Professors of accountancy are in high demand at American universities. A random sample of 28 new accounting professors found the average salary was $135 thousand with a standard deviation of $16 thousand. Assume the distribution is normally distributed. Construct a 90% confidence interval for the salary of new accounting professors. Answers are in thousands of dollars.

A. [129.8497, 140.1503]

B. [130.0260, 139.9740]

C. [107.7520, 162.2480]

D. [131.0268, 138.9732]

Step-by-step explanation:

The confidence interval formula is:

I (1-alpha) (μ) = mean + - [ (t (n-1)) * S/sqrt (n) ]

alpha = is the proportion of the distribution tails that are outside the confidence interval. In this case, 10% because 100-90%

t (n-1) = is the critical value of the t distribution with n-1 degrees of freedom for an area of alpha/2 (5%). In this case is 1.7033

S = sample standard deviation. In this case $16,000

mean = $135,000

n = number of observations

=28

Then, the confidence interval (90%):

I 90% (μ) = 135000 + - [1.7033 * (16,000/sqrt (28))

I 90% (μ) = 135,000 + - [5150.29)

I 90% (μ) = [135,000-5150.29; 135,000-5150.29]

I 90% (μ) = [129,849.71; 140,150.29]