Question

An existing inventory for a test measuring self-esteem indicates that the scores have a standard deviation of 8. A psychologist gave the self-esteem test to a random sample of 100 individuals, and their mean score was 60. Construct a 99% confidence interval for the true mean of all test scores. Then complete the table below Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. What is the lower limit of the 99% confidence interval? What is the upper limit of the 99% confidence interval? 0 0 E X $ ?

Answer 1

population std dev ,    σ =    8.0000
Sample Size ,   n =    100
Sample Mean,    x̅ =   60.0000

Level of Significance ,    α =    0.01          
'   '   '          
z value=   z α/2=   2.576   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   8/√100=   0.8000          
margin of error, E=Z*SE =   2.5758   *   0.8000   =   2.061
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    60.00   -   2.0607   =   57.9393 ~ 57.9
Interval Upper Limit = x̅ + E =    60.00   -   2.0607   =   62.0607 ~ 62.1

99%   confidence interval is (   57.9   < µ <   62.1   )

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