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 1.37. The yards per carry of 25 randomly selected running backs are shown below. Assume the yards per carry are normally distributed.

2.4 4.5 3.6 5.3 4.8 6.8 5.9 4.9 4.9 1.7 2.1 6.4 4.5 5.1 4.1 7.4 6.9 4.4 6.8 6.6 4.8 7.1 6.1 7.2 6.6

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

A. Use a? t-distribution because nless than<30 and ? is unknown.

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

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

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

E. Cannot use the standard normal distribution or the? t-distribution because sigma? is? unknown, n less than n<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. With 90?% ?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 90?% of them will have yards per carry 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.

Answer 1

Solution 1:

Option D is correct.

Since population standard deviation is known and sample size, n is greater than 30, we use normal distribution or z-distribution.

Solution 2:

We first calculate the sample mean, for confidence interval.

Sample mean, =

Sample mean, = 130.9/25

Sample mean, = 5.236

Using Z-tables, the critical value at = 0.10/2 = 0.05 is 1.645

90% confidence interval is given by:-

5.236 1.645*1.37/25

5.236 1.645*0.274

5.236 0.451

4.79, 5.69

A. The 90% confidence interval is 4.79 to 5.69

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