Answers to Selected Exercises

1.

  • 244
  • 15
  • 50

3. N(244,[latex]\frac{{15}}{{\sqrt{50}}}[\latex]) 5. As the sample size increases, there will be less variability in the mean, so the interval size decreases. 7. X is the time in minutes it takes to complete the U.S. Census short form. [latex]\overline{X}[/latex]is the mean time it took a sample of 200 people to complete the U.S. Census short form.

9.

CI: (7.9441, 8.4559)

This is a normal distribution curve. The peak of the curve coincides with the point 8.2 on the horizontal axis.  A central region is shaded between points 7.94 and 8.46.

EBM = 0.26

11. The level of confidence would decrease because decreasing n makes the confidence interval wider, so at the same error bound, the confidence level decreases.

13. a. [latex]\overline{x}[/latex]= 2.2, b. σ = 0.2, c. n = 20

14.[latex]\overline{X}[/latex] is the mean weight of a sample of 20 heads of lettuce.

16.

EBM = 0.07

CI: (2.1264, 2.2736)

This is a normal distribution curve. The peak of the curve coincides with the point 2.2 on the horizontal axis. A central region is shaded between points 2.13 and 2.27.

18. The interval is greater because the level of confidence increased. If the only change made in the analysis is a change in confidence level, then all we are doing is changing how much area is being calculated for the normal distribution. Therefore, a larger confidence level results in larger areas and larger intervals.

20. The confidence level would increase.

22. 30.4

24. σ

26. μ

28. normal

30. 0.025

32. (24.52,36.28)

34. We are 95% confident that the true mean age for Winger Foothill College students is between 24.52 and 36.28.

36. The error bound for the mean would decrease because as the CL decreases, you need less area under the normal curve (which translates into a smaller interval) to capture the true population mean.

37.

71
3
48
X is the height of a Swiss male, and is the mean height from a sample of 48 Swiss males.
Normal. We know the standard deviation for the population, and the sample size is greater than 30.
CI: (70.151, 71.49)
EBM = 0.849
The confidence interval will decrease in size, because the sample size increased. Recall, when all factors remain unchanged, an increase in sample size decreases variability. Thus, we do not need as large an interval to capture the true population mean.
39.

a. [latex]\overline{x}[/latex] = 23.6, b. σ = 7, c. n = 100

  1. X is the time needed to complete an individual tax form.[latex]\overline{X}[/latex] is the mean time to complete tax forms from a sample of 100 customers.
  2. N(23.6,[latex]\frac{{7}}{{\sqrt{100}}}) because we know sigma.
    1. (22.228, 24.972)
    2. EBM = 1.372
  3. It will need to change the sample size. The firm needs to determine what the confidence level should be, then apply the error bound formula to determine the necessary sample size.
  4. The confidence level would increase as a result of a larger interval. Smaller sample sizes result in more variability. To capture the true population mean, we need to have a larger interval.
  5. According to the error bound formula, the firm needs to survey 206 people. Since we increase the confidence level, we need to increase either our error bound or the sample size.

41.

7.9
2.5
20

X is the number of letters a single camper will send home.[latex]\overline{X}[/latex]is the mean number of letters sent home from a sample of 20 campers.
N 7.9([latex]\frac{{2.5}}{{\sqrt{20}}}[/latex])
CI: (6.98, 8.82)
EBM: 0.92
The error bound and confidence interval will decrease.

45.

Use the formula for EBM, solved for n:

n = [latex]\frac{{{z}^{2}{\sigma}^{2}}}{{{EBM}^{2}}}[/latex]

From the statement of the problem, you know that σ = 2.5, and you needEBM = 1.

z = z0.035 = 1.812

(This is the value of z for which the area under the density curve to the right of z is 0.035.)

n = [latex]\frac{{{z}^{2}{\sigma}^{2}}}{{{EBM}^{2}}}[/latex] = [latex]\frac{{{1.812}^{2}{2.5}^{2}}}{{{1}^{2}}}\approx{20.52}[/latex]

You need to measure at least 21 male students to achieve your goal.

