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.
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter4: Equations Of Linear Functions
Section4.6: Regression And Median-fit Lines
Problem 4GP
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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.
y = -1.41 +0.0235x1 +0.00486x2
where
X1 high-school grade point average
X2 = 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
Regression Statistics
Observations
ANOVA
Regression
Residual
Total
Intercept
Yes
0.9681
0.9373
0.9194
0.1296
df
10
2
7
9
Coefficients
-1.4053
0.023467
SS
1.76209
0.00486
0.1179
1.88
Standard Error
0.4848
0.0086666
0.001077
X1
X2
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.
because the R value is lower
d. Use the t test and a = 0.05 to test Ho: ₁
between 0.05 and 0.10 ✓
For B₁, the p-value is
For B2, the p-value is
between 0.05 and 0.10 ✓
MS
0.8810
0.0168
t Stat
-2.8987
2.7078
, So reject
SO
= 0 and Ho: B₂ = 0. Use t table.
4.5125
do not reject
F
0.0007
P-value
0.0268
0.002
than 0.50.
Significance F
Ho B₁ = 0.
:
Ho: B₂ = 0.
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