Question

You are interested in finding a 90% confidence interval for the average number of days of class that college students miss each year. The data below show the number of missed days for 14 randomly selected college students. Round answers to 3 decimal places where possible. 1 8 10 9 1 3 1 8 1 0 2 1 5 8 a. To compute the confidence interval use a distribution. b. With 90% confidence the population mean number of days of class that college students miss is between and days. c. If many groups of 14 randomly selected non-residential college students are surveyed, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population mean number of missed class days and about percent will not contain the true population mean number of missed class days.

Answer 1

sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   3.6765
Sample Size ,   n =    14
Sample Mean,    x̅ = ΣX/n =    4.1429

a)

to compute the confidence interval use a t- distribution

because sample size is small and also std dev of population is unknown

b)

Level of Significance ,    α =    0.1          
degree of freedom=   DF=n-1=   13          
't value='   tα/2=   1.7709   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   3.676   / √   14   =   0.9826
margin of error , E=t*SE =   1.7709   *   0.983   =   1.740
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    4.14   -   1.740   =   2.4028
Interval Upper Limit = x̅ + E =    4.14   -   1.740   =   5.8829
90%   confidence interval is (   2.403 < µ <   5.883 )

c)

About 90 percent of these confidence intervals will contain the true population mean number of missed class days and about 10 percent will not contain the true population mean number of missed class days