A Single Population Mean using the Normal Distribution – Practice

1.

• • • 244
    • 15
    • 50

3. [latex]N\left(244,\frac{15}{\sqrt{50}}\right)[/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)

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. 1. [latex]\overline{x}[/latex]= 2.2, 2. σ = 0.2, 3. n = 20

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

17.

EBM = 0.07

CI: (2.1264, 2.2736)

19. 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.

21. The confidence level would increase.

23. 30.4

25. σ

27. μ

29. normal

31. 0.025

33. (24.52,36.28)

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

37. 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.

A Single Population Mean using the Student t Distribution — Practice

39. 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.

41.

CI: (1.3808, 1.6192)

EBM = 0.12

 43.

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

45. [latex]\overline{X}[/latex] the mean number of hours spent watching television per month from a sample of 108 Americans.

47. CI: (142.92, 159.08)

EBM = 8.08

49.

1. 3.26
2. 1.02
3. 39

51. [latex]\mu[/latex]

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

55. 0.25

57. (2.93, 3.59)

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

60. 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.

A Population Proportion — Practice

63. The sample size needed would increase. As the confidence level increases, 

$\mathrm{\alpha }$

decreases and [latex]z_{\frac{\alpha}{2}}[/latex] increases. To maintain the same error bound, the size of the sample needs to increase.

65. 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.

67. CI: (0.5321, 0.6679)

EBM: 0.0679

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

71. CI: (0.19432, 0.33068)

EBM: 0.0707

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

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

79. The number of girls, ages 8 to 12, in the 5 p.m. Monday night beginning ice-skating class.

81.

• • 1. x = 64
    2. n = 80
    3. p′ = 0.8

83. p

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

87. 0.04

89. (0.72; 0.88)

91. 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%.

93. 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.

1. 71
2. 3
3. 48

b. X is the height of a Swiss male, and is the mean height from a sample of 48 Swiss males.

c. Normal. We know the standard deviation for the population, and the sample size is greater than 30.

d. CI: (70.151, 71.49)

EBM = 0.849

e. 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.

101.

a. [latex]\overline{x}[/latex] = $568,873

b. [latex]CL=0.95 \alpha = 1-0.95 = 0.05 z_{\frac{\alpha}{2}} = 1.96[/latex]

[latex]EBM = z_{0.025} \frac{\sigma}{\sqrt{n}} = 1.96 \frac{909200}{\sqrt{40}} = \$ 281,764[/latex]

c. [latex]\overline{x} - EBM = 568,873 - 281,764 = 287,109[/latex]

[latex]\overline{x} + 568,873 + 281,764 = 850,637[/latex]

d. We estimate with 95% confidence that the mean amount of contributions received from all individuals by House candidates is between $287,109 and $850,637.

103.

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 = z_0.035 = 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.

A Single Population Mean using the Student t Distribution — Homework

105.

a.

111.

a.

• • 1. 1. 1. [latex]\overline{x}[/latex]=
          2. s_x =
          3. n =
          4. n – 1 =

b. 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.

c. We will use a Student’s t-distribution, because we do not know the population standard deviation.

d.

1. CI: (11.12 , 12.08)
2. Check student’s solution.
3. EBM: 0.48

113.

a.

• • 1. 1. CI: (7.64 , 9.36)

       2. 

       3. EBM: 0.86

b. The sample should have been increased.

c. Answers will vary.

d. Answers will vary.

e. Answers will vary.

115. (12.32, 14.29)

A Population Proportion

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

121.

a.

1. (0.72, 0.82)
2. (0.65, 0.76)
3. (0.60, 0.72)

b. Yes, the intervals (0.72, 0.82) and (0.65, 0.76) overlap, and the intervals (0.65, 0.76) and (0.60, 0.72) overlap.

c. We can say that there does not appear to be a significant difference between the proportion of Asian adults who say that their families would welcome a White person into their families and the proportion of Asian adults who say that their families would welcome a Latino person into their families.

d. We can say that there is a significant difference between the proportion of Asian adults who say that their families would welcome a White person into their families and the proportion of Asian adults who say that their families would welcome a Black person into their families.

123.

131.

a. p’ = [latex]\frac{{(0.55+0.49)}}{{2}}=0.52[/latex],  EBP = 0.55 – 0.52 = 0.03

b. 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.

c. CL = 0.75, so α = 1 – 0.75 = 0.25 and [latex]\frac{\alpha}{2} = 0.125 \ z_{\frac{\alpha}{2}} = 1.150[/latex]. (The area to the right of this z is 0.125, so the area to the left is 1 – 0.125 = 0.875.)

(p′ – EBP, p′ + EBP) = (0.52 – 0.018, 0.52 + 0.018) = (0.502, 0.538)

Alternate Solution