Question

ANOVA 07003 07003 01138 esidual otal r 9596 81861 01138 (Report answer to 2 decimal places.) m Tables Dollars Dollars Upper 01:44:54 American Express company has long believed that its cardhoiders tend to travel more extensively than others-both on busines and for pleasure. As part of a comprehensive research effort undertaken by a New York marketing research firm on behalf c American Express, a study was conducted to determine the relationship between miles traveled (x) and charges made on th American Express card (y). Eight cardholders were randomly selected and their total charges recorded for a specified period. questionnaire was then mailed to these five cardholders requesting information on the number of miles traveled during this tim period. The data follow. Sick Days & Age iles 1,000's) ($1,000's) harges ardholder 1.2 2.0 1.5 2.1 1.3 3.1 1.3 2.8 3.2 1.1 3.8 1.7 0.5 2.1 0.8 A regression Analysis has been performed to estimate the model and the output is given. Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations 0.82669 68341 63065 62566

Answer 1

X		(x-x̅)²
1.2		0.10
2		0.24
1.5		0.00
2.3		0.62
1.7		0.04
0.5		1.03
2.1		0.35
0.8		0.51

sample size ,   n =   8      
here, x̅ = Σx / n=   1.51
              
SSxx =    Σ(x-x̅)² =    2.8688      

step 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=   0.63                      
                          
Predicted Y at X=   2   is                  
Ŷ =   4.348   +   -1.329   *   2   =   1.689
                          
standard error, S(ŷ)=Se*√(1/n+(X-X̅)²/Sxx) =    0.285                      
margin of error,E=t*Std error=t* S(ŷ) =   2.4469   *   0.2852   =   0.6980      
                          
Confidence Lower Limit=Ŷ +E =    1.689   -   0.6980   =   0.99 thousand dollars
Confidence Upper Limit=Ŷ +E =   1.689   +   0.6980   =   2.39 thousand dollars