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.
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.
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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Question
![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:
\[
\hat{y} = -1.41 + 0.0235x_1 + 0.00486x_2
\]
where
\(x_1 =\) high-school grade point average
\(x_2 =\) SAT mathematics score
\(\hat{y} =\) 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: 0.9681
- R Square: 0.9373
- Adjusted R Square: 0.9194
- Standard Error: 0.1296
- Observations: 10
---
**ANOVA**
| | df | SS | MS | F | Significance F |
|--------------|----|--------|--------|---------|----------------|
| Regression | 2 | 1.76209| 0.8810 | 52.31 | 0.0007 |
| Residual | 7 | 0.1179 | 0.0168 | | |
| Total | 9 | 1.88 | | | |
---
**Coefficients**
| Coefficient | Standard Error | t Stat | P-value |
|-------------|----------------|--------|---------|
| Intercept | -1.4053 | 0.4848 | -2.8987 | 0.0268 |
| X1 | 0.023467 | 0.0086666 | 2.7078 | 0.0293 |
| X2 | 0.00486 | 0.001077 | 4.5125 | 0.0031 |
**b. Using \(\alpha = 0.05\), test for overall significance.**
- There exists no significant relationship.
**c. Did the estimated regression equation provide a good fit to the data? Explain.**
- Yes, \(R^2\) is significantly](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F631d52d2-d8ca-4fc4-a52f-45ccb5125001%2F91064c5f-3a01-462e-8ee9-9a42d120431f%2F77tgu3j_processed.png&w=3840&q=75)
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:
\[
\hat{y} = -1.41 + 0.0235x_1 + 0.00486x_2
\]
where
\(x_1 =\) high-school grade point average
\(x_2 =\) SAT mathematics score
\(\hat{y} =\) 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: 0.9681
- R Square: 0.9373
- Adjusted R Square: 0.9194
- Standard Error: 0.1296
- Observations: 10
---
**ANOVA**
| | df | SS | MS | F | Significance F |
|--------------|----|--------|--------|---------|----------------|
| Regression | 2 | 1.76209| 0.8810 | 52.31 | 0.0007 |
| Residual | 7 | 0.1179 | 0.0168 | | |
| Total | 9 | 1.88 | | | |
---
**Coefficients**
| Coefficient | Standard Error | t Stat | P-value |
|-------------|----------------|--------|---------|
| Intercept | -1.4053 | 0.4848 | -2.8987 | 0.0268 |
| X1 | 0.023467 | 0.0086666 | 2.7078 | 0.0293 |
| X2 | 0.00486 | 0.001077 | 4.5125 | 0.0031 |
**b. Using \(\alpha = 0.05\), test for overall significance.**
- There exists no significant relationship.
**c. Did the estimated regression equation provide a good fit to the data? Explain.**
- Yes, \(R^2\) is significantly
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