Question

and for pleasure. As part of a comprehensive research effort undertaken by a New York marketing research firm on behalf of American Express, a study was conducted to determine the relationship between miles traveled (x) and charges made on the American Express card (y). Eight cardholders were randomly selected and their total charges recorded for a specified period. A questionnaire was then mailed to these five cardholders requesting information on the number of miles traveled during this time 2.0 1.5 0.5 2.1 0.8 3.2 74115 ANOVA df SS MS ression idual 37232 6 0.67644 7 3.04876 37232 11274 1.04240 74 Standard Error Stat lue r 95% ntercept 3.76292 49 0.00002 587220.00374139445 0.19824 Step 4 of 4: (Report answer to 2 decimal places.) the 95% confidence interval for the average credit card charges, given that the cardholder traveled 2 thousand miles. m Tables Keypad Answer 1 Point Dollars

Answer 1

4)

X Value=   2                      
Confidence Level=   95%                      
                          
                          
Sample Size , n=   8                      
Degrees of Freedom,df=n-2 =   6                      
critical t Value=tα/2 =   2.447   [excel function: =t.inv.2t(α/2,df) ]                  
                          
X̅ =    1.51                      
Σ(x-x̅)² =Sxx   2.9                      
Standard Error of the Estimate,Se=   0.34                      
                          
Predicted Y at X=   2   is                  
Ŷ =   3.763   +   -0.909   *   2   =   1.944

For Individual Response Y                  
standard error, S(ŷ)=Se*√(1+1/n+(X-X̅)²/Sxx) =   0.3690              
margin of error,E=t*std error=t*S(ŷ)=    2.4469   *   0.37   =   0.9029
                  
Prediction Interval Lower Limit=Ŷ -E =   1.944   -   0.9029   =   1.04
Prediction Interval Upper Limit=Ŷ +E =   1.944   +   0.9029   =   2.85