Question

A shareholders' group is lodging a protest against your company. The shareholders group claimed that the mean tenure for a chief exective office (CEO) was at least 9 years. A survey of 91 companies reported in The Wall Street Journal found a sample mean tenure of 7.7 years for CEOs with a standard deviation of 4.5 years (The Wall Street Journal, January 2, 2007).

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 0.001. Your hypotheses are:

      H0:μ≥9H0:μ≥9
      H1:μ<9H1:μ<9

What is the p-value for this sample?
(Report answer accurate to 4 decimal places.)

Answer 1

Solution :

Given that,

Population mean = = 9

Sample mean = = 7.7

Sample standard deviation = s = 4.5

Sample size = n = 91

Level of significance = = 0.001

This is a left (One) tailed test,

The null and alternative hypothesis is,  

Ho: 9

Ha: 9

The test statistics,

t = ( - )/ (s/)

= ( 7.7 - 9 ) / ( 4.5 /91 )

= -2.756

P- Value = 0.0035

  

The p-value is p = 0.0035 > 0.001, it is concluded that the null hypothesis is fail to reject.