Question

(10 pts) Use the following information for all parts:
On a particular golf course, a sample of 40 golfers have a mean golf score of 79. Suppose the population standard deviation for this course is 3.6605.

(a) Using the formula a 90% confidence interval as presented in lecture, fill in the blanks with the appropriate values for this problem for calculating the confidence interval below. To enter $\sqrt x$ where x is any number, type sqrt(x). For example, should be typed as sqrt(2).

(b) Using the formula from part (a), find a 90% confidence interval for the mean golf score of all golfers. Round each answer to the nearest hundredth (2 decimal places).

(c) Calculate the margin of error for the answer in part (b). Round the answer to the nearest hundredth (2 decimal places).



(d) Using the formula a 99% confidence interval as presented in lecture, fill in the blanks with the appropriate values for this problem for calculating the confidence interval below. To enter $\sqrt x$ where x is any number, type sqrt(x). For example, should be typed as sqrt(2).

(e) Using the formula from part (d), find a 99% confidence interval for the mean golf score of all golfers. Round each answer to the nearest hundredth (2 decimal places).

(f) What is the correct interpretation of the confidence interval found in part (e)?
A. We are 99% confident that the sample mean golf score for a sample of 40 golfers lies in the interval found in part (e).
B. The true population mean golf score for all golfers lies in the interval found in part (e).
C. There is a 1% chance that the sample mean golf score for a sample of 40 golfers is not in the interval found in part (e).
D. We are 99% confident that the true population mean golf score for all golfers lies in the interval found in part (e).

Answer 1

A) At 90% confidence interval the critical value is z_0.05 = 1.645

The formula for 90% confidence interval is

$\bar x$ +/- z_0.05 * $\sigma/\sqrt n$​​​​​

B) The 90% confidence interval for population mean is

79 +/- 1.645 * 3.6605/$\sqrt 40$

= 79 +/- 0.95

= 78.05, 79.95

C) Margin of error = z_0.05 * $\sigma/\sqrt n$

= 1.645 * 3.6605/$\sqrt 40$

= 0.95

D) At 99% confidence interval the critical value is z_0.005 = 2.58

The formula for 99% confidence interval is

$\bar x$ +/- z_0.005 * $\sigma/\sqrt n$

E) The 99% confidence interval for population mean is

79 +/- 2.58 * 3.6605/$\sqrt 40$

= 79 +/- 1.49

= 77.51, 80.49

F) Option - D) We are 99% confident that the true population mean golf score for all golfers lies in the interval found in part(e) .