Use the standard normal distribution or the t-distribution to construct a 90 % c

Question

Use the standard normal distribution or the t-distribution to construct a 90 % 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 1.27 . The yards per carry of 25 randomly selected running backs are shown below. Assume the yards per carry are normally distributed. 1.5 4.8 3.1 3.5 3.5 6.7 6.2 4.9 5.1 1.9 1.6 4.3 5.7 6.3 5.6 4.5 6.9 3.7 7.2 5.6 2.5 6.8 4.8 5.9 3.6

Which distribution should be used to construct the confidence interval?

A. Use a normal distribution because n less than 30 , the miles per gallon are normally distributed and sigma is unknown.

B. Use a normal distribution because sigma is known and the data are normally distributed.

C. Use a t-distribution because nless than 30 and sigma is known.

D. Use a t-distribution because nless than 30 and sigma is unknown.

E. Cannot use the standard normal distribution or the t-distribution because sigma is unknown, n less than 30 , and the yards are not normally distributed.

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

A. The 90 % 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. If a large sample of players are taken approximately 90% of them will have yards per carry between the bounds of the confidence interval.

B. With 90% confidence, it can be said that the population mean yards per carry is between the bounds of the confidence interval.

C. It can be said that 90% 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.

Explanation / Answer

Sigma is known and given to be 1.27. Population is normal. So for the first question answer is "B. Use a normal distribution because sigma is known and the data are normally distributed."

Sample mean=4.648

z(0.95)=1.645 z(0.05)=-1.645

So Confidence interval is (4.23,5.07)

Interpretation of a confidence interval is that 90% of the samples will have sample mean within the interval if we take a large number of samples. So correct option is "A. If a large sample of players are taken approximately 90% of them will have yards per carry between the bounds of the confidence interval."

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