Question

A sample is selected to find a 90% confidence interval for the average starting salary. Here are the sample statistics: n = 31, x¯ = $43, 780, s = $1, 600. a). Find the t− score used in the calculation of the confidence interval.

b). Build a 90% confidence interval for the mean starting salary.

c). Based on the result of part b), could we make a conclusion that the mean staring salary is below $45, 000? Explain your reason.

Answer 1

a)

t critical value at 0.10 significance level with 30 df = 1.697

b)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
43780 ± t(0.1/2, 31 -1) * 1600/√(31)
Lower Limit = 43780 - 1.697 * 1600/√(31)
Lower Limit = 43292
Upper Limit = 43780 + 1.697 *1600/√(31)
Upper Limit = 44268
90% Confidence interval is ( 43292 , 44268 )

c)

Since 45000 not contained in confidnece interval and all values in confidence interval

are less than 45000,

We have sufficient evidence to support the claim that the mean staring salary is below $45, 000