Answer:
Step-by-step explanation:
a) We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean
For the null hypothesis,
H0:µ ≥ 9
For the alternative hypothesis,
Ha:µ < 9
This is a left tailed test.
b) Since the population standard deviation is not given, the distribution is a student's t.
Since n = 85,
Degrees of freedom, df = n - 1 = 85 - 1 = 84
t = (x - µ)/(s/√n)
Where
x = sample mean = 7.67
µ = population mean = 9
s = samples standard deviation = 6.38
t = (7.67 - 9)/(6.38/√85) = - 1.92
We would determine the p value using the t test calculator. It becomes
p = 0.029
c) α = 0.01
Since α, 0.01 < the p value, 0.029, then we would fail to reject the null hypothesis. Therefore, At a 1% level of significance, there was no significant evidence that the the mean tenure for a CEO is shorter than nine years.