The Humber Student Federation conducted a survey to determine the mean study time/week of students at...

The Humber Student Federation conducted a survey to determine the mean study time/week of students at the Lakeshore Campus. A sample of 18 students had a mean of 7.5 hours and a standard deviation of 2.2 hours. Assuming that the sampling distribution is about normal, calculate a 99% confidence interval for the mean study time all students at the Lakeshore Campus.

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Expert Solution

Level of Significance ,    α =    0.01
sample std dev ,    s =    2.2000
Sample Size ,   n =    18
Sample Mean,    x̅ =   7.5000

Level of Significance ,    α =    0.01
degree of freedom=   DF=n-1=   17
't value='   tα/2=   2.8982   [Excel formula =t.inv(α/2,df) ]

Standard Error , SE = s/√n =   2.200   / √   18   =   0.5185
margin of error , E=t*SE =   2.8982   *   0.519   =   1.503

confidence interval is
Interval Lower Limit = x̅ - E =    7.50   -   1.503   =   5.9971
Interval Upper Limit = x̅ + E =    7.50   -   1.503   =   9.0029
99%   confidence interval is (   6.00   < µ <   9.00   )

milcah answered 9 months ago

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