Elementary statistics 6th edition larson solutions manual 1 by jorge.yang756

Solution Manual for Elementary Statistics 6th Edition by Larson Farber ISBN

0321911210 9780321911216

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Confidence Intervals

6.1 CONFIDENCE INTERVALS

6.1 Try It Yourself Solutions

1a. x   x  867  28.9

FOR THE MEAN (LARGE SAMPLES)

n 30

b. A point estimate for the population mean number of hours worked is 28.9. 2a. z

c. You are 95% confident that the margin of error for the population mean is about 2.8 hours. 3a. b. x  28.9, E  2.8 x E  28.9 2.8  26.1

28.9  2.8  31.7

c. With 95% confidence, you can say that the population mean number of hours worked is between 26.1 and 31.7 hours. This confidence interval is wider than the one found in Example 3.

4a. Enter the data.

b. 75% CI: (28.2, 31.0) 85% CI: (27.8, 31.4) 90% CI: (27.5, 31.7)

c. As the confidence level increases, so does the width of the interval.

5a. n  30, x  22.9,   1.5, zc  1.645

1.5 30

E  22.9 0.5  22.4

 E  22.9  0.5  23.4

c. With 90% confidence, you can say that the mean age of the students is between 22.4 and 23.4

Pearson Education, Inc.

2015

CHAPTER

7.9  b. E  zc  1.96 7.9 30  2.8

c  1.96, n  30,  

x  E





b. E  zc  1.645  0.5

years. Because of the larger sample size, the confidence interval is slightly narrower.

c. You should have at least 60 employees in you sample. Because of the larger margin of error, the sample size needed is much smaller.

6a. zc  1.96, E  2,   7.9  z  2  1.96  7.9 2 b. n   c      59.94  60  E   2 

212

6.1 EXERCISE SOLUTIONS

1. You are more likely to be correct using an interval estimate because it is unlikely that a point estimate will exactly equal the population mean.

2. b

3. d; As the level of confidence increases, zc increases, causing wider intervals.

4. No, the 95% confidence interval means that with 95% confidence you can say that the population mean is in this interval. If a large number of samples is collected and a confidence interval created for each, approximately 95% of these intervals will contain the population mean.

17. Because c  0.88is the lowest level of confidence, the interval associated with it will be the narrowest. Thus, this matches (c).

18. Because c  0.90 is the second lowest level of confidence, the interval associated with it will be the second narrowest. Thus, this matches (d).

19. Because c  0.95is the third lowest level of confidence, the interval associated with it will be the third narrowest. Thus, this matches (b).

20. Because c  0.98is the highest level of confidence, the interval associated with it will be the widest. Thus, this matches (a).

5. 1.28 6. 1.44 7. 1.15 8. 2.17 9. 11. x   3.8 4.27  0.47 x   26.43 24.67  1.76 10. 12. x   9.5 8.76  0.74 x   46.56 48.12  1.56 13. 15.  E  zc  E  zc  1.96  1.28  1.861  0.192 14. 16.  E  zc  E  zc  1.645  2.24  0.675  1.030

21. 22.  x  zc  x  zc  12.3  1.645  31.39  1.96  12.3  0.349  (12.0, 12.6)  31.3  0.173  (31.22, 31.56) 23.  x  zc  10.5  2.575 2.14  10.5  0.821  (9.7, 11.3) 5.2 30 2.9 50 1.3 75 4.6 100 1.5 50 0.8 82

214

Copyright © 2015 Pearson Education, Inc. 24.  x  zc  20.6  1.28  20.6  0.602  (20.0, 21.2) 25. 26. 27. 28. (12.0,14.8)  x  14.8 12.0  13.4, E 14.8 13.4 1.4 2 (21.61, 30.15)  x  30.15  21.61  25.88, E  30.15 25.88  4.27 2 (1.71, 2.05)  x  2.05 1.71  1.88, E  2.05 1.88  0.17 2 (3.144, 3.176)  x  3.176  3.144  3.16, E  3.176 3.16  0.016 2 29. c  0.90  zc  1.645  z  2  (1.645)(6.8) 2 n   c     125.13  126  E   1  30. c  0.95  zc  1.96  z  2  (1.96)(2.5) 2 n   c      24.01 25  E   1  31. c  0.80  zc  1.28  z  2  (1.28)(4.1) 2 n   c      6.89  7  E   2  32. c  0.98  zc  2.33  z  2  (2.33)(10.1) 2 n   c     138.45  139  E   2  33. (26.2, 30.1)  2E  30.1 26.2  3.9  E  1.95and x  26.2  E  26.2  1.95  28.15 34. (44.07, 80.97  2E  80.97 44.07  36.9  E  18.45and x  44.07  E  44.07  18.45  62.52 35. 90% CI: 95%CI:  x  zc  x  zc  3.63  1.645 0.21  3.63  0.0499  (3.58, 3.68) 48  3.63  1.96 0.21  3.63  0.0594  (3.57, 3.69) 48 4.7 100

CHAPTER 6 │ CONFIDENCE INTERVALS

With 90% confidence, you can say that the population mean price is between $3.58 and $3.68. With 95% confidence, you can say that the population mean price is between $3.57 and $3.69. The 95% CI is wider.

CHAPTER 6 │ CONFIDENCE INTERVALS 215

With 90% confidence and with 95% confidence, you can say that the population mean concentration is between 21 and 25 cubic centimeters per cubic meter. When rounded to the nearest whole number, both confidence intervals have the same width.

With 95% confidence, you can say that the population mean cost is between $2532.20 and $2767.80.

With 99% confidence, you can say that the population mean repair cost is between $144.85 and $155.15.

