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 11 years. A survey of 69 companies reported in The Wall Street Journal found a sample mean tenure of 8.4 years for CEOs with a standard deviation of s=s= 5.6 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=0.002α=0.002. Your hypotheses are:

      Ho:μ=11Ho:μ=11
      Ha:μ<11Ha:μ<11

What is the test statistic for this sample?
test statistic =  (Report answer accurate to 3 decimal places.)

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

The p-value is...

• less than (or equal to) αα
• greater than α

This test statistic leads to a decision to...

• reject the null
• accept the null
• fail to reject the null

As such, the final conclusion is that...

• There is sufficient evidence to warrant rejection of the claim that the population mean is less than 11.
• There is not sufficient evidence to warrant rejection of the claim that the population mean is less than 11.
• The sample data support the claim that the population mean is less than 11.
• There is not sufficient sample evidence to support the claim that the population mean is less than 11.

Answer 1

Given that the shareholder's group claimed that the mean tenure for a chief exective office (CEO) was at least 11 years.

To test the claim a survey of n = 69 companies reported in The Wall Street Journal found a sample mean tenure of = 8.4 years for CEOs with a standard deviation of s= 5.6 years.

Since the population standard deviation is not known but the sample size is greater than 30 hence we can assume it as normal distribution and a t-distribution is applicable for hypothesis testing.

Based on the claim the hypotheses are:

Ho: μ=11

Ha: μ<11

Based on the hypothesis it will be a left tailed test.

Test statistic:

Rejection region:

Based on the type of hypothesis and given significance level the critical score for rejection region is calculated using the excel formula for t-distribution which takes the significance level and the degree of freedom df = n-1 as parameters.

Thus the formula used is =NORM.S.DIST(0.02, 68) the tc computed as −2.094. So, reject Ho if t < -2.094

P-value:

The P-value is computed using the excel formula for t-distribution which is =T.DIST(-3.857, 68, TRUE), thus the P-value computed as 0.0001

The p-value is less than (or equal to) α

Decision:

Since the test statistic is less than tc hence this test statistic leads to a decision to reject the null hypothesis.

Conclusion:

Since we are able to reject the null hypothesis hence at 0.02 level of significance we conclude that there is sufficient evidence to warrant rejection of the claim that the population mean is at least 11.

Note: Feel free to ask if any query remains.