will reject if the .

Step 2

Step 3 -

Step 4 The is . Therefore we reject . There is evidence at level of significance that the population mean is not equal to 20.

Question 9.236

40. JMP output: Choose the left-tailed test.

APPLYING THE CONCEPTS

Question 9.237

41. Health Care Costs. The U.S. Agency for Healthcare Research and Quality (www.ahrq.gov) reports that, in 2010, the mean cost of a stay in the hospital for American women ages 18–44 was $15,200. A recent random sample of 400 hospital stays of women ages 18–44 showed a mean cost of $16,000, with a standard deviation of $5000. Test whether the population mean cost has increased since 2010, using level of significance .

9.4.41

Critical-value method: . . Reject if . . Since is , the conclusion is reject . There is evidence at the 0.05 level of significance that the population mean cost of a stay in the hospital for women aged 18–44 is greater than $15,200. Therefore we can conclude at level of significance 0.05 that the population mean cost of a stay in the hospital for American women aged 18–24 has increased since 2010. method: . Reject if the . . . Since the is , the conclusion is reject . There is evidence at the 0.05 level of significance that the population mean cost of a stay in the hospital for women aged 18–44 is greater than $15,200. Therefore we can conclude at level of significance 0.05 that the population mean cost of a stay in the hospital for American women aged 18-24 has increased since 2010.

Question 9.238

42. iPhone Apps. According to a 2010 Nielsen survey,⁹ the mean number of apps downloaded by iPhone users is 40. Suppose a recent sample of 36 iPhone users downloaded an average of 45 apps, with a standard deviation of 24. Test whether the population mean number of apps is greater than 40, using level of significance .

Question 9.239

43. Facebook Friends. According to Facebook.com, the mean number of Facebook friends is 130. Suppose a sample of 100 Facebook users has a mean number of 110 Facebook friends, with a standard deviation of 50. Test whether the population mean number of Facebook friends is less than the reported 130, using level of significance .

9.4.43

Critical-value method: . . Reject if . . Since is , the conclusion is reject . There is evidence at the 0.05 level of significance that the population mean number of Facebook friends is less than 130. method: . Reject if . . . Since the is , the conclusion is reject . There is evidence at the 0.05 level of significance that the population mean number of Facebook friends is less than 130.

Question 9.240

44. Small Business Employees. The U.S. Census Bureau reports that the average number of employees in a small business is 16.1. Suppose a sample of 49 small businesses showed a mean of 15 employees, with a standard deviation of 25. Test whether the population mean number of employees in a small business is different from the reported 16.1, using level of significance .

Internet Response Times. Use the following information for Exercises 45 and 46: The Web site www.internettrafficreport.com monitors Internet traffic worldwide and reports on the response times of randomly selected servers.

Question 9.241

45. On June 6, 2011, the Web site reported the following response times to Asia, in milliseconds:

165

175

2221

872

311

127

195

1801

769

225

261

249

421

We want to test whether the population mean response time is slower than 180 milliseconds, using a test and level of significance . A boxplot of the data is provided.

(Hint: The boxplot is right-skewed and the normal distribution is symmetric.) Can we proceed with the test? Explain.

9.4.45

No. The distribution of the variable is not normal and the sample size is less than 30.

Question 9.242

46. On June 6, 2011, the Web site reported the following response times to Asia, in milliseconds:

The normal probability plot of the data is also shown. We want to perform a test.

Are the conditions for performing the test satisfied? Explain how you know.
Test, using level of significance , whether the population mean response time is less than 60 milliseconds.
Explain why we can't use a test for this problem.

Deepwater Horizon Cleanup Costs. The following table represents the amount of money disbursed by BP to a random sample of six Florida counties, for cleanup of the Deepwater Horizon oil spill, in millions of dollars.¹⁰ Use the following information for Exercises 47–49.

