Question

1. The number of vehicles on a highway link is counted on each of 40 randomly chosen days. The mean number of vehicles is found to be 135, and the standard deviation is 90.

(a) Find a 90% confidence interval for the sample mean.

(b) Find a 90% confidence interval for the mean, assuming it had been based upon a sample of 15 days, instead of a sample 40 days.

Thank you very much!

Answer 1

sample std dev ,    s =    90.0
Sample Size ,   n =    40.0
Sample Mean,    x̅ =   135.0
Level of Significance ,    α =    0.1          
degree of freedom=   DF=n-1=   39          
't value='   tα/2=   1.6849   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   90.0000   / √   40   =   14.230249
margin of error , E=t*SE =   1.6849   *   14.23025   =   23.976193
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    135.00   -   23.976193   =   111.023807
Interval Upper Limit = x̅ + E =    135.00   -   23.976193   =   158.976193
90%   confidence interval is (   111.024   < µ <   158.976   )

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

sample std dev ,    s =    90.0
Sample Size ,   n =    15.0
Sample Mean,    x̅ =   135.0

Level of Significance ,    α =    0.1          
degree of freedom=   DF=n-1=   14          
't value='   tα/2=   1.7613   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   90.0000   / √   15   =   23.237900
margin of error , E=t*SE =   1.7613   *   23.23790   =   40.929149
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    135.00   -   40.929149   =   94.070851
Interval Upper Limit = x̅ + E =    135.00   -   40.929149   =   175.929149
90%   confidence interval is (   94.071   < µ <   175.929   )

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

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