Question

Suppose the sample mean forearm lengths for a randomly selected group of eleven men turns out to be 25. 5 cm with a corresponding standard deviation of 1. 52 cm. Find the 99% confidence interval for the true mean forearm 2 length of all men.

2) Refer to the last problem. Consider repeating the task of finding similar confidence intervals 4000 times (each of these 4000 intervals has the same confidence level of 99% and each interval is based on the same sample size eleven). Approximately how many of these intervals do you expect to include the true forearm length of all men? (Your answer must be given as an integer. Also note that there is no partial credit given for this problem).

Answer 1

a)

sample std dev ,    s =    1.5200
Sample Size ,   n =    11
Sample Mean,    x̅ =   25.5000
Level of Significance ,    α =    0.01          
degree of freedom=   DF=n-1=   10          
't value='   tα/2=   3.1693   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   1.5200   / √   11   =   0.458297
margin of error , E=t*SE =   3.1693   *   0.45830   =   1.452469
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    25.50   -   1.452469   =   24.047531
Interval Upper Limit = x̅ + E =    25.50   -   1.452469   =   26.952469
99%   confidence interval is (   24.05   < µ <   26.95   )

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from 4000 repeatations there are (0.99*4000) 3960    intervals are expected to include the true forearm length of all men

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Please revert back in case of any doubt.

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