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The registrar's office at State University would like to determine a 95% confidence interval
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# Question : The registrar's office at State University would like to determine a 95% confidence interval : 2150491

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Solve the problem.

6) What is zα/2 when α = 0.01?

A) 2.575

B) 1.96

C) 2.33

D) 1.645

7) What is the confidence level of the following confidence interval for μ?

overbar(x) ± 0.99((σ/√(n)))

A) 99%

B) 90%

C) 80%

D) 67%

8) The registrar's office at State University would like to determine a 95% confidence interval for the mean commute time of its students. A member of the staff randomly chooses a parking lot and surveys the first 200 students who park in the chosen lot on a given day. The confidence interval is

A) meaningful because the sample size exceeds 30 and the Central Limit Theorem ensures normality of the sampling distribution of the sample mean.

B) meaningful because the sample is representative of the population.

C) not meaningful because of the lack of random sampling.

D) not meaningful because the sampling distribution of the sample mean is not normal.

9) A 90% confidence interval for the mean percentage of airline reservations being canceled on the day of the flight is (3.9%, 7.3%). What is the point estimator of the mean percentage of reservations that are canceled on the day of the flight?

A) 3.65%

B) 3.4%

C) 5.60%

D) 1.70%

10) A 95% confidence interval for the average salary of all CEOs in the electronics industry was constructed using the results of a random survey of 45 CEOs. The interval was (\$130,771, \$146,241). To make more useful inferences from the data, it is desired to reduce the width of the confidence interval. Which of the following will result in a reduced interval width?

A) Increase the sample size and increase the confidence level.

B) Decrease the sample size and increase the confidence level.

C) Increase the sample size and decrease the confidence level.

D) Decrease the sample size and decrease the confidence level.

11) Suppose a large labor union wishes to estimate the mean number of hours per month a union member is absent from work. The union decides to sample 375 of its members at random and monitor the working time of each of them for 1 month. At the end of the month, the total number of hours absent from work is recorded for each employee. Which of the following should be used to estimate the parameter of interest for this problem?

A) A large sample confidence interval for p.

B) A small sample confidence interval for μ.

C) A large sample confidence interval for μ.

D) A small sample confidence interval for p.

12) Explain what the phrase 95% confident means when we interpret a 95% confidence interval for μ.

A) 95% of the observations in the population fall within the bounds of the calculated interval.

B) The probability that the sample mean falls in the calculated interval is 0.95.

C) In repeated sampling, 95% of similarly constructed intervals contain the value of the population mean.

D) 95% of similarly constructed intervals would contain the value of the sampled mean.

14) A retired statistician was interested in determining the average cost of a \$200,000.00 term life insurance policy for a 60-year-old male non-smoker. He randomly sampled 65 subjects (60-year-old male non-smokers) and constructed the following 95 percent confidence interval for the mean cost of the term life insurance: (\$850.00, \$1050.00). What value of alpha was used to create this confidence interval?

A) 0.05

B) 0.10

C) 0.01

D) 0.025

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

15) Suppose (1,000, 2,100) is a 95% confidence interval for μ. To make more useful inferences from the data, it is desired to reduce the width of the confidence interval. Explain why an increase in sample size will lead to a narrower interval of the estimate of μ.

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Answer the question True or False.

23) Find zα/2 for the given value of α.

α = 0.05

A) 2.81

B) 0.33

C) 1.96

D) 1.645

24) Determine the confidence level for the given confidence interval for μ.

overbar(x) ± 1.34((σ/√(n)))

A) 95.5%

B) 91%

C) 41%

D) 82%

25) A random sample of n measurements was selected from a population with unknown mean μ and known standard deviation σ. Calculate a 95% confidence interval for μ for the given situation. Round to the nearest hundredth when necessary.

n = 160, overbar(x) = 68, σ = 15

A) 68 ± 2.32

B) 68 ± 0.18

C) 68 ± 29.4

D) 68 ± 1.95

26) A 90% confidence interval for the average salary of all CEOs in the electronics industry was constructed using the results of a random survey of 45 CEOs. The interval was (\$124,443, \$140,235). Give a practical interpretation of the interval.

A) We are 90% confident that the mean salary of the sampled CEOs falls in the interval \$124,443 to \$140,235.

B) 90% of all CEOs in the electronics industry have salaries that fall between \$124,443 to \$140,235.

C) We are 90% confident that the mean salary of all CEOs in the electronics industry falls in the interval \$124,443 to \$140,235.

