Question

A shareholder's group is lodging a protest against your company. The shareholder's group claimed that the mean tenure for the chief executive office (CEO) was at least 10 years. A survey of 21 companies reported in the Wall Street Journal found a sample mean tenure of 9 years for CEOs with a standard deviation of s=5.3 years. Test whether this is significant evidence that the mean tenure is less. (Please use calculator not table)

Answer 1

Solution :

Given that,

Population mean = = 10

Sample mean = = 9

Sample standard deviation = s = 5.3

Sample size = n = 21

Level of significance = = 0.05

This is a left (One) tailed test,

The null and alternative hypothesis is,  

Ho: 10

Ha: 10

The test statistics,

t = ( - )/ (s/)

= ( 9 - 10 ) / ( 5.3 /21)

= -0.865

Critical value of  the significance level is α = 0.05, and the critical value for a left-tailed test is

= -1.725

Since it is observed that t = -0.865 = -1.725, it is then concluded that the null hypothesis is fails to reject.

P- Value = 0.1987

The p-value is p = 0.1987 0.05, it is concluded that the null hypothesis is fails to reject.

Conclusion :

It is concluded that the null hypothesis Ho is fails to reject. Therefore, there is not significant to claim that the population

mean μ is less than 10, at the 0.05 significance level.