The n = 50 CI is wider because a smaller sample is taken, giving less information about the population.

The n = 40 CI is wider because a smaller sample is taken, giving less information about the population.

The

CI is wider because of the increased variability within the sample.

The   19.5CI is wider because of the increased variability within the sample.

43. (a) An increase in the level of confidence will widen the confidence interval.

(b) An increase in the sample size will narrow the confidence interval.

44. Answers will vary.

216 CHAPTER 6 │ CONFIDENCE INTERVALS Copyright © 2015 Pearson Education, Inc. 36. 90% CI:  x  zc  23  1.645 6.7 36  23  1.837  (21, 25) 95%CI:  x  zc  23  1.96 6.7 36  23  2.189  (21, 25)

37.  x  zc  2650  1.96 425  2650  117.80  (2532.20, 2767.80) 50

38.  x  zc  150  2.575 15.5  150  5.15  (144.85, 155.15) 60

39.  x  zc  2650  1.96 425  2650  93.13  (2556.9, 2743.1) 80

40.  x  zc  150  2.575 15.5  150  6.31  (143.69, 156.31) 40

41.  x  zc  2650  1.96 375  2650  103.94  (2546.06, 2753.94) 50

  425

42.  x  zc  150  2.575 19.5  150  6.48  (143.52, 156.48) 60

With 90% confidence, you can say that the population mean length of time is between 22.5 and 25.7 minutes. With 99% confidence, you can say that the population mean length of time is between 21.6 and 26.6 minutes. The 99% CI is wider.

With 90% confidence, you can say that the population mean closing stock price is between $19.22 and $20.70. With 99% confidence, you can say that the population mean closing stock price is between $18.80 and $21.12. The 99% CI is wider.

218 CHAPTER 6 │ CONFIDENCE INTERVALS

46. x   x  678.67 19.96 n 34 90%CI:  x  zc  19.96  1.645 2.62  19.96  0.739  (19.22, 20.70) 34 99%CI:  x  zc  19.96  2.575 2.62  19.96  1.157  (18.80, 21.12) 34

 z  2  1.96  4.8 2 47. n   c      88.510  89  E   1   z  2  2.575 1.4 2 48. n   c      3.249  4  E   2   z  2  1.96  2.8 2 49. (a) n   c     120.473  121servings  E   0.5   z  2  2.575  2.8 2 (b) n   c      207.936  208servings  E   0.5 

 z  2  1.645 1.2 2 50. (a) n   c      3.897  4students  E   1   z  2  2.575 1.2 2 (b) n   c      9.548 10students  E   1 

CHAPTER 6 │ CONFIDENCE INTERVALS 219 Copyright © 2015 Pearson Education, Inc.  z  2  1.645  0.85 2 51. (a) n   c      31.282  32cans  E   0.25   z  2  1.645 0.85 2 (b) n   c      86.893  87cans  E   0.15 

(c) E = 0.15 requires a larger sample size. As the error size decreases, a larger sample must be taken to obtain enough information from the population to ensure the desired accuracy.

(c) E = 1 requires a larger sample size. As the error size decreases, a larger sample must be taken to obtain enough information from the population to ensure the desired accuracy.

(c)   0.3 requires a larger sample size. Due to the increased variability in the population, a larger sample is needed to ensure the desired accuracy.

(c)   0.2 requires a larger sample size. Due to the increased variability in the population, a larger sample is needed to ensure the desired accuracy.

55. (a) An increase in the level of confidence will increase the minimum sample size required.

(b) An increase (larger E) in the error tolerance will decrease the minimum sample size required.

56. Sample answer: A 99% CI may not be practical to use in all situations. It may produce a CI so wide that is has no practical application.

 z  2  1.96 3 2 52. (a) n   c      34.574  35bottles  E   1   z  2  1.96 3 2 (b) n   c      8.644  9bottles  E   2 

 z  2  2.575 0.25 2 53. (a) n   c      41.441  42soccer balls  E   0.1   z  2  2.575 0.30 2 (b) n   c      59.676  60soccer balls  E   0.1 

 z  2  2.575 0.20 2 54. (a) n   c     11.788 12soccer balls  E   0.15   z  2  2.575 0.10 2 (b) n   c      2.947  3soccer balls  E   0.15 

57. (a)   0.707 (b)   0.949 N n N 1 N n N 1

(e) The finite population correction factor approaches 1 as the sample size decreases and the population size remains the same.

N n N 1 N n N 1

(e) The finite population correction factor approaches 1 as the population size increases and the sample size remains the same.

Write original equation.

Multiply each side by n.

Divide each side by E Square each side.

6.2 CONFIDENCE INTERVALS FOR THE MEAN (SMALL SAMPLES)

6.2 Try It Yourself Solutions

1a. d.f.  n 1  22 1  21

b. c  0.90

c. tc  1.721

d. For a t-distribution curve with 21 degrees of freedom, 90% of the area under the curve lies between t  1.721.

