Question

A shareholders’ group, in lodging a protest, claimed that the mean tenure for a chief executive office (CEO) was at least nine years. A survey of 85 companies reported in The Wall Street Journal found a sample mean tenure of 7.27 years for CEOs with a standard deviation of 6.38 years (The Wall Street Journal, January 2, 2007).

Compute the value of the test statistic. (Round to two decimal places)

What is the p-value? (Round to three decimal places).

At α=0.01, what is your conclusion?

Answer 1

Solution :

Given that,

Population mean = = 9

Sample mean = = 7.27

Sample standard deviation = s = 6.38

Sample size = n = 85

Level of significance = = 0.01

This is a left (One) tailed test,

The null and alternative hypothesis is,  

Ho: 9

Ha: 9

The test statistics,

t = ( - )/ (s/)

= ( 7.27 - 9) / ( 6.38 / 85 )

= -2.50

P- Value = 0.007

The p-value is p = 0.007 < 0.01, it is concluded that the null hypothesis is rejected.

Conclusion :

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the mean tenure for a chief executive office (CEO) was at least nine years, at the 0.01 significance level.