Question

A publisher is interested in the effects on sales of college texts that include more than 100 data files. The publisher plans to produce 20 texts in the business area and randomly choses 10 to have more than 100 data files. The remaining 10 are produced with at most 100 data files. For those with more than 100, first-year sales averaged 9254, and the sample standard deviation was 2107. For the books with at most 100, average first-year sales were 8167, and the standard deviation was 1681.

a. Assuming that the two population distibutions are normal with the same variance, at the 5% significance level, test the null hypothesis that the population means are equal against the alternative that the true mean is higher for books with more than 100 data files.

b. Assuming that the two population distibutions are normal with the same variance, find a 95% confidence interval for the difference between two population means.

Answer 1

a)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 >   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   sample 1                  
mean of sample 1,    x̅1=   9254.00                  
standard deviation of sample 1,   s1 =    2107.00                  
size of sample 1,    n1=   10                  
                          
Sample #2   ---->   sample 2                  
mean of sample 2,    x̅2=   8167.00                  
standard deviation of sample 2,   s2 =    1681.00                  
size of sample 2,    n2=   10                  
                          
difference in sample means =    x̅1-x̅2 =    9254.0000   -   8167.0   =   1087.00  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    1905.9394                  
std error , SE =    Sp*√(1/n1+1/n2) =    852.3620                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   1087.0000   -   0   ) /    852.36   =   1.2753
                          
Degree of freedom, DF=   n1+n2-2 =    18                  
  
p-value =        0.109212   [excel function: =T.DIST.RT(t stat,df) ]              
Conclusion:     p-value>α , Do not reject null hypothesis              

          there is not enough evidence the true mean is higher for books with more than 100 data files.

..........

b)

Degree of freedom, DF=   n1+n2-2 =    18              
t-critical value =    t α/2 =    2.1009   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    1905.9394              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    852.3620              
margin of error, E = t*SE =    2.1009   *   852.3620   =   1790.7461  
                      
difference of means =    x̅1-x̅2 =    9254.0000   -   8167.000   =   1087.0000
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    1087.0000   -   1790.7461   =   -703.7461
Interval Upper Limit=   (x̅1-x̅2) + E =    1087.0000   +   1790.7461   =   2877.7461

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