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.y = -1.41 +0.0235x1 +0.00486x2whereX1 high-school grade point averageX2 = SAT mathemathics scorey = final college grade point averageRound 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 OUTPUTMultiple RR SquareAdjusted R SquareStandard ErrorRegression StatisticsObservationsANOVARegressionResidualTotalInterceptYes0.96810.93730.91940.1296df10279Coefficients-1.40530.023467SS1.762090.004860.11791.88Standard Error0.48480.00866660.001077X1X2b. Using a = 0.05, test for overall significance.There exists no significant relationship.c. Did the estimated regression equation provide a good fit to the data? Explain.because the R value is lowerd. Use the t test and a = 0.05 to test Ho: ₁between 0.05 and 0.10 ✓For B₁, the p-value isFor B2, the p-value isbetween 0.05 and 0.10 ✓MS0.88100.0168t Stat-2.89872.7078, So rejectSO= 0 and Ho: B₂ = 0. Use t table.4.5125do not rejectF0.0007P-value0.02680.002than 0.50.Significance FHo B₁ = 0.:Ho: B₂ = 0.

ANOVA: F = MSReg/MSres = 0.8810/0.0168 =52.4405 p-value at F= 52.4405 and 2,7 df is 0.000061

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 Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations ANOVA Regression Residual Total 0.9681 0.9373 0.9194 Intercept 0.1296 df 10 2 7 9 ✔ SS 1.76209 0.1179 1.88 Coefficients -1.4053 X1 0.023467 X2 0.00486 b. Using a = 0.05, test for overall significance. There exists no significant relationship. c. Did the estimated regression equation provide a good fit to the data? Explain. Yes because the R2 value is lower d. Use the t test and a = 0.05 to test Ho: B1 = 0 and For B₁, the p-value is between 0.05 and 0.10 For B2, the p-value is between 0.05 and 0.10 Standard Error 0.4848 0.0086666 0.001077 MS 0.8810 0.0168 So reject t Stat -2.8987 2.7078 4.5125 Ho: B2 = 0. Use t table. ✔ ✔ , so do not reject F 0.0007 P-value 0.0268 0.002 than 0.50. Ho X Ho Significance F B₁ = 0. B₂ = 0.

1st Edition

ISBN:9781938168383

Author:Jay Abramson

Publisher:Jay Abramson

Chapter4: Linear Functions

Section: Chapter Questions

Problem 40RE: For the following exercises, consider the data in Table 5, which shows the percent of unemployed ina...