Could you please help me solve Problem J - please see the attached files. Please
ID: 3306954 • Letter: C
Question
Could you please help me solve Problem J - please see the attached files. Please help to do it step-by-step with all the workings and solutions. Thank you very much!
Problem J Consumer Research is an independent agency that is collecting data on annual income (INCOME) and household size (SIZE), to predict annual credit card charges. It runs a regression analysis on the data and an incomplete Excel output is shown below. Regression Statistic:s Multiple R R Square Adjusted R Square Standard Error Observations 510.49549.3 ANOVA MS 179 Significance E-O Regression Resi 20 60605.648 Coefficients t Stat Lower 95% tandar Error 52.694714 4.15578994 0.00048872730.01720 P-value Intercept INCOME SIZE 25.0629568.471472852.958512230.00776734 7.391781505 776 7 27. The sample size is: 28. The coefficient of determination is 29. The Sum of Squares for Error (i.e., Residual) is, 30. The Sum of Squares for Total (SST is 31. The Mean Square for Regression is 32.The observed or computed F-value is: 33. The hypothesis to be tested at the 5% level of significance is: Ho: Ha: B1 = B2 = 0 At least one of the B is not equal to 0 Is the null hypothesis rejected or not rejected; justify your answer? 34. The hypothesis to be tested at the 1% level of significance is Ho: Bi=0 35. The est : 36. The observed or computed t-stat (i.e., t-value) for the independent variable SIZE is 37. What is the estinated annual credit charges if INCOME-20, and SIZE = 3? imate of the intercept bo isExplanation / Answer
27. The total number of observations can be obtained from the "degrees of freedom(df)" column. The total degrees of freedom is n-1 where n is the total number of observations in the sample. Thus, n-1=22 which gives n=23.Thus, the sample size is n=23
28. The coefficient of determination is nothing but R2. This can be obtained by SSregression/SStotal.
From the table SSregression=17960368.3
SStotal can be obtained as SStotal = SSregression + SSresidual
The mean sum of squares of residual is 260605.648 and the degrees of freedom is 20.
So, SSresidual =20 x 260605.648 = 5212112.96
Thus, SStotal = 17960368.3 + 5212112.96 = 23172481.26
Thus, the coefficient of determination is given by R2=17960368.3 / 23172481.26 = 0.77507
29. The mean sum of squares is given by Sum of squares of errors/Degrees of freedom
From the above table, the mean sum of squares of error is given by MSresidual =260605.648 and the degrees of freedom of residuals is 20. So, the sum of squares of errors(residual) is 20 x 260605.648 = 5212112.96
30. The sum of Sum of squares for total is given by SStotal = SSregression + SSresidual
From the table, SSregression =17960368.3 and from part (29) SSresidual = 5212112.96
So, SStotal = 17960368.3+521212.96.3 = 23173481.26
31. The mean sum of squares of regression is given by Sum of Squares of regression/Degrees of freedom of regression, that is 17960368.3 / 2 = 8980184.15. Thus, the mean sum of squares of regression is 8980184.15
32. The F-Value for the given regression analysis is given by Mean sum of squares of Regression / Mean sum of squares of residuals From part (31.) MSregression = 8980184.15 and from the table MSresidual=260605.648
Thus, F = 8980184.15 / 260605.648 = 34.4589
Thus, the computed F-value is 34.4589
33.Since, the value of our F-statistic as computed in part (32.) is greater than the F-significant (given in the table) So, we have enough evidence against our null hypothesis and the test is significant. therefore,the null hypothesis will be rejected at 5% level of significance.
34.From the table we see that the T-statistic for the coeffcient for the predictor variable income (B1) is 2.9585 as compared t0 the t-critical at 21 degrees of freedom 2.0796, thus, this value lies in the rejection region and the p-value is 0.007767 which is less than our level of significance 0.01 So, we conclude that the the test is significant and hence we reject our null hypothesis.
35. The estimate of the intercept parameter B0 can be obtained by multiplying the value of the t-statistic with the standard error. This value is 1465.725.
36. The t-statistic for the independent variable size is given by estimated coefficient/Standard error, that gives 408.400776/71808401 = 5.68734. Thus, the value of the t-statistic for the independent variable size is 5.68734
37. The regression equation from the above calculations can be stated as
Annual Credit Charges = 1465.725 + 25.062956INCOME + 408.400776SIZE
When INCOME = 20 and SIZE = 3, the estimate value of the annula credit charge is given by:
Annual Credit Charges = 1465.725 + 25.062956 x 20 + 408.400776 x 3
Annual Credit Charges = 1465.725 + 501.25912 + 1225.202328
Annual Credit Charges = 3192.186448