47. X is the number of hours a patient waits in the emergency room before being called back to be examined. [latex]\overline{X}[/latex] is the mean wait time of 70 patients in the emergency room.

49.

CI: (1.3808, 1.6192)

This is a normal distribution curve. The peak of the curve coincides with the point 1.5 on the horizontal axis.  A central region is shaded between points 1.38 and 1.62.

EBM = 0.12

 51.

  1. [latex]\overline{x}[/latex]= 151
  2. [latex]{x}_{s}[/latex] = 32
  3. n = 108
  4. n – 1 = 107

53. the mean number of hours spent watching television per month from a sample of 108 Americans.

55.

CI: (142.92, 159.08)

This is a normal distribution curve. The peak of the curve coincides with the point 151 on the horizontal axis.  A central region is shaded between points 142.92 and 159.08.

EBM = 8.08

57.

  1. 3.26
  2. 1.02
  3. 39

59. [latex]\mu[/latex]

61. [latex]{t}_{38}[/latex]

63. 0.25

65. (2.93, 3.59)

67. We are 95% confident that the true mean number of colors for national flags is between 2.93 colors and 3.59 colors.

68. The error bound would become EBM = 0.245. This error bound decreases because as sample sizes increase, variability decreases and we need less interval length to capture the true mean.

71.

    1. 629
    2. 6944
    3. 35
    4. 34
  1. t34
    1. CI: (6244, 11,014)
    2. EB = 2385
  2. It will become smaller
75. [latex]\overline{x}[/latex]= $,854.23, s = $521,130.41

Note that we are not given the population standard deviation, only the standard deviation of the sample.

There are 30 measures in the sample, so n = 30, and df = 30 – 1 = 29

CL = 0.96, so α = 1 – CL = 1 – 0.96 = 0.04

[latex]\frac{{\alpha}}{{2}}=0.02={t}_{0.02}-2.150[/latex]

EBM = [latex]{t}_{\frac{{\alpha}}{{2}}}\left(\frac{{s}}{{\sqrt{n}}}\right=2.150\left\frac{{521,130.41}}{{\sqrt{30}}}\right)[/latex] ~ $204,561.66

[latex]\overline{x}[/latex]-EBM = $251,854.23 – $204,561.66 = $47,292.57

[latex]\overline{x}[/latex]+EBM = $251,854.23+ $204,561.66 = $456,415.89

We estimate with 96% confidence that the mean amount of money raised by all Leadership PACs during the 2011–2012 election cycle lies between $47,292.57 and $456,415.89.

Alternate Solution

Enter the data as a list.

Press STAT and arrow over to TESTS.

Arrow down to 8:TInterval.

Press ENTER.

Arrow to Data and press ENTER.

Arrow down and enter the name of the list where the data is stored.

Enter Freq: 1

Enter C-Level: 0.96

Arrow down to Calculate and press Enter.

The 96% confidence interval is ($47,262, $456,447).

The difference between solutions arises from rounding differences.

79.

      1. [latex]\overline{x}[/latex]=
      2. sx =
      3. n =
      4. n – 1 =
    1. X is the number of unoccupied seats on a single flight.[latex]\overline{X}[/latex] is the mean number of unoccupied seats from a sample of 225 flights.
    2. We will use a Student’s-t distribution, because we do not know the population standard deviation.
      1. CI: (11.12 , 12.08)
      2. Check student’s solution.
      3. EBM: 0.48

81.

    1. CI: (7.64 , 9.36)
    2. EBM: 0.86
  1. The sample should have been increased.
  2. Answers will vary.
  3. Answers will vary.
  4. Answers will vary.

83. (12.32, 14.29)

 

A Population Proportion

87. It would decrease, because the z-score would decrease, which reducing the numerator and lowering the number.

89. X is the number of “successes” where the woman makes the majority of the purchasing decisions for the household. P′ is the percentage of households sampled where the woman makes the majority of the purchasing decisions for the household.