D) 90% of the sampled CEOs have salaries that fell in the interval \$124,443 to \$140,235.

27) A random sample of 250 students at a university finds that these students take a mean of 15.2 credit hours per quarter with a standard deviation of 2.3 credit hours. Estimate the mean credit hours taken by a student each quarter using a 95% confidence interval. Round to the nearest thousandth.

A) 15.2 ± .018

B) 15.2 ± .188

C) 15.2 ± .012

D) 15.2 ± .285

28) A random sample of 250 students at a university finds that these students take a mean of 15 credit hours per quarter with a standard deviation of 2.5 credit hours. The 99% confidence interval for the mean is 15 ± 0.407. Interpret the interval.

A) The probability that a student takes 14.593 to 15.407 credit hours in a quarter is 0.99.

B) 99% of the students take between 14.593 to 15.407 credit hours per quarter.

C) We are 99% confident that the average number of credit hours per quarter of the sampled students falls in the interval 14.593 to 15.407 hours.

D) We are 99% confident that the average number of credit hours per quarter of students at the university falls in the interval 14.593 to 15.407 hours.

29) The director of a hospital wishes to estimate the mean number of people who are admitted to the emergency room during a 24-hour period. The director randomly selects 81 different 24-hour periods and determines the number of admissions for each. For this sample, overbar(x) = 15.6 and s2 = 16. Estimate the mean number of admissions per 24-hour period with a 99% confidence interval.

A) 15.6 ± .440

B) 15.6 ± .127

C) 15.6 ± 1.144

D) 15.6 ± 4.578

30) Suppose a large labor union wishes to estimate the mean number of hours per month a union member is absent from work. The union decides to sample 301 of its members at random and monitor the working time of each of them for 1 month. At the end of the month, the total number of hours absent from work is recorded for each employee. If the mean and standard deviation of the sample are overbar(x) = 9.3 hours and s = 2.3 hours, find a 95% confidence interval for the true mean number of hours a union member is absent per month. Round to the nearest thousandth.

A) 9.3 ± .126

B) 9.3 ± .015

C) 9.3 ± .171

D) 9.3 ± .260

31) Parking at a large university can be extremely difficult at times. One particular university is trying to determine the location of a new parking garage. As part of their research, officials are interested in estimating the average parking time of students from within the various colleges on campus. A survey of 338 College of Business (COBA) students yields the following descriptive information regarding the length of time (in minutes) it took them to find a parking spot. Note that the "Lo 95%" and "Up 95%" refer to the endpoints of the desired confidence interval.

Variable N Lo 95% CI Mean Up 95% CI SD

Parking Time 338 9.1944 10.466 11.738 11.885

Give a practical interpretation for the 95% confidence interval given above.

A) 95% of the COBA students had parking times that fell between 9.19 and 11.74 minutes.

B) We are 95% confident that the average parking time of the 338 COBA students surveyed falls between 9.19 and 11.74 minutes.

C) 95% of the COBA students had parking times of 10.466 minutes.

D) We are 95% confident that the average parking time of all COBA students falls between 9.19 and 11.74 minutes.

32) Parking at a large university can be extremely difficult at times. One particular university is trying to determine the location of a new parking garage. As part of their research, officials are interested in estimating the average parking time of students from within the various colleges on campus. A survey of 338 College of Business (COBA) students yields the following descriptive information regarding the length of time (in minutes) it took them to find a parking spot. Note that the "Lo 95%" and "Up 95%" refer to the endpoints of the desired confidence interval.

Variable N Lo 95% CI Mean Up 95% CI SD

Parking Time 338 9.1944 10.466 11.738 11.885

Explain what the phrase "95% confident" means when working with a 95% confidence interval.

A) In repeated sampling, 95% of the population means will fall within the interval created.

B) 95% of the observations in the population will fall within the endpoints of the interval.

C) In repeated sampling, 95% of the intervals created will contain the population mean.

D) In repeated sampling, 95% of the sample means will fall within the interval created.

35) How much money does the average professional football fan spend on food at a single football game? That question was posed to 60 randomly selected football fans. The sampled results show that the sample mean was \$70.00 and prior sampling indicated that the population standard deviation was \$17.50. Use this information to create a 95 percent confidence interval for the population mean.

A) 70 ± 1.645((17.50/√(60)))

B) 70 ± 1.671((17.50/√(60)))

C) 70 ± 1.833((17.50/√(60)))

D) 70 ± 1.960((17.50/√(60)))

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