222 CHAPTER 6 │ CONFIDENCE INTERVALS Copyright © 2015 Pearson Education, Inc. 100 50 100 1 400 50 400 1 700 50 700 1 1000 50 1000 1 n n    c 58. (a)   0.711 (b)   0.937 (c)   0.964 (d)   0.975

59. (a) 99% CI: x  z  N n  8.6  2.575 4.9 200 25  8.6  2.366  (6.2, 11.0) c n N 1 25 200 1 (b) 90%CI: x  z  N n 10.9 1.645 2.8 500 50 10.9  0.619  (10.3, 11.5) c n N 1 50 500 1 (c) 95%CI: x  z  N n  40.3 1.96 0.5 300 68  40.3  0.105  (40.2, 40.4) c n N 1 68 300 1 (d) 80%CI: x  z  N n  56.7 1.28 9.8 400 36  56.7 1.997  (54.7, 58.7) c n N 1 36 400 1 60. Sample answer: E  zc E  zc n = zc E  z  2 n  E 

N n N 1 N n N 1 N n N 1 N n N 1

c. With 90% confidence, you can say that the population mean temperature of coffee sold is between 157.6F and 166.4F. With 99% confidence, you can say that the population mean temperature of coffee sold is between 154.6F and 169.4F.

c. With 90% confidence, you can say that the population mean number of days the car model sits on the lot is between 9.08 and 10.42 days. With 95% confidence, you can say that the population mean number of days the car model sits on the lot is between 8.94 and 10.56 days. The 90% confidence interval is slightly narrower.

4a. Is  known? No

b. Is n  30? No Is the population normally distributed? Yes

c. Use the t-distribution because is not known and the population is normally distributed.

6.2 EXERCISE SOLUTIONS

224 CHAPTER 6 │ CONFIDENCE INTERVALS Copyright © 2015 Pearson Education, Inc. 36 E  tc  1.753  4.4 99% CI: tc  2.947 E  tc  2.947 10  7.4 16 b. 90% CI: 99%CI: x  E 162  4.4  (157.6, 166.4) x  E 162  7.4  (154.6, 169.4)

3a. d.f.  n 1  36 1  35 90% CI: tc  1.690 E  tc  1.690 2.39  0.67 95% CI: tc  2.030 E  tc  2.030 2.39  0.81 36 b 90% CI: 95%CI: x  E  9.75  0.67  (9.08, 10.42) x  E  9.75  0.81  (8.94, 10.56)

1. tc  1.833 5. E  tc  2.131 2. tc  2.201  2.664 3. tc  2.947 6. E  tc  4.032 4. tc  2.426  4.938 7. E  tc  1.691  0.686 8. E  tc  2.896 4.7  4.537 10 16 16

41.5)

With 95% confidence, you can say that the population mean commute time to work is between 29.5 and 41.5 minutes.

(15.0, 29.4)

With 95% confidence, you can say that the population mean driving distance to work is between 15.0 and 29.4 miles.

8.16  (71.84, 88.16)

With 95% confidence, you can say that the population mean repair cost is between $71.84 and $88.16.

41.16

226 CHAPTER 6 │ CONFIDENCE

Copyright © 2015 Pearson Education, Inc. 14 10. 11. x  tc x  tc  13.4  2.365 0.85  13.4  0.711  (12.7, 14.1)  4.3  2.650 0.34  4.3  0.241  (4.1, 4.5) 12. x  tc  24.7  2.678  24.7  1.742  (23.0, 26.4) 13. 14. 15. 16. (14.7, 22.1)  x  14.7  22.1  18.4  E  22.1 18.4  3.7 2 (6.17, 8.53)  x  6.17  8.53  7.35  E  8.53 7.35 1.18 2 (64.6, 83.6)  x  64.6  83.6  74.1 E  83.6 74.1  9.5 2 (16.2, 29.8)  x  16.2  29.8  23  E  29.8 23  6.8 2 17. E  tc  2.365 7.2  6.02 8 x  E  35.5  6.02  (29.5,

INTERVALS

18. E  tc  2.776 5.8  7.2 5 x  E 



22.2  7.2

19. E  tc  2.179



x  E 



13.5

8.16 13

20. E  tc  2.447



7 x  E 110  41.16 

4.6 50

44.5

(68.84, 151.16)

With 95% confidence, you can say that the population mean repair cost is between $68.84 and $151.16.

CHAPTER 6 │ CONFIDENCE INTERVALS 227

6.44

(29.1, 41.9)

With 95% confidence, you can say that the population mean commute time to work is between 29.1 and 41.9 minutes. This confidence interval is slightly wider than the one found in Exercise

(17.6, 26.8)

With 95% confidence, you can say that the population mean driving distance to work is between 17.6 and 26.8 miles. This confidence interval is narrower than the one found in Exercise 18.

8.15  (71.85, 88.15)

With 95% confidence, you can say that the population mean repair cost is between $71.85 and $88.15. This confidence interval is slightly narrower than the one found in Exercise 19.

(72.96, 147.04)

With 95% confidence, you can say that the population mean repair cost is between $72.96 and $147.04. This confidence interval is narrower than the one found in Exercise 20.



22.  E  zc  1.96 5.2  4.56 5 x  E  22.2  4.56 

17.

23.  E  zc  1.96 15  8.15 13

 E  80 

24.  E  zc  1.96 50  37.04 7 x  E 110  37.04 

25. (a) (b) x  1764.2 s  252.4 (c) x  tc  1764.2  3.106 252.35  1764.2  226.26  (1537.9, 1990.5) 26. (a) (b) (c) 27. (a) (b) (c) 28. (a) (b) (c) x  2.35 s  1.03 x  tc x  7.49 s  1.64 x  tc x  12.19 s  1.75 x  tc  2.35  2.977 1.03  2.35  0.792  (1.56, 3.14)  7.49  2.947 1.64  7.49  1.21  (6.28, 8.70)  12.19  2.898 1.755  12.19  1.20  (10.99, 13.39)

With 95% confidence, you can say that the population mean BMI is between 26.0 and 29.4.

32. Use a t-distribution because is unknown and the interest rates are normally distributed.

With 95% confidence, you can say that the population mean interest rate is between 3.37% and 3.77%.

33. Use a t-distribution because unknown and n

With 95% confidence, you can say that the population mean gas mileage is between 20.8 and 22.7.

