Question

eBook 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=-1.4053+.023541 +.004922 where 21 = high-school grade point average 22 = SAT mathematics score y-final college grade point average Round your answers to 4 decimal places. a. Complete the missing entries in this Excel Regression tool output. Enter negative values as negative numbers. ANOVA df F I Significance F SS LMS 1.7621 Regression 1 A1 Residual Total 9 1 .88 Error Stat P-value Intercept Coefficients -1.4053 0.0235 0.0049 Standard Error 0.4848 0.0087 0.0011 X2 b. Using a = .05, test for overall significance. There exists significant relationship. c. Did the estimated regression equation provide a good fit to the data? Explain. Yes d. Use the t test and a = .05 to test Ho: B10 and Ho: Ba 0. Use t table. The input in the box below will not be graded, but may be reviewed and considered by your instruction. blank

Answer 1

a) The table is filled as follows (with formulas given):

	df	SS	MS	F	Significance F
Regression	p = 2	1.7621	1.7621/2 = 0.8810	0.88105/0.0168 = 52.4434	6.12516E-05
Residual	n - p - 1 = 10 - 2 - 1 = 7	1.88 - 1.7621 = 0.1179	0.1179/7 = 0.0168
Total	9 = n - 1	1.88

	Coefficients	SE	t-stat	p-value
Intercept	-1.4053	0.4848	-2.8987	0.023
x1	0.0235	0.0087	2.7011	0.03059
x2	0.0049	0.0011	4.4545	0.003

b) As we can see the F-statistic, it is significant with a p-value < 0.01. Hence, we can say that there exists significant relationship.

c) R-square = SSR/SST = 1.7621/1.88 = 0.9372
93.72% variatino can be explained by the regression equation. Yes, it is a good fit to the data.

d) We will use