Question

Use the standard normal distribution or the​ t-distribution to construct a 99​% confidence interval for the population mean. Justify your decision. If neither distribution can be​ used, explain why. Interpret the results.

In a recent​ season, the population standard deviation of the yards per carry for all running backs was 1.21. The yards per carry of 25 randomly selected running backs are shown below. Assume the yards per carry are normally distributed.

1.6 3.4 2.8 6.9 6.1 5.8 7.3 6.5 2.8  2.4 2.2 4.7 7.4 7.1 5.7 2.9 2.6 6.8 3.9 5.4 2.8   6.1 4.2 3.9 7.4  

Which distribution should be used to construct the confidence​ interval?

A. Use a normal distribution because sigmaσ is known and the data are normally distributed.

B. Use a​ t-distribution because n <30 and σ is known.

C. Use a​ t-distribution because n <30 and σ is unknown.

D. Use a normal distribution because n <30​, the data are normally distributed and σ is unknown.

E. Cannot use the standard normal distribution or the​ t-distribution because σ is​ unknown, n <30​, and the data are not normally distributed.

Select the correct choice below​ and, if​ necessary, fill in any answer boxes to complete your choice.

A. The 99​% confidence interval is ​ . ​(Round to two decimal places as​ needed.)

B. Neither distribution can be used to construct the confidence interval.

Interpret the results. Choose the correct answer below.

A. With 99​% ​confidence, it can be said that the population mean yards per carry is between the bounds of the confidence interval.

B. If a large sample of players are taken approximately 99​% of them will have yards per carry between the bounds of the confidence interval.

C. It can be said that 99​% of players have a yards per carry between the bounds of the confidence interval.

D. Neither distribution can be used to construct the confidence interval.

Answer 1

(a) Option (A) Use a normal distribution because sigmaσ is known and the data are normally distributed.

(b) Confidence interval:

n = 22     

x-bar = 4.691     

s = 1.21     

% = 99     

Standard Error, SE = σ/√n =    1.21 /√22 = 0.257972867

z- score = 2.575829304     

Width of the confidence interval = z * SE =     2.57582930354892 * 0.257972866790289 = 0.66449407

Lower Limit of the confidence interval = x-bar - width =      4.691 - 0.664494069798946 = 4.02650593

Upper Limit of the confidence interval = x-bar + width =      4.691 + 0.664494069798946 = 5.35549407

Option (A) The 99% confidence interval is [4.03, 5.36]