HW#2

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Apr 3, 2024

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Farzaneh Hashemabadi HW#2 STAT 652-Spring 2024 UIN: 233005386 Email: Farzaneh.hashemabadi@tamu.edu Problem 1. (1) Write out the regression equation: x 1 = METAL , x 2 = TEMP, x 3 = WATTS , x 1 x 2 = METxTEMP Y = 7.204 + 1.363 x 1 + 0.306 x 2 + 0.01 x 3 – 0.002 x 1 x 2 (2) Is there strong evidence that each independent variable is adding predictive value, given the others? Be 95% confident. H0 = Each independent has no effect on the dependent variable, given the other variables in the model. H1 = Each independent variable is adding predictive value, given the other variables in the model. - P-value to test B 1 = 0.15 > 0.05 → We cannot reject the null, so X 1 could be assumed, with no effect on response value, given the other variables in the model. - P-value to test B 2 = 0.85 > 0.05 → We cannot reject the null, so X 2 could be assumed, with no effect on response value, given the other variables in the model. - P-value to test B 3 = 0.04< 0.05 → We can reject the null, so X 3 variable is adding predictive value, given the other variables in the model. - P-value to test B 4 = 0.9717 > 0.05 → We cannot reject the null, so X 4 could be assumed, with no effect on response value, given the other variables in the model. (3) Calculate and report the 95% confidence interval for the coefficient of metxtemp.

Farzaneh Hashemabadi HW#2 STAT 652-Spring 2024 UIN: 233005386 Email: Farzaneh.hashemabadi@tamu.edu df = n- (K+1) = 25- (4+1) = 20 t α /2,df = t 0.025,20 = 2.086 S Bj = 0.077 CI 95% = [ – 0.00277- 2.086*0.077, – 0.00277+2.086*0.077] = [-0.1638, 0.1583] (4) What does the VIF column of the output indicate about collinearity problems? • none of the variables indicate multicollinearity • all variables indicate multicollinearity • two variables indicate multicollinearity (Correct) • three variables indicate multicollinearity • only one variable indicates multicollinearity Problem 2. (1) Please look at the output in Exercise 12.27 (page 692-693, 7th edition) your textbook. a) How much percent of the variation in ratings is accounted for by the three predictors? Re: R-squared = 0.9795. This indicates that approximately 97.95% of the variation in the rating scores can be explained by the three predictors combined: age, monthly family income, and debt payments as a fraction of income. b) Does the F-test clearly show that the three independent variables have predictive value for the rating score? Be 95% confident. H0: B 0 =…= B k = 0 Ha: B 0 =…= B k ≠ 0 F-test = 7925.829, and P>F =0.0000 << 0.05 So, we can reject the null hypothesis. It means at least one of the independent variables has significant predictive value for the rating score.

Farzaneh Hashemabadi HW#2 STAT 652-Spring 2024 UIN: 233005386 Email: Farzaneh.hashemabadi@tamu.edu - For age: F = 0.2344, and p-value = 0.6285 > 0.05, No significant predictive value given the other predictors. - For monthly income: F = 4112.023, and p-value = 0.000 < 0.05, has significant predictive value given the other predictors. - For debt: F = 2044.05, and p-value = 0.000 < 0.05, has significant predictive value given the other predictors. (2) Look at the additional output in Exercise 12.28 (page 693, 7th edition) your textbook. a) By how much percent has the R square been reduced by eliminating age and debt as predictors? Percent Reduction = Rsquare (full model)− Rsquare (reduced model) R−square (full model) ×100% = 0.979− 0.895 0.979 ×100% = 8.58% b) You want to answer if age and debt add statistically significant predictive value once income is given. Carry out the test. What is the value of the test statistics and p-value? Do age and debt add statistically significant predictive value once income is given? Be 95% confident. H 0 : B age = B debt = 0 H a : B age = B debt ≠ 0 MS drop = 𝑆𝑆 (??𝑔????𝑖??,????????)−𝑆𝑆(??𝑔????𝑖??,???????) ?−𝑔 = 97348.339−(99379.032∗ 0.895261) 3−1 = 4189.06 F = MSdrop 𝑀𝑆𝐸 (????????) = 4189.06/4.1 = 1021.72 → p-value ≈ 0.000 F = 4189.06 > F 0.05, 2, 496 = 3 → we can reject the null Therefore, age and debt add significant predictive value once income is given. Problem 3 (1) Please read Exercise 12.30 (page 695-696, 7th edition) your textbook and use the data and related output to answer the following questions. a) Provide an estimate for the mean systolic blood pressure for an infant of age 8 days weighing 5 kg.

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Farzaneh Hashemabadi HW#2 STAT 652-Spring 2024 UIN: 233005386 Email: Farzaneh.hashemabadi@tamu.edu b) Provide a 95% confidence interval for the mean systolic blood pressure for an infant of age 8 days weighing 5 kg. CI 95% = [42.43, 89.27] (2) Please read Exercise 12.46 (page 700-701,7th edition) your textbook and use the given output to answer the following questions. a) Which independent variable is the best predictor of loan volume? The best predictor of loan volume is the one that hast the highest correlation value with loan volume. Given the table in the text book, the number of accounts correlation value is 0.9403 which is the highest one among the independent values. _ none _ Transactions _ Deposit volume _ Number of accounts (correct) _ Employees b) Is there a substantial collinearity problem? Re: we can see that there are high correlations between several pairs of independent variables. One of those pairs is "Deposit volume (millions)" and "Number of accounts" have a correlation of 0.975. Such high correlations indicate a substantial collinearity problem. d) Use the additional output in Exercise 12.47 (page 701, 7th edition) your textbook. Use the R square value to compute an overall F statistic. i. What is the value of F statistics and p value?

Farzaneh Hashemabadi HW#2 STAT 652-Spring 2024 UIN: 233005386 Email: Farzaneh.hashemabadi@tamu.edu F = 0.89/4 1−0.89/(43−(4+1)) = 76.98 > 3.83 The corresponding p-value is extremely small, essentially 0. ii. Is there clear evidence that there is predictive value in the model? Be 99% confident. Since the F > F 0.01, 4, 38 = 3.83, there is significant evidence that there is predictive value in the model. iii. Which individual predictors have been shown to have unique, predictive value? Be 99% confident. _ Number of accounts: P-value = 0.0569 > 0.01 (not unique) _ Deposit volume: P-value = 0.0999 > 0.01 (not unique) _ none (Correct) _ Transactions: P-value = 0.5954 >> 0.01 (not unique) _ Employees: P-value = 0.3648 >> 0.01 (not unique) Therefore, based on the provided information and a 99% confidence level (p-value = 0.01), none of the individual predictors have shown unique predictive value. e) Do you have a contradiction in answers between the two previous two parts? If there is a contradiction, then explain the reason. Re: The regression analysis provided reveals a significant F-test, indicating that the overall model is predictive of loan volume, yet none of the individual predictors show significance at a 99% confidence level. This apparent contradiction can be explained by the presence of multicollinearity among the predictors, which implies the apparent unique contribution of each variable when considered separately. The significant overall model suggests that the predictors collectively explain the dependent variable's variance, but their interconnected relationships make it challenging to isolate the impact of each one.