541

County	Cleanup costs ($ millions)
Broward	0.85
Escambia	0.70
Franklin	0.50
Pinellas	1.15
Santa Rosa	0.50
Walton	1.35

Question 9.243

deepwaterclean

47. The normality of the data was confirmed in the Section 8.1 exercises. Test, at level of significance , whether the population mean amount of cleanup money exceeds $500,000.

9.4.47

Step 1 We will reject if the .

Step 2

Step 3

Step 4 The is . Therefore we reject . There is evidence at level of significance that the population mean amount of cleanup money is greater than $500,000.

Question 9.244

deepwaterclean

48. Answer the following:

Repeat your test from Exercise 47, this time using level of significance .
How do you think we should resolve the apparent contradiction in Exercise 47 and part (a) of this exercise?
Assess the strength of the evidence against the null hypothesis. Does this change depending on which level of you use?

Question 9.245

deepwaterclean

49.What if we changed to some larger value (though still smaller than )? Otherwise, everything else remains unchanged. Describe how this change would affect the following, if at all:

The -value
The conclusion from Exercise 47
The conclusion from Exercise 48(a)
The strength of the evidence against the null hypothesis

9.4.49

(a) Decrease (b) Stay the same (c) Increase (d) Depends on new value of (e) Stay the same (f) Depends on new value of

Question 9.246

wiisales

50. Wii Game Sales. The following table represents the number of units sold in the United States for the week ending March 26, 2011, for a random sample of eight Wii games.¹¹ The normality of the data was confirmed in the Section 8.1 exercises.

Construct and interpret a 95% interval for the population mean number of units sold. (See Section 8.2.)
Use your confidence interval to test, using level of significance , whether differs from the following values:
1. 30,000 units
2. 31,000 units
3. 0 units
4. 79,000 units

Game	Units (1000s)	Game	Units (1000s)
Wii Sports Resort	65	Zumba Fitness	56
Super Mario All Stars	40	Wii Fit Plus	36
Just Dance 2	74	Michael Jackson	42
New Super Mario Bros.	16	Lego Star Wars	110

Question 9.247

georgiarain

51. A Rainy Month in Georgia? The following table represents the total rainfall (in inches) for the month of February 2011 for a random sample of 10 locations in Georgia.¹² The normality was checked in the Section 8.1 exercises. Test whether the population mean amount of rainfall differs from 4 inches, using level of significance .

Location	Rainfall (inches)	Location	Rainfall (inches)
Athens	4.72	Atlanta	4.25
Augusta	4.31	Cartersville	3.03
Dekalb	2.96	Fulton	4.36
Gainesville	4.06	Lafayette	3.75
Marietta	3.20	Rome	3.26

9.4.51

Step 1 We will reject if the .

Step 2

Step 3

Step 4 The is not . Therefore we do not reject . There is insufficient evidence at level of significance that the population mean amount of rainfall differs from 4 inches.

Question 9.248

electricmiles

52. Electric Cars. The accompanying table shows the miles-per-gallon equivalent (MPGe) for five electric cars, as reported by www.hybridcars.com in 2014. Assume the data are drawn from a normal distribution. Test whether the population mean mileage is greater than 90 MPGe, using level of significance .

Electric Vehicle	Mileage (MPGe)
Tesla Model S	89
Nissan Leaf	99
Ford Focus	105
Mitsubishi i-MiEV	112
Chevrolet Spark	119

BRINGING IT ALL TOGETHER

Community College Tuition. Use the following information for Exercises 53–63. The College Board reported that the mean tuition and fees at community colleges nationwide for the 2013–2014 academic year was $3264. Data were gathered on the total tuition and fees for a random sample of 10 community colleges in 2015. The normal probability plot is shown here.

Normal probability plot.

542

Question 9.249

53. Is it appropriate to apply the test for the mean? Why or why not?

9.4.53

Yes. The normal probability plot indicates acceptable normality.

Question 9.250

54. Find the critical value for a right-tailed hypothesis test using level of significance .

Question 9.251

55. For the sample of 10 community colleges, the sample mean tuition and fees was $3541 with a sample standard deviation of $438. Test whether the population mean tuition and fees have increased using level of significance .

