Question

(1 point) A class survey in a large class for first-year college students asked, "About how many minutes do you study on a typical weeknight?" The mean response of the 253 students was x⎯⎯⎯x¯ = 134 minutes. Suppose that we know that the study time follows a Normal distribution with standard deviation σσ = 65 minutes in the population of all first-year students at this university.

Use the survey result to give a 90% confidence interval for the mean study time of all first-year students.

Answer 1

population std dev ,    σ =    65.0000
Sample Size ,   n =    253
Sample Mean,    x̅ =   134.0000

Level of Significance ,    α =    0.1          
'   '   '          
z value=   z α/2=   1.6449   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   65.0000   / √   253   =   4.086515
margin of error, E=Z*SE =   1.6449   *   4.08651   =   6.721719
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    134.00   -   6.721719   =   127.278281
Interval Upper Limit = x̅ + E =    134.00   -   6.721719   =   140.721719
90%   confidence interval is (   127.28   < µ <   140.72   )

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

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