Question

A market analyst wants to know if the new website he designed is showing increased page views per visit. A customer is randomly sent to one of two different​ websites, offering the same​ products, but with different designs. Assume that the data come from a distribution that is Normally distributed. The data is shown in the table to the right. Complete parts a through c below. Website 1 Website 2 n 80 95 ybar 6.5 6.3 s 4.9 5.4 Calculate the test statistic. Let the difference of the sample means be y1-y2 Calculate the​ P-value. State the conclusion. b) Find a 95​% confidence interval for the mean difference in page views from the two websites using the pooled degrees of freedom. ​c) The confidence interval for the unpooled variance is (-1.34, 1.74). Are your answers different from the interval where the variance is not​ pooled? Explain briefly why or why not.

Answer 1

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   1                  
mean of sample 1,    x̅1=   6.500                  
standard deviation of sample 1,   s1 =    4.9000                  
size of sample 1,    n1=   80                  
                          
Sample #2   ---->   2                  
mean of sample 2,    x̅2=   6.300                  
standard deviation of sample 2,   s2 =    5.4000                  
size of sample 2,    n2=   95                  
                          
difference in sample means =    x̅1-x̅2 =    6.5000   -   6.3   =   0.20  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    5.1777                  
std error , SE =    Sp*√(1/n1+1/n2) =    0.7857                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   0.2000   -   0   ) /    0.79   =   0.2546
                          
Degree of freedom, DF=   n1+n2-2 =    173                  
t-critical value , t* =        1.974   (excel formula =t.inv(α/2,df)              
Decision:   | t-stat | < | critical value |, so, Do not Reject Ho                      
p-value =        0.7994   (excel function: =T.DIST.2T(t stat,df) )              
Conclusion:     p-value>α , Do not reject null hypothesis                      

b)

Degree of freedom, DF=   n1+n2-2 =    173              
t-critical value =    t α/2 =    1.9738   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    5.1777              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    0.7857              
margin of error, E = t*SE =    1.9738   *   0.79   =   1.55  
                      
difference of means =    x̅1-x̅2 =    6.5000   -   6.300   =   0.2000
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    0.2000   -   1.5508   =   -1.3508
Interval Upper Limit=   (x̅1-x̅2) + E =    0.2000   +   1.5508   =   1.7508

c)

No, because both interval contain null hypothesis zero, so, null hypothesis is not rejected