9.4.55

Step 1

Step 2 Reject if .

Step 3

Step 4 is . Therefore we reject . There is evidence at level of significance that the population mean tuition and fees for community colleges has increased from the previous level of $3264.

Question 9.252

56. Estimate the -value for the hypothesis test in the previous exercise.

Question 9.253

57. Assess the strength of evidence against the null hypothesis.

9.4.57

There is solid evidence against the null hypothesis that the population mean is $3264.

Question 9.254

58. A 95% confidence interval for the population mean tuition and fees is given by lower bound = $3227, upper bound = $3854. Use this confidence interval to test, at level of significance , whether differs from the following values.

$4000
$3500
$3264

Question 9.255

59. A data analyst, in attempting to use technology to test whether the population mean tuition and fees have increased, obtained the following Minitab output. However, it appears that the analyst asked for the wrong hypothesis test. How can you tell?

Minitab test output.

9.4.59

The line “Test of vs. .”

Question 9.256

60. How can we use the -value on the Minitab printout to find the -value needed for the right-tailed hypothesis test we performed in Exercise 55?

Question 9.257

61. Suppose, when we did Exercise 55, we also asked for the wrong hypothesis test, and obtained this Minitab output. For level of significance , what would have been our conclusion in that case? Why would this have been in error? What are some of the possible consequences of making an error of this sort?

9.4.61

Do not reject . Fees have increased. Answers will vary.

Question 9.258

62. Based on your experiences in these exercises, write a sentence about the importance of understanding the statistics behind the “point and click” power of statistical software.

Question 9.259

63. Challenge Exercise. Note that we have concluded that there is insufficient evidence that the population mean cost has changed, but evidence exists that the population mean cost has increased. How can the mean cost have increased without changing? Explain what is going on here, in terms of either critical regions or -values.

9.4.63

Since the for the two-tailed test is twice the for the one-tailed test, it is possible to conclude that there is insufficient evidence that the population mean cost has changed, but there is evidence that the population mean cost has increased if is between the two .

WORKING WITH LARGE DATA SETS

New York Towns. Work with the New York data set for Exercises 64 and 65.

Question 9.260

newyork

64. Use technology to find the summary statistics for the variable tot_pop, which lists the population for each of the towns and cities in New York with at least 1000 people.

Question 9.261

newyork

65. Suppose we are using the data in this data set as a sample of the population of all the towns and cities in the northeastern United States with at least 1000 people. Use technology to test at level of significance whether the population mean of these towns differs from 50,000.

9.4.65

WORKING WITH LARGE DATA SETS

Fast Food versus Full Service Restaurants. Open the data set, Restaurants. Here, we will look at the variable FFR per 1000, which refers to the number of fast food restaurants per 1000 residents of the county. For example, a value of 0.6 would mean that there are 0.6 fast food restaurants per 1000 residents in the county. We will perform a hypothesis test for the population mean FFR per 1000. We will then see whether this hypothesis test made the correct decision. Use technology to do the following:

Question 9.262

restaurants

66. Obtain a random sample of size 100 from the data set.

Question 9.263

restaurants

67. Using your sample, test whether the population mean number of fast food restaurants per 1000 residents differs from 0.4, using level of significance .

Question 9.264

restaurants

68. Find the actual value of the population mean number of fast food restaurants per county. Did your hypothesis test in Exercise 67 make the right decision? Explain.

WORKING WITH LARGE DATA SETS

Chapter 9 Case Study: Clothing Store Sales. Open the Chapter 9 Case Study data set, Clothing Store. Retail stores want you to come back again and again, with a short amount of time between purchases. The Clothing Store data set tracks the number of days since the last purchase for each customer. The marketing manager wants to make sure that the mean number of days since the last purchase is less than 150 days. Use technology to do the following:

Question 9.265

clothingstore

69. Obtain a random sample of size 100 from the data set.

Question 9.266

clothingstore

70. Using your sample, test whether the population mean number of days since the last purchase is less than 150, using level of significance .