34. Use the standard normal distribution because  is known and the yards per carry are normally distributed.

With 95% confidence, you can say that the population mean yards per carry is between 3.80 and 4.88 yards.

35. Cannot use the standard normal distribution or the t-distribution because  unknown, n  30 , and we do not know if the times are normally distributed.

36. Use the standard normal distribution because  is known and the lengths of stay are normally distributed.

With 95% confidence, you can say that the population mean length of stay is between 5.3 and 7.1 days.

222 CHAPTER 6 │ CONFIDENCE INTERVALS Copyright © 2015 Pearson Education, Inc. 35 40 (b) s 15,426.35 (c) x  tc  71,968.06  2.44115,426.35  71,968.06  6364.98  (65,603.08, 78,333.04) 30. (a) (b) x  65,588.73 s 11,828.21 (c) x  tc  65,588.73  2.42611,828.21  65,588.73  4537.12  (61,051.61, 70,125.85)



. x  tc  27.7  2.009 6.12  27.7  1.74  (26.0, 29.4) 50

31. Use a t-distribution because unknown and n

x  tc  3.57  2.145 0.36  3.57  0.20  (3.37, 3.77) 15



x  tc  21.76  2.014 3.17  21.76  0.95  (20.8, 22.7) 45

 x  zc  4.34  1.96 1.21  4.34  0.53  (3.81, 4.87) 20

 x  zc  6.2  1.96 1.7 13  6.2  0.924  (5.3, 7.1)

are making good light bulbs because for this sample the t-value is t

6.3 CONFIDENCE INTERVALS FOR POPULATION PROPORTIONS

6.3 Try It Yourself Solutions

0 007

(0 043, 0.057) e. With 90% confidence, you can say that the proportion of U.S. teachers who say that “all or almost all” of the information they find using search engines online is accurate or trustworthy is between 4.3% and 5.7%.



between t0.99  2.797 and t0.99  2.797 38. n  16, x  1015, s  25 t  x   1015 1000  2.4 s 25 n 16 t0.99  2.947, t0.99  2.947 They



t0.99  2.947 and t0.99  2.947

are not making good tennis balls because for this sample the t-value is t

10 , which is not

2.4 , which is between

1a. x 123, n  2462 b. p ˆ  123 2462  5.0% 2a. b. p ˆ  0.050, q ˆ  0.950 np ˆ  (2462)(0.050)  123 1  5 nq ˆ  (2462)(0.950)  2338.9  5 Distribution of p ˆ c. zc  1.645 is approximately

E  zc 1.645  0.007 d. p ˆ  E  0.050 

3a. b. n  498, p ˆ  0.25 q ˆ  1  p ˆ 1  0.25  0.75 np ˆ  498 0.25  124.5  5 nq ˆ  498 0.75  373.5  5 Distribution of p ˆ

p ˆ q ˆ n 0.050  0.950 2462

normal.



is approximately normal.

e. With 99% confidence, you can say that the proportion of U.S. adults who think that people over 65 are the more dangerous drivers is between 20% and 30%.

c. (1) At least 1692 adults should be included in the sample. (2) At least 1448 adults should be included in the sample.

6.3 EXERCISE SOLUTIONS

1. False. To estimate the value of p, the population proportion of successes, use the point estimate

4a. (1) (2) p ˆ  0.5, q ˆ  0.5, zc  1.645, E  0.02 p ˆ  0 31, q ˆ  0.69, zc  1 645, E  0.02  z 2  1.645 2 b. (1) n  p ˆ q ˆ  c     (0.5)(0.5) 0.02  1691.266 1692  z 2  1.645 2 (2) n  p ˆ q ˆ  c     (0.31)(0.69) 0.02  1447.05  1448

p ˆ  x . n 2. True 3. p ˆ  x  662  0.661, q ˆ 1 p ˆ  0.339 4. p ˆ  x  2439  0.830, q ˆ  1 p ˆ  0.170 n 1002 n 2939 5. p ˆ  x  4912  0.423, q ˆ  1 p ˆ  0.577 6. p ˆ  x  110  0.110, q ˆ 1 p ˆ  0.890 n 11,605 n 1003 7. 0.905, 0.933  p ˆ  0.905  0.933  0.919  E  0.933 0.919  0.014 8. 0.245, 0.475  p ˆ  0.245  0.475  0.360  E  0.475 0.360  0.115 9. 0.512, 0.596  p ˆ  0.512  0.596  0.554  E  0.596 0.554  0.042 10. 0.087, 0.263  p ˆ  0.087  0.263  0.175  E  0.263 0.175  0.088 p ˆ q ˆ n 0.25(0.75) 498

With 90% confidence, you can say that the population proportion of U.S. males ages 18-64 who say they have gone to the dentist in the past year is between 55.7% and 61.9%. With 95% confidence, you can say it is between 55.1% and 62.5%. The 95% confidence interval is slightly wider.

With 90% confidence, you can say that the population proportion of U.S. females ages 18-64 who say they have gone to the dentist in the past year is between 62.6% and 70.2%. With 95% confidence, you can say it is between 61.9% and 70.9%. The 95% confidence interval is slightly wider.

With 99% confidence, you can say that the population proportion of U.S. adults who say they have started paying bills online in the past year is between 43.8% and 48.4%.

With 99% confidence, you can say that the population proportion of U.S. adults who say they have seen a ghost is between 16.4% and 19.6%.

