Answer to
Homework for Statistics Overview
BSNS5230, Jinchang Wang
Do the following problems. Mark your answers with problem numbers and question
numbers. Show your solution steps.
1. In a study of the wages earned by auto mechanics in a city, 17 randomly selected auto
mechanics were interviewed. Their hourly wages are as follows (in $):
25, 26, 21, 22, 25, 24.5, 19.5, 21.5, 26.5, 20.5, 17.5, 20, 21.5, 22.5, 20.5, 23, 25.5.
The sample mean = 22.47 ($); σ = 2.58, calculated by using either your calculator or Excel.
(a) Determine the 95% confidence interval for the average hourly wage of the city’s auto
mechanics.
Run Excel function, E =confidence.t(0.05,2.58,17) = 1.33.
Confidence interval: (22.47−1.33, 22.47+1.33) = (21.14, 23.80)
(b) Determine the 99% confidence interval for the average hourly wage of the city’s auto
mechanics.
Run Excel function, E =confidence.t(0.01,2.58,17) = 1.83.
Confidence interval: (22.47−1.83, 22.47+1.83) = (20.64, 24.30)
(c) If another city’s average hourly wage is $24, can we draw a conclusion that the two cities’
average hourly wages are different? Explain briefly.
Yes, with 95% confidence level. That is because 24 is not in the 95% confidence interval
(21.14, 23.80). Since the range is below 24, we conclude with 95% confidence level that this
city’s hourly wage of auto mechanics is lower than that of the other city. That is, we would have
a chance of 0.05 to be wrong if we conclude that this city’s hourly wage is lower than the other
city.
No, with 99% confidence level. That is because 24 is within the 99% confidence
interval (20.64, 24.30). That is, with 99% confidence level, we cannot conclude from the result
of the sample that the average hourly wage of the auto mechanics is different from that of the
other city.
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2. The following are the lifetimes (in hours) of a random sample of a type of electrical
component:
1428.4, 970.4, 724.6, 691.4, 856.7, 1144.3, 817.6, 1630.7, 1082.2, 932.8.
Sample size n=10.
Sample mean = 1,027.91 (hours), σ = 303.93 (calculated by calculators or Excel).
(a) Determine the 95% confidence interval for the average lifetime of the component.
E = confidence.t(0.05, 303.93, 10) = 217.42.
95% confidence interval: (1027.91−217.42, 1027.91+217.42) = (810.49, 1245.33)
(b) Determine the 99% confidence interval for the average lifetime of the component.
E = confidence.t(0.01, 303.93, 10) = 312.35.
99% confidence interval: (1027.91−312.35, 1027.91+312.35) = (715.56, 1340.26)
(c) In the last year, the average lifetime of the component was 833.7 hours. According to the
above sampling results, can we say that the average lifetime of the component is improved
compared to the last year? Briefly explain.
No, we cannot say after the study of the sample data that the average lifetime of the
component is improved compared to the last year, since 833.7 is within the 99% confidence
interval (715.56, 1340.26), as well as within the 95% confidence interval (810.49, 1245.33).
That is, our sample does not show a sufficiently significant evidence to support an improvement
even though the sample mean 1027.91 is higher than last year’s average 833.7.
(d) If you want the sample result more accurately reflect the population, what would you do?
We need to increase the sample size to make the sample result more accurate.
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3. A poll result shows that, among 150 randomly selected adults, 57 approved the job of the
president of United States.
Sample proportion p = 57/150 = 0.38.
(a) Can we say that exactly 38% of Americans approved the job of the president? Explain
briefly.
No, 38% is just the proportion of this particular sample. Another sample’s proportion
would be very much likely different from 38%. Neither is the proportion of the population that
includes all American people. The population proportion can be higher or lower than 38%.
(b) Determine the 95% confidence interval for the approval rate of the president’s job in United
States.
Calculate σ = √𝑝(1 − 𝑝) = √0.38(1 − 0.38) = 0.485
Error of sample margin = E = confidence.t(0.05, 0.485, 150) = 0.078.
95% confidence interval: (0.38−0.078, 0.38+0.078) = (0.302, 0.458) = (30.2%, 45.8%).
(c) Suppose 400 adults are randomly selected and interviewed, and 38% of them approved the
president’s job. Determine the 95% confidence interval of the approval rate.
Error of sample margin = E = confidence.t(0.05, 0.485, 400) = 0.048
95% confidence interval: (0.38−0.048, 0.38+0.048) = (0.332, 0.428) = (33.2%, 42.8%).
(d) Suppose 1,000 adults are randomly selected and interviewed, and 38% of them approved the
president’s job. Determine the 95% confidence interval of the approval rate.
Error of sample margin = E = confidence.t(0.05, 0.485, 1000) = 0.030
95% confidence interval: (0.38−0.030, 0.38+0.030) = (0.35, 0.41) = (35%, 41%).
(e) If you wish to reduce the 95% confidence interval to 4% (i.e., 38%±2%), at least how many
people should be interviewed?
Try different sample size n in Excel, and find the n that makes the value of function
confidence.t(0.05,0.485,n) less than 0.02. We have found that when n=2265,
confidence.t(0.05,0.485,2265) = 0.02. Therefore, the sample size should be at least 2,265 to
bring the 95% confidence interval to the level of ±2%.
Check: If n=2,265, then E=0.02 for 95% confidence, and 95% confidence interval is
(0.38−0.02, 0.38+0.02) = (0.36, 0.40) = (36%, 40%).
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