Question

A shareholders' group is lodging a protest against your company. The shareholders group claimed that the mean tenure for a chief exective officer (CEO) was at least 11 years. A survey of 62 companies reported in The Wall Street Journal found a sample mean tenure of 8.7 years for CEOs with a standard deviation of s = 4.7 years (The Wall Street Journal, January 2, 2007). You don't know the population standard deviation but can assume it is normally distributed.

You want to formulate and test a hypothesis that can be used to challenge the validity of the claim made by the group, at a significance level of α = 0.02 . Your hypotheses are: H 0 : μ = 11 H A : μ < 11

What is the test statistic for this sample? t =

(Report answer accurate to 2 decimal places.)

What is the degree of freedom for this test?

d f =

Answer 1

Solution :

= 11

= 8.7

s = 4.7

n = 62

Degrees of freedom (df) = n - 1 = 62 -1 = 61

This is the left tailed test .

The null and alternative hypothesis is

H0 :   = 11

Ha : < 11

Test statistic = t

= ( - ) / s/ n

= (8.7 -11) /4.7 / 62

= -3.85

P (Z < -3.85 ) = 0.0001

P-value = 0.0001

= 0.02

0.0001< 0.02

Reject the null hypothesis .

There is sufficient evidence to suggest that