Education, Inc. 90%CI: p ˆ  zc  0.588 1.645  0.588  0.031  (0.557, 0.619) 95% CI: p ˆ  zc  0.588 1.96 (0.588)(0.412)  0.588  0.037  (0.551, 0.625) 674

Pearson

12. p ˆ  x  279  0.664, q ˆ  1 p ˆ  0.336 n 90%CI: 420 p ˆ  zc  0.664 1.645  0.664  0.0379  (0.626, 0.702) 95% CI: p ˆ  zc  0.664 1.96 (0.6643)(0.3357)  0.664  0.045  (0.619, 0.709) 420

13. p ˆ  x  1435  0.461, q ˆ 1 p ˆ  0.539 n p ˆ  zc 3110  0.461 2.575 (0.461)(0.539)  0.461 0.023  (0.438, 0.484) 3110

14. p ˆ  x  n 722 4013  0.180, q ˆ  1 p ˆ  0.820 p ˆ  zc  0.180  2.575 (0.180)(0.820)  0.180  0.016  (0.164, 0.196) 4013

15. p ˆ  x  1272  0.570, q ˆ 1 p ˆ  0.430 n p ˆ  zc 2230  0.570 1.96  0.570  0.021  (0.549, 0.591) p ˆ q ˆ n (0.588)(0.412) 674 p ˆ q ˆ n p ˆ q ˆ n (0.6643)(0.3357) 420 p ˆ q ˆ n p ˆ q ˆ n p ˆ q ˆ n p ˆ q ˆ n (0.570)(0.430) 2230

CHAPTER 6 │ CONFIDENCE INTERVALS 227 Copyright © 2015 Pearson Education, Inc. 16. p ˆ  x  n 734 2303  0.319, q ˆ  1 p ˆ  0.681 p ˆ  zc  0.319 1.645  0.319  0.016  (0.303, 0.335) p ˆ q ˆ n (0.319)(0.681) 2303

228 CHAPTER 6 │ CONFIDENCE INTERVALS

Education, Inc. E   E   E   E   E   E    z 2  1.96 2 17. (a) n  p ˆ q ˆ  c     0.5  0.5 0.04   600.25  601adults  z 2  1.96 2 (b) n  p ˆ q ˆ  c     0.48 0.52 0.04   599.3  600 adults

 z 2  2.575 2 18. (a) n  p ˆ q ˆ  c   0.5 0.5   4144.14  4145 adults  E   0.02   z 2  2.575 2 (b) n  p ˆ q ˆ  c   0.87 0.13  1874.8 1875 adults  E   0.02 

the population proportion reduces the minimum sample size needed.  z 2  1.645 2 19. (a) n  p ˆ q ˆ  c     0.5 0.5 0.03   751.67  752 adults  z 2  1.645 2 (b) n  p ˆ q ˆ  c     0.43 0.57 0.03   736.94  737 adults

 z 2  1.96 2 20. (a) n  p ˆ q ˆ  c     0.5  0.5 0.05   384.16  385 adults  z 2  1.96 2 (b) n  p ˆ q ˆ  c     0.28 0.72 0.05   309.79  310 adults

21. (a) p ˆ  0.69, q ˆ  0.31, n  1044 p ˆ  zc  0.69  2.575  0.69  0.037  (0.653, 0.727) (b) p ˆ  0 72, q ˆ  0.28, n  871 p ˆ  zc  0.72  2.575  0.72  0.039  (0.681, 0.759) (c) p ˆ  0.62, q ˆ  0 38, n  1097 p ˆ q ˆ n (0.69)(0.31) 1044 p ˆ q ˆ n (0.72)(0.28) 871 p ˆ q ˆ n (0.62)(0.38) 1097

CHAPTER 6 │ CONFIDENCE INTERVALS 229 Copyright © 2015 Pearson Education, Inc. p ˆ  zc  0.62  2.575  0.62  0.038  (0.582, 0.658) (d) p ˆ  0.75, q ˆ  0.25, n  1003 p ˆ  zc  0.75  2.575  0.75  0.035  (0.715, 0.785) p ˆ q ˆ n (0.75)(0.25) 1003

22. It is possible that the population proportion for the United States is the same as the population proportion for Great Britain, France, or Spain because the confidence intervals overlap. It is possible that the population proportion for Great Britain is the same as the population proportion for Spain because the confidence intervals overlap.

25. No, it is unlikely that the two proportions are equal because the confidence intervals estimating the proportions do not overlap. The 99% confidence intervals are (0.260, 0.380) and (0.496, 0.624). Although these intervals are wider, they still do not overlap.

26. No, it is unlikely that the two proportions are equal because the confidence intervals estimating the proportions do not overlap. The 99% confidence intervals are (0.298, 0.422) and (0.204, 0.316). Using these intervals, it is possible that the two proportions are equal because the confidence intervals overlap.

23. (a) p ˆ  0.32, q ˆ  1 p ˆ  0.68 p ˆ  zc  0.32 1.96  0.32  0.046  (0.274, 0.366) (b) p ˆ  0.56, q ˆ  1 p ˆ  0.44 p ˆ  zc  0.56 1.96  0.56  0.049  (0.511, 0.609) 24. (a) p ˆ  0.36, q ˆ  1 p ˆ  0.64 p ˆ  zc  0.36 1.96  0.36  0.047  (0.313, 0.407) (b) p ˆ  0.26, q ˆ  1 p ˆ  0.74 p ˆ  zc  0.26 1.96  0.26  0.043  (0.217, 0.303)

