Question

A company claims its employees' mean age is 36 years old. A simple random sample of 40 employees is taken, and is found to have a sample mean of 35.8 years, and sample deviation of 2.2 years.

a. construct and interpret a 98% confidence interval estimate of the true population mean age.

b. do you think the company's claimed mean is reasonable, based on your answer to part A? Why?

c. if we assume population mean is 36, as claimed, and the standard population to be 2.5, what is the probability of a sample of 40 people having a mean age of 35.9 years or less?

Answer 1

A)

Level of Significance ,    α =    0.02          
degree of freedom=   DF=n-1=   39          
't value='   tα/2=   2.4258   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   2.2000   / √   40   =   0.3479
margin of error , E=t*SE =   2.4258   *   0.34785   =   0.8438
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    35.80   -   0.843830   =   34.9562
Interval Upper Limit = x̅ + E =    35.80   -   0.843830   =   36.6438
98%   confidence interval is (   34.96   < µ <   36.64   )

..............

B)

36 is in the interval, so company's claim is reasonable

...............

c)

µ =    36                                  
σ =    2.5                                  
n=   40                                  
                                      
X =   35.9                                  
                                      
Z =   (X - µ )/(σ/√n) = (   35.9   -   36.000   ) / (   2.500   / √   40   ) =   -0.253
                                      
P(X ≤   35.9   ) = P(Z ≤   -0.253   ) =   0.4001                  
.....................

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