Question

Answer the question. CHAPTER 8: ESTIMATION AND CONFIDENCE INTERVALS 1. As degrees of freedom increase, the...

Answer the question.

CHAPTER 8: ESTIMATION AND CONFIDENCE INTERVALS

1. As degrees of freedom increase, the t-distribution approaches the:

A. binomial distribution

B. exponential distribution

C. standard normal distribution

D. None of the above

2. Given a t-distribution with 14 degrees of freedom, the area left of - 1.761 is

A. 0.025

B. 0.05

C. 0.10

D. 0.90

E. None of the above

3. 100 samples of size fifty were taken from a population with population mean 72. The sample mean and sample standard deviation were recorded and the 95% confidence interval for the population mean was recorded for each sample. Of the 100 confidence intervals, how many would we expect to have 72 between the endpoints?

A. 0

B. 5

C. 95

D. 100

4. You are given a confidence interval for the population mean of 26 to 42. The sample mean used to construct this confidence interval was:

A. 1.96

B. 26

C. 34

D. Cannot be determined with the given information

5. A sample of five price/earnings ratios for companies in the Services sector follows. 15 11 14 17 12 A confidence interval for the population mean is requested. In order to construct the confidence interval one must assume:

A. that the sample came from a normal distribution

B. that the population standard deviation is known

C. No assumptions are needed

D. None of choices are correct

6. A sample of five price/earnings ratios for companies in the Services sector follows. 15 11 14 17 12 A confidence interval for the population mean is requested. Assuming that the sample comes from a normal population, the appropriate distribution used in constructing this confidence interval is the:

A. F

B. t

C. z

D. ? 2

E. None of the choices are correct

7. A sample of five price/earnings ratios for companies in the Services sector follows. 15 11 14 17 12 A confidence interval for the population mean is requested. Assuming that the sample comes from a normal population, the appropriate degrees of freedom used for this confidence interval is:

A. 2

B. 3

C. 4

D. 5

E. None of the choices are correct

8. A sample of five price/earnings ratios for companies in the Services sector follows. 15 11 14 17 12 A confidence interval for the population mean is requested. Assuming that the sample comes from a normal population, the 95% confidence interval is

A. 6.39 to 21.21

B. 9.77 to 26.63

C. 10.84 to 16.76

D. 11.71 to 15.89 E.

None of the choices are correct

9. A sample of size 36 is taken from a population with standard deviation of the population is 12. The sample mean is found to be 116. Construct a 95% confidence interval.

A. - 25.89 to 49.89

B. 95.63 to 136.37

C. 111.94 to 120.06

D. 112.08 to 119.92

E. None of the above

10. The useful life of a certain type of light bulb is known to have a standard deviation of σ = 40 hours. How large a sample should be taken if it is desired to have a margin of error of 10 hours or less at a 95% level of confidence?

A. 8

B. 37

C. 44

D. 62

E. None of the choices are correct

11. A random sample of 300 voters showed 47% in favor of a certain ballot proposal. A 90% confidence interval estimate for the population proportion of voters favoring the proposal is:

A. 0.38 to 0.56

B. 0.4226 to 0.5174

C. 0.4412 to 0.4988

D. 0.4136 to 0.5264

E. None of the above

12. In choosing a sample size for a public-opinion survey, what hypothesized value of the population proportion will lead to the largest sample size when the confidence level and the maximum sample error are specified?

A. p = .

1 B. p = .5

C. p = .99

D. The confidence level must be known before an answer can be given

Answer question 13 - 14 based on the information below: Restoran Lazatlah has 1500 retail outlets throughout the Sabah. The owner, Mr. Rashid, is evaluating a potential location for a new outlet, based in part, on the mean annual income of the individuals in the marketing are of the new location. A sample of size n = 50 was taken; the sample mean income is RM24,000. The population is not believed to be highly skewed. The population standard deviation is estimated to be RM3000, and the confidence coefficient to be used in the interval estimate is .90.

13. Compute the margin of error to the point of estimate:

A. RM543.91

B. RM697.91

C. RM24000 + RM543.91

D. RM24000 + RM697.91

14. What is the interval estimate of the population mean:

A. RM543.91

B. RM697.91

C. RM24000 + RM543.91

D. RM24000 + RM697.91 Answer question 15 - 17 based on the information below: Public Says is an online survey firm that is doing a research to identify the opinion of the people in the Republic of Wakanda on who they would vote for in the coming presidential election. There are only two competing candidates for the office, T’Challa and Erik Killmonger that From the data retrieved, 7300 out of 10000 online respondents who are registered voters favor T’Challa.

15 Based on 95% confidence interval, what is the proportion of the population that favors T’Challa:

A. 0.8 + 0.0248

B. 0.8 + 0.0208

C. 0.8 + 0.0277

D. Cannot be determined with the given information

16. How large is the sample needed if Public Says would like a .99 probability that the sample proportion is within + 0.01 of the population proportion:

A. 8656

B. 13,526

C. 15,277

D. 23,870

17. If the population proportion is unknown, what is the highest possible sample size recommended:

A. 8656

B. 13,526

C. 15,277

D. 23,870

Homework Answers

Answer #1

1. C.  standard normal distribution

Explanation: As degrees of freedom increases, the t-distribution approaches normality. This means that the area near the center increases while the area near the tails decreases. When df decreases, the area near the center decreases while the area near the tails increases.

2. B. 0.05

Explanation: R code-

> pt(- 1.761, df = 14, lower.tail = TRUE)
[1] 0.05002709

4. C. 34

Explanation: given a confidence interval for the population mean of 26 to 42. Since t distribution is symmetric, confidence interval is symetrically spread. So answer is (26+42)/2 = 34. Note that, here we assumed that the data is from Nornal distribution (or assymtotically). Otherwise is not possible.

5. A. that the sample came from a normal distribution.

6. B. t

Explanation: T = . Note that, T is the statistics to test the hypothesis.

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