91.

CI: (0.5321, 0.6679)

This is a normal distribution curve. The peak of the curve coincides with the point 0.6 on the horizontal axis.  A central region is shaded between points 0.5321 and 0.6679.

EBM: 0.0679

 

92. X is the number of “successes” where an executive prefers a truck. P′ is the percentage of executives sampled who prefer a truck.

94.

CI: (0.19432, 0.33068)

This is a normal distribution curve. The peak of the curve coincides with the point 0.26 on the horizontal axis.  A central region is shaded between points 0.1943 and 0.3307.

EBM: 0.0707

 

96.

The sampling error means that the true mean can be 2% above or below the sample mean.

 

98. P′ is the proportion of voters sampled who said the economy is the most important issue in the upcoming election.

102. The number of girls, ages 8 to 12, in the 5 P.M. Monday night beginning ice-skating class.

103.

  1. x = 64
  2. n = 80
  3. p′ = 0.8

105. p

107. P’~N[latex]\left(0.8,\sqrt{\frac{{(0.8)(0.2)}}{{80}}}\right[/latex]. (0.72171, 0.87829).

109. 0.04

111. (0.72; 0.88)

113. With 92% confidence, we estimate the proportion of girls, ages 8 to 12, in a beginning ice-skating class at the Ice Chalet to be between 72% and 88%.

105. The error bound would increase. Assuming all other variables are kept constant, as the confidence level increases, the area under the curve corresponding to the confidence level becomes larger, which creates a wider interval and thus a larger error.

117.

  1. 1,068
  2. The sample size would need to be increased since the critical value increases as the confidence level increases.

119.

  1. X = the number of people who feel that the president is doing an acceptable job;P′ = the proportion of people in a sample who feel that the president is doing an acceptable job.

  2. N[latex]\left(0.61,\sqrt{\frac{{(0.61)(0.39)}}{{1200}}}\right)[/latex]
    1. CI: (0.59, 0.63)
    2. Check student’s solution
    3. EBM: 0.02

123.

X = the number of adult Americans who feel that crime is the main problem; P′ = the proportion of adult Americans who feel that crime is the main problem
Since we are estimating a proportion, given P′ = 0.2 and n = 1000, the distribution we should use is N[latex]\left(0.2,\sqrt{\frac{{(0.2)(0.8)}}{{1000}}}\right)[/latex]. 

CI: (0.18, 0.22)
Check student’s solution.
EBM: 0.02
One way to lower the sampling error is to increase the sample size.
The stated “± 3%” represents the maximum error bound. This means that those doing the study are reporting a maximum error of 3%. Thus, they estimate the percentage of adult Americans who feel that crime is the main problem to be between 18% and 22%.
125. 0.79
127. 0.030
129. 0.6614

131.

  1. p’ = [latex]\frac{{(0.55+0.49)}}{{2}}=0.52[/latex],  EBP = 0.55 – 0.52 = 0.03
  2. No, the confidence interval includes values less than or equal to 0.50. It is possible that less than half of the population believe this.
  3. EBP=(1.150)[latex]\sqrt{\frac{{0.52(0.48)}}{{1000}}[/latex]0.018
    (p′ – EBP, p′ + EBP) = (0.52 – 0.018, 0.52 + 0.018) = (0.502, 0.538)

    Alternate Solution

    STAT TESTS A: 1-PropZinterval with x = (0.52)(1,000), n = 1,000, CL = 0.75.

    Answer is (0.502, 0.538) Yes – this interval does not fall less than 0.50 so we can conclude that at least half of all American adults believe that major sports programs corrupt education – but we do so with only 75% confidence.CL = 0.75, so α = 1 – 0.75 = 0.25 and [latex][/latex]=1.150. (The area to the right of this z is 0.125, so the area to the left is 1 – 0.125 = 0.875.)