27. 31.4% 1%  (30.4%, 32.4%)  (0.304, 0.324) E  zc  zc  E  0.01 8451 (0.314)(0.686) 1.981 zc 1.98 P( 1.98  z 1.98)  0.9762 0.0238  0.9524  c (30.4%, 32.4%)is approximately a 95.2% CI. 28. 19%  3%  (16%, 22%)  (0.16, 0.22) E  zc  zc  E  0.03 1000 (0.19)(0.81)  2.418  zc  2.42 p ˆ q ˆ n (0.32)(0.68) 400 p ˆ q ˆ n (0.56)(0.44) 400 p ˆ q ˆ n (0.36)(0.64) 400 p ˆ q ˆ n (0.26)(0.74) 400 n p ˆ q ˆ n p ˆ q ˆ

29. If np

5, the sampling distribution of p ˆ may not be normally distributed, so zc cannot be used to calculate the confidence interval.

a 98.4%

approximately

CI.



nq ˆ

6.4 CONFIDENCE INTERVALS FOR VARIANCE AND STANDARD

6.4 Try It Yourself Solutions

d. For a chi-square distribution curve with 29 degrees of freedom, 90% of the area under the curve lies between 17.708 and 42.557.

232 CHAPTER 6 │ CONFIDENCE INTERVALS Copyright © 2015 Pearson Education, Inc. E R L 30. Sample answer: E  zc E    z 2 Write original equation. Multiplyeach side by n Divide each side by E 31. n  p ˆ q ˆ  c    Squareeachside. p ˆ q ˆ  1 p ˆ p ˆ q ˆ p ˆ q ˆ  1 p ˆ p ˆ q ˆ 0.0 1.0 0.00 0.45 0.55 0.2475 0.1 0.9 0.09 0.46 0.54 0.2484 0.2 0.8 0.16 0.47 0.53 0.2491 0.3 0.7 0.21 0.48 0.52 0.2496 0.4 0.6 0.24 0.49 0.51 0.2499 0.5 0.5 0.25 0.50 0.50 0.2500 0.6 0.4 0.24 0.51 0.49 0.2499 0.7 0.3 0.21 0.52 0.48 0.2496 0.8 0.2 0.16 0.53 0.47 0.2491 0.9 0.1 0.09 0.54 0.46 0.2484 1.0 0.0 0.00 0.55 0.45 0.2475 p ˆ  0 5 gives the maximum value of p ˆ q ˆ

d.f.  n 1 

1a.

29 level of confidence = 0.90

Area

of 

 2  42.557 ,  2 

R L

b. Area to the right of  2 is 0.05.

to the right

2 is 0.95. c.

17.708

2a. 90% CI:  2  42.557 ,  2  17.708 R L 95%CI:  2  45.722 ,  2  16.047 R L 2  (n 1)s2 (n 1)s2   29 (1.2)2 29 (1.2)2 

DEVIATION

d. With 90% confidence, you can say that the population variance is between 0.98 and 2.36 and that the population standard deviation is between 0.99 and 1.54. With 95% confidence, you can say that the population variance is between 0.91 and 2.60, and that the population standard deviation is between 0.96 and 1.61.

6.4 EXERCISE SOLUTIONS

1. Yes.

2. It approaches the shape of the normal curve.

234 CHAPTER 6 │ CONFIDENCE INTERVALS Copyright © 2015 Pearson Education, Inc. 2  (n 1)s2 (n 1)s2   29 (1.2)2 29 (1.2)2  95% CI for  :   2 ,  2    , 45.722 16.047   (0.91, 2.60)  R L    c. 90% CI for  : 95% CI for  : 0.981, 2.358  0.99, 1.54 0.913, 2.602  0.96, 1.61

3.  2  14.067 ,  2  2.167 4.  2  31.319,  2  4.075 5.  2  32.852,  2  8.907 R L R L R L 6.  2  44.314 ,  2  11.524 7.  2  52.336,  2  13.121 8.  2  63.167 ,  2  37.689 R L R L R L  (n 1)s2 (n 1)s2   29  (11.56) 29  (11.56)  9. (a)   2 ,  2    , 45.722 16.047   (7.33, 20.89)  R L    (b)  7.3321, 20.8911  (2.71, 4.57)  (n 1)s2 (n 1)s2   6 (0.64) 6 (0.64)  10. (a)   2 ,  2    , 18.548 0.676   (0.21, 5.68)  R L    (b)  0.2070, 5.6805  (0.46, 2.38)  (n 1)s2 (n 1)s2  17  (35)2 17 (35)2  11. (a)   2 ,  2    , 27.587 8.672   (755, 2401)  R L    (b)  754.885, 2401.407   (27, 49)  (n 1)s2 (n 1)s2   40  (278.1)2 40  (278.1)2  12. (a)   2 ,  2    , 63.691 22.164   (48,571.8, 139,577.0)  R L    (b)  48,571.77, 139,576.99   (220.4, 373.6)

CHAPTER 6 │ CONFIDENCE INTERVALS 235 Copyright © 2015 Pearson Education, Inc. 13. (a) s2  0.0793  (n 1)s2 (n 1)s2   16  (0.0793) 16  (0.0793)    2 ,  2    , 28.845 6.908   (0.0440, 0.1837)  R L   

(b)  0.04399, 0.18367   (0.2097, 0.4286)

With 95% confidence, you can say that the population variance is between 0.0440 and 0.1837, and the population standard deviation is between 0.2097 and 0.4286 inch.

With 90% confidence, you can say that the population variance is between 0.0006 and 0.0022, and the population standard deviation is between 0.0247 and 0.0469 fluid ounce.

With 99% confidence, you can say that the population variance is between 0.0305 and 0.1914, and the population standard deviation is between 0.1747 and 0.4375 hour.

With 95% confidence, you can say that the population variance is between 0.00001 and 0.000007, and the population standard deviation is between 0.0038 and 0.0082 inch.

With 99% confidence, you can say that the population variance is between 6.63 and 55.46, and the population standard deviation is between $2.58 and $7.55.

