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5 The admissions officer for Clearwater College developed the following estimated regression equation relating the final college GP to the student's SAT Mathematics score and
5 The admissions officer for Clearwater College developed the following estimated regression equation relating the final college GP to the student's SAT Mathematics score and high school GPA ---1.41 +0.023521 +0.004862) where =high-school grade point average D-SAT mathemathics score V-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 Square Adjusted R Square Standard Error Observations ANOVA MS F Sigricance dy SS 1.76209 Regression Residual Total 9 1.88 Star A.value Coefficients -1.4053 Standard Error 0.4848 bateret dr SS MS F Significance 1.76209 Regression Residual Total 9 1.88 Coefficients Standard Error t Stat P-value Intercept -1.4053 0.4848 X1 0.023467 0.0086666 X2 0.00486 0.001077 b. Using a = 0.05, test for overall significance. - Select your answer - c. Did the estimated regression equation provide a good fit to the data? Explain. - Select your answer - V because the R2 value is Select your answer than 0.50. d. Use the t test and a = 0.05 to test H: B = 0 and H : B2 = 0. Use t table. For B1, the p-value is Select your answer SO Selist your answer H: B, = 0. For B2, the p-value is Select your answer - SO Select your answer H: B2 = 0 . # LOGUE 1.76209 Residual Total 9 1.88 X1 Coefficients Standard Error t Stat P-value Intercept -1.4053 0.4848 0.023467 0.0086666 X2 0.00486 0.001077 b. Using a = 0.05, test for overall significance. Select your answer - - Select your answer - There exists significant relationship. provide a good fit to the data? Explain. There exists no significant relationship. value is - Select your answer than 0.50. d. Use the t test and a = 0.05 to test Ho : B1 = 0 and Ho: B2 = 0. Use t table. For B., the p-value is SO Select your answer - Select your answer - H : B1 = 0. For B2, the p-value is Select your answer - - Select your answer - H: B2 = 0. . V V SO MS Significance F 1.76209 Regression Residual Total 9 1.88 Coefficients Standard Error t Stat P-value Intercept -1.4053 0.4848 X1 0.023467 0.0086666 X2 0.00486 0.001077 b. Using a = 0.05, test for overall significance. Select your answer - c. Did the estimated regression equation provide a good fit to the data? Explain. Select your answer - because the Rvalue is Select your answer than 0.50 - Select your answer - 0.05 to test H: B1 O and H :B2 = 0. Use t table. Yes No For Pi, the p-value is Select your answer - , SO - Select your answer - HB = 0. For B2, the p-value is Select your answer SO Select your answer. H: B2 = 0. F = V Residual Total 9 1.88 Coefficients Standard Error I Star P-value Intercept -1.4053 0.4848 X1 0.023467 0.0086666 X2 0.00486 0.001077 b. Using a 0.05, test for overall significance, Select your answer c. Did the estimated regression equation provide a good fit to the data? Explain. Select your answer because the R value is Select your answer than 0.50 Select your answer d. Use the t test and a = 0.05 to test H: Bi = 0 Tower table. greater For B1, the p-value is Select your answer - SOT senect your answer H:8=0. For B2, the p-value is - Select your answer - Select your answer. He: B, -0. V SO 1.88 Coefficients Standard Error Star P-value Intercept -1.4053 0.4848 X1 0.023467 0.0086666 X2 0.00486 0.001077 b. Using a 0.05, test for overall significance. Select your answer c. Did the estimated regression equation provide a good fit to the data? Explain. - Select your answer. because the value is Select your answer than 0.50 d. Use the t test and a = 0.05 to test Ho: Az = 0 and Ho : B; = 0. Use e table. For B1, the p-value is Select your answer Select your answer H: B = 0 -Select your answer For B2, the p-value is less than 0.01 50 - Select your answer v Ho: B20 between 0.01 and 0.02 between 0.02 and 0.05 between 0.05 and 0.10 between 0.10 and 0.20 between 0.20 and 0.40 greater than 0.40 so O-Icon Key Tips ps Coefficients Standard Error Star P.value Intercept -1.4053 0.4848 X1 0.023467 0.0086666 X2 0.00186 0.001077 b. Using a 0.05 test for overall significance. Select your answer c. Did the estimated regression equation provide a good fit to the data? Explain. Select your answer because the R& value is Select your answer than 0.50 d. Use the t test and a = 0.05 to test Hi=0 and HB -0. Use t table. For Bu, the p-value is Select your answer 50 Selock your answer H: B=0 Select your answer For Bs, the p-value is Select your answer SO roject H: B20 do not reject V O-Icon Key Exercise 15.47 (Testing for Significance) Coefficients Standard Error t Stat Intercept P value -1.4053 0.4848 X1 0.023467 0.0086666 X2 0.00486 0.001077 b. Using a 0.05, test for overall significance. Select your answer c. Did the estimated regression equation provide a good fit to the data? Explain. - Select your answer, because the value is Select your answer than 0.50 d. Use the t test and a = 0.05 to test Ho: B -0 and Ho: B20. Un table. For Bu, the p-value is Select your answer SO Select your answer Ho: -0. For B2, the p-value is Select your answer 50 Select your answer Ho : M -0. - Select your answer less than 0.01 between 0.01 and 0.02 between 0.02 and 0.05 between 0.05 and 0.10 between 0.10 and 0.20 between 0.20 and 0.40 0-Icon Key greater than 0.40 v
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