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he admissions officer for Clearwater College developed the following estimated regression equation relating the final college GPA to the student's SAT mathematics score and high-school GPA. ŷ = -1.41 +0.0235₁ +0.00486x₂ where ₁=high-school grade point average 2 SAT mathemathics score y = final college grade point average Cound test statistic values to 2 decimal places and all other values to 4 decimal places. Do not round your intermediate calculations. . Complete the missing entries in this Excel Regression tool output. Enter negative values as negative numbers. SUMMARY OUTPUT Multiple R Regression Statistics 0.9681

he admissions officer for Clearwater College developed the following estimated regression equation relating the final college GPA to the student's SAT mathematics score and high-school GPA. ŷ = -1.41 +0.0235₁ +0.00486x₂ where ₁=high-school grade point average 2 SAT mathemathics score y = final college grade point average Cound test statistic values to 2 decimal places and all other values to 4 decimal places. Do not round your intermediate calculations. . Complete the missing entries in this Excel Regression tool output. Enter negative values as negative numbers. SUMMARY OUTPUT Multiple R Regression Statistics 0.9681

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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The admissions officer for Clearwater College developed the following estimated regression equation relating the final college GPA to the student's SAT mathematics score and high-school GPA. ŷ = -1.41 +0.0235x1 +0.00486x2 where *1 = high-school grade point average *2 = SAT mathemathics score y = final college grade point average Round test statistic values to 2 decimal places and all other values to 4 decimal places. Do not round your intermediate calculations. a. Complete the missing entries in this Excel Regression tool output. Enter negative values as negative numbers. SUMMARY OUTPUT Multiple R R Square Adjusted R Square Standard Error Observations ANOVA Regression Statistics Regression Residual Total Intercept 0.9681 0.9373 0.9194 0.1296 df 10 2 7 9 Coefficients -1.4053 X1 0.023467 X2 0.00486 b. Using a = 0.05, test for overall significance. There exists significant relationship. SS 1.76209 0.1179 1.88 Standard Error 0.4848 0.0086666 0.001077 MS 0.8810 0.0168 t Stat -2.8987 2.7076 4.5125 F 52.4405 P-value c. Did the estimated regression equation provide a good fit to the data? Explain. Yes ✓, because the R value is greater ✓than 0.50. d. Use the t test and a = 0.05 to test Ho: ₁ = 0 and Ho: B₂ = 0. Use t table. For B₁, the p-value is between 0.02 and 0.05 ✓ so reject For B₂, the p-value is less than 0.01 , so reject 0.0230 0.0303 0.0028 ✓ Ho: B₁ = 0. ✓ Ho: B₂ = 0. Significance F 0.000061
Transcribed Image Text:The admissions officer for Clearwater College developed the following estimated regression equation relating the final college GPA to the student's SAT mathematics score and high-school GPA. ŷ = -1.41 +0.0235x1 +0.00486x2 where *1 = high-school grade point average *2 = SAT mathemathics score y = final college grade point average Round test statistic values to 2 decimal places and all other values to 4 decimal places. Do not round your intermediate calculations. a. Complete the missing entries in this Excel Regression tool output. Enter negative values as negative numbers. SUMMARY OUTPUT Multiple R R Square Adjusted R Square Standard Error Observations ANOVA Regression Statistics Regression Residual Total Intercept 0.9681 0.9373 0.9194 0.1296 df 10 2 7 9 Coefficients -1.4053 X1 0.023467 X2 0.00486 b. Using a = 0.05, test for overall significance. There exists significant relationship. SS 1.76209 0.1179 1.88 Standard Error 0.4848 0.0086666 0.001077 MS 0.8810 0.0168 t Stat -2.8987 2.7076 4.5125 F 52.4405 P-value c. Did the estimated regression equation provide a good fit to the data? Explain. Yes ✓, because the R value is greater ✓than 0.50. d. Use the t test and a = 0.05 to test Ho: ₁ = 0 and Ho: B₂ = 0. Use t table. For B₁, the p-value is between 0.02 and 0.05 ✓ so reject For B₂, the p-value is less than 0.01 , so reject 0.0230 0.0303 0.0028 ✓ Ho: B₁ = 0. ✓ Ho: B₂ = 0. Significance F 0.000061
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