236 CHAPTER 6 │

CONFIDENCE INTERVALS

14. (a) s2  0.00103  (n 1)s2 (n 1)s2   14  (0.00103) 14  (0.00103)    2 ,  2    , 23.685 6.571   (0.0006, 0.0022)  R L   



 

(b)

0.000609, 0.002194

(0.0247, 0.0469)

15. (a) s2  0.06414  (n 1)s2 (n 1)s2   17  (0.06414) 17  (0.06414)    2 ,  2    , 35.718 5.697   (0.0305, 0.1914)  R L    (b)  0.030527,

  (0.1747,

0.191395

0.4375)

16. (a) s2 

 (n 1)s2 (n 1)s2   14  (0.000027209) 14  (0.000027209)    2 ,  2    , 26.119 5.629   (0.00001, 0.00007)  R L    (b) 

 

0.000027209

0.0000146, 0.0000677

(0.0038, 0.0082)

 (n 1)s2 (n 1)s2   13 (3.90)2 13 (390)2  17. (a)   2 ,  2    , 29.819 3.565   (6.63, 55.46)  R L    (b)  6.631,

 

55.464

(2.58, 7.45)

 (n 1)s2 (n 1)s2   10  (109)2 10  (109)2  18. (a)   2 ,  2    , 15.987 4.865   (7432, 24,421)

(b)  7431.663, 24,421.377   (86, 156)

With 80% confidence, you can say that the population variance is between 7432 and 24,421, and the population standard deviation is between $86 and $156.

CHAPTER 6 │ CONFIDENCE INTERVALS 237

 R L   

 (n 1)s2 (n 1)s2  18 (15)2 18 (15)2  19. (a)   2 ,  2    , 31.526 8.231   (128, 492)  R L    (b)  128.465, 492.042   (11, 22)

With 95% confidence, you can say that the population variance is between 128 and 492, and the population standard deviation is between 11 and 22 grains per gallon.

21,224,305

With 90% confidence, you can say that the population variance is between 8,831,450 and 21,224,305, and the population standard deviation is between $2972 and $4607.

9,104,741, 25,615,326

With 80% confidence, you can say that the population variance is between 9,104,741 and 25,615,326, and the population standard deviation is between $3017 and $5061.

With 98% confidence, you can say that the population variance is between 39.13 and 136.11, and the population standard deviation is between 6.26 and 11.67 inches.

With 98% confidence, you can say that the population variance is between 7.0 and 30.6, and the population standard deviation is between 2.6 and 5.5 minutes.

28,564,792

5345)

With 90% confidence, you can say that the population variance is between 9,586,982 and 28,564,792, and the population standard deviation is between $3096 and $5345. 25. 95% CI for

:(0.2097, 0.4286)

Yes, because all of the values in the confidence interval are less than 0.5.

238 CHAPTER 6 │ CONFIDENCE INTERVALS

 (n 1)s2 (n 1)s2   29(3600)2 29(3600)2  20. (a)   2 ,  2    , 42.557 17.708   (8,831,450,

 R L    (b)  8,831,450,

  (2972,

21,224,305)

4607)

 (n 1)s2 (n 1)s2  13 (3725)2 13 (3725)2  21. (a)   2 ,  2    , 19.812 7.042   (9,104,741,

 R L   



  (3017,

25,615,326)

(b)

5061)

 (n 1)s2 (n 1)s2   29  (8.18)2 29 (8.18)2  22. (a)   2  2    49.588 14.256   (39.13, 136.11)  R L    (b)  39.13, 136.11  (6.26, 11.67)

 (n 1)s2 (n 1)s2   (21)(3.6)2 (21)(3.6)2  23. (a)   2   2     38.932 8.897   (7.0, 30.6)  R L    (b)  6.99, 30.59  

(2.6, 5.5)

 (n 1)s2 (n 1)s2   19(3900)2 19(3900)2  24. (a)   2   2    30.144  10.117   (9,586,982, 28,564,792)  R L   

 9,586,982,

  (3096,

(b)



27. Answers will vary. Sample answer: Unlike a confidence interval for a population mean or proportion, a confidence interval for a population variance does not have a margin of error. The left and right endpoints must be calculated separately.

CHAPTER 6 REVIEW EXERCISE SOLUTIONS

With 90% confidence, you can say that the population mean waking time is between 91.8 and 115.2 minutes past 5:00 A.M.

With 95% confidence, you can say that the population mean driving distance is between 6.6 and 12.4 miles.

Education, Inc. 2 2

2015 Pearson

1. (a) x  103.5 (b)  E  zc  1.645  11.7 2. (a) (b) x  9.5 E  zc  1.96  2.9 3. x  zc 103.5 1.645 45 103.5 11.7  (91.8, 115.2) 40

4. x  zc  9.5  1.96 8  9.5  2.9  (6.6, 12.4) 30

5. 20.75, 24.10  x  20.75  24.10  22.425  E  24.10 22.425 1.675 6. 7.428, 7.562  x  7.428  7.562  7.495  E  7.562 7.495  0.067  z  2  1.96 45 2 7. n   c      77.79  78 people  E   10   z  2  2.575 8 2 8. n   c     106.09  107people  E   2  9. tc  1.383 10. tc  2.069 11. tc  2.624 12. tc  2.756 13. E  tc 1.753 25.6 11.2 14. E  tc  2.064  0.5 30 1.1 25   

With 90% confidence, you can say that the population mean annual fuel cost is between $2676 and $3182.

With 99% confidence, you can say that the population mean annual fuel cost is between $2517 and $3341.

