Use the standard normal distribution or the t-distribution to construct a 95% co

Question

Use the standard normal distribution or the t-distribution to construct a 95% confidence interval for the population mean. Justify your decision. If neither distribution can be used, explain why. Interpret the results. In a random sample of 12 mortgage institutions, the mean interest rate was 3.52% and the standard deviation was 0.46%. Assume the interest rates are normally distributed.

Which distribution should be used to construct the confidence interval?

A.Use a normal distribution because n<30 and the interest rates are normally distributed.

B.Use a normal distribution because the interest rates are normally distributed and is known.

C.Use a t-distribution because the interest rates are normally distributed and is known.

D.Use a t-distribution because it is a random sample, is unknown, and the interest rates are normally distributed.

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

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

A.The 95% 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.It can be said that 95% of institutions have an interest rate between the bounds of the confidence interval.

B.With 95% confidence, it can be said that the population mean interest rate is between the bounds of the confidence interval.

C.If a large sample of institutions are taken approximately 95% of them will have an interest rate between the bounds of the confidence interval.

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

Explanation / Answer

D.Use a t-distribution because it is a random sample, is unknown, and the interest rates are normally distributed.

n = 12     

x-bar = 3.52     

s = 0.46     

% = 95     

Standard Error, SE = s/Ön =    0.46/12 = 0.132790562

Degrees of freedom = n - 1 =   12 -1 = 11   

t- score = 2.200985159     

Width of the confidence interval = t * SE =     2.20098515872184 * 0.132790561913614 = 0.292270056

Lower Limit of the confidence interval = x-bar - width =      3.52 - 0.292270055990198 = 3.227729944

Upper Limit of the confidence interval = x-bar + width =      3.52 + 0.292270055990198 = 3.812270056

The 95% confidence interval is [3.23, 3.81]

B.With 95% confidence, it can be said that the population mean interest rate is between the bounds of the confidence interval.

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