Question

Use the standard normal distribution or he t-distribution to construct a 90% confidence nterval for the population mean. Justity your decision, f nether distrbuion can be used, explain wihy Inberpnet the resuts In a recent season, the population standard deviation of the yards per carry for all running backs was 1 24 normally distributed yards per carry of 25 randomly selected running backs are shown below Assume the yards per carry are 7.4 4.7 3.7 59 5.6 65 3.8 73 3.7 62 4.9 4.7 3 1 Which distribution should be used to construct the confidence interval? O A. Uso a t-distribution because n 30 and ? is unknown. OB. Use a normal distribution because n<30,the mies per gallon are norm any distributed and?isunknown ? ?. Use a 1-distrbution because n< 30 and is known 0 D. Use a normal distrbution because ? is known and the data are normaly dstbuted ? E. Cannot use the standard normal dstrit ution or te t-distubon because, is unknown, n< 30, and the yards are not normaly distrbuted

Answer 1

D. Use a Normal distribution because $\sigma$ is known and the data are normaly distributed.

The 90% confidence for the population mean ? is computed using the following expression

$CI = \left(\bar X - \frac{z_c \times \sigma}{\sqrt{n}} , \bar X + \frac{z_c \times \sigma}{\sqrt{n}} \right)$

$CI = \left(4.752 - \frac{ 1.645 \times 1.24}{\sqrt{ 25}} , 4.752 + \frac{ 1.645 \times 1.24}{\sqrt{ 25}} \right)$

$=(4.752-0.408,4.752+0.408)$

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