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]

Answer: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 1.7033S= sample standard deviation. In this case $16,000mean= $135,000n= number of observations=28Then, 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]