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 10 years. A survey of 71 companies reported in The Wall Street Journal found a sample mean tenure of 8.9 years for CEOs with a standard deviation of 5.9 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 0.10. Your hypotheses are:

H 0 : μ ≥ 10

H 1 : μ < 10

What is the test statistic for this sample? -1.571 Correct (Report answer accurate to 3 decimal places.)

What is the p-value for this sample?

*Can you explain how to find P-value? I don't understand when its just written as t_0.10;70 . My notes don't go in depth for how to go with that method or any of the matter.

I really appreciate it, thanks*

Answer 1

Here we have х 38.9, s %35.9, п %371 Hypotheses are Ho u10 На:и «10 Level of significance: a 0.1 Test is one tailed (left tailed) df n-1-70 Degree of freedom: Test Statistics: =(8.9-10)/(5.9/sqrt(71)) s/n = -1.571 Critical value: CV = -1.294 Rejection Region: Ift<-1.294, Reject HO Decision: Since test statistics lies in rejection region so we reject the null hypothesis P-value 0.0603 P-value: Decision: Since p-value is less than level of significance so we reject the null hypothesis Excel function for critical value: =TINV(0.2,70) =TDIST ( 1.571,70,1) Excel function for p-value:

The p-value is:

p-value = 0.0603

Since p-value is less than 0.10 so we reject the null hypothesis.