Question

HW 44. 2

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 83 companies reported in The Wall Street Journal found a sample mean tenure of 9 years for CEOs with a standard deviation of s=s= 4.4 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=0.005α=0.005. 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 αα

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

Ho :   µ =   11                  
Ha :   µ <   11       (Left tail test)          
                          
Level of Significance ,    α =    0.01                  
sample std dev ,    s =    4.4000                  
Sample Size ,   n =    83                  
Sample Mean,    x̅ =   9.0000                  
                          
degree of freedom=   DF=n-1=   82                  
                          
Standard Error , SE = s/√n =   4.4000   / √    83   =   0.4830      
t-test statistic= (x̅ - µ )/SE = (   9.000   -   11   ) /    0.48296   =   -4.141
                                
p-Value   =   0.0000   [Excel formula =t.dist(t-stat,df) ]              

p-value<α, Reject null hypothesis   

• The sample data support the claim that the population mean is less than 11.
• ...................

THANKS

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