With 95% confidence, you can say that the population proportion of U.S. adults who say the economy is the most important issue facing the country today is between 42.7% and 49.5%.

With 99% confidence, you can say that the population proportion of U.S. adults who say they would trust doctors to tell the truth is between 80.9% and 89.1%.

242 CHAPTER 6 │ CONFIDENCE INTERVALS Copyright © 2015 Pearson Education, Inc. 16 0.9 12 20     17. x  tc  72.1  1.753 25.6  72.1  11.2  (60.9, 83.3) 18. x  tc  3.5  2.064  3.5  0.5  (3.0, 4.0) 19. 20. x  tc x  tc  6.8  2.718   6.8  0.7  (6.1, 7.5)    25.2  2.861 16.5   25.2  10.6  (14.6, 35.8)   21. x  tc  2929  1.703  2929  252.96  (2676.0, 3182.0)

22. x  tc  2929  2.771 786  2929  411.6  (2517, 3341)

23. p ˆ  x  375  0.461, q ˆ  1 p ˆ  0 539 24. p ˆ  x  425  0.85, q ˆ  1 p ˆ  0.15 25. n p ˆ  x  n 814 552 1023  0.540, q ˆ  1 p ˆ  0.460 26. n p ˆ  x  n 500 90 800  0.113, q ˆ  1 p ˆ  0.887 27. p ˆ  zc  0.4611.96 0.461 0.539  0.461 0.034  (0.427, 0.495) 814

28. p ˆ  zc  0.85  2.575 0.85 0.15  0.85  0.041  (0.809, 0.891) 500

29. p ˆ  zc  0.540 1.645 0.540  0.460  0.540  0.026  (0.514, 0.566) p ˆ q ˆ n p ˆ q ˆ n

CHAPTER 6 │ CONFIDENCE INTERVALS 243

10 2 3

p ˆ q ˆ n

With 90% confidence, you can say that the population proportion of U.S. adults who say they have worked the night shift at some point in their lives is between 51.4% and 56.6%.

With 98% confidence, you can say that the population proportion of U.S. adults who say they are making the minimum payment(s) on their credit card(s) is between 8.7% and 13.9%.

With 95% confidence we can say that the population variance is between 27.2 and 113.6, and the population standard deviation is between 5.2 and 10.7 ounces.

With 98% confidence, you can say that the population variance is between 0.80 and 3.07, and the population standard deviation is between 0.89 and 1.75 seconds.

244 CHAPTER 6 │ CONFIDENCE

INTERVALS

 z 2  1.96 2 31. (a) n  p ˆ q ˆ  c     0.50  0.50 0.05   384.16  385 adults  z 2  1.96 2 (b) n  p ˆ q ˆ  c     0.63 0.37 0.05   358.19  359 adults

 z 2  2.575 2 32. n  p ˆ q ˆ  c     0.63 0.37 0.025   2472.96  2473 adults 33. The sample size is much larger.  2  23.337 ,  2  4.404 34.  2  42.980 ,  2  10.856 R L R L 35.  2  24.996 ,  2  7.261 36.  2  23.589 ,  2  1.735 R L R L 37. s2  49.0294 2  (n 1)s2 (n 1)s2   16  (49.0294) 16  (49.0294)  (a) 95% CI for  :   2 ,  2    , 28.845 6.908   (27.2, 113.6)  R L    (b) 95% CI for  : 27.195, 113.560  (5.2,

10.7)

38. s2  1.4171 2  (n 1)s2 (n 1)s2   (25)(1.4171) (25)(1.4171)  (a) 98% CI for  :   2 ,  2    , 44.314 11.524   (0.80, 3.07)  R L    (b) 98% CI for  :

 

0.7995, 3.0742

(0.89, 1.75)

p ˆ q ˆ n

CHAPTER 6 QUIZ SOLUTIONS

With 95% confidence, you can say that the population mean amount of time is between 5.989 and 7.707 minutes.

With 90% confidence you can say that the population mean amount of time is between 4.65 and 8.57 minutes.

With 90% confidence you can say that the population mean amount of time is between 4.79 and 8.43 minutes. This confidence interval is narrower than the one found in part (b).

With 95% confidence you can say that the population mean annual earnings is between $28,379 and $35,063.

With 90% confidence you can say that the population proportion of U.S. adults who think that the United States should not put more emphasis on producing domestic energy from solar power is between 74.0% and 78.4%.

1. (a) x  6.848 (b)  E  zc  1.96  0.859 (c)  x  zc  6.848  1.96 2.4 30  6.848  0.859  (5.989, 7.707)

 z  2  2.575 2.4 2 2. n   c      38.18  39 students  E   1  3. (a) (b) x  6.61, x  tc s  3.376  6.61  1.833 3.376  6.61  1.957  (4.65, 8.57) 10

4. x  tc  31,721  2.201 5260  31,721  3342  (28,379, 35,063) 12

5. (a) p ˆ  x  n 779 1022  0.762 (b) p ˆ  zc  0.762 1.645 0.762 0.238  0.762  0.022  (0.740, 0.784) 1022

 z 2  2.575 2 (c) n  p ˆ q ˆ  c   0.762  0.238   751.56  752 adults  E   0.04  p ˆ q ˆ n

With 95% confidence you can say that the population standard deviation is between 2.32 and 6.17 minutes.

246 CHAPTER 6 │ CONFIDENCE INTERVALS Copyright © 2015 Pearson Education, Inc.  (n 1)s2 (n 1)s2   9  (3.38)2 9 (3.38)2  6. (a)   2 ,  2    , 19.023 2.700   (5.41, 38.08)  R L    (b)  5.4050, 38.0813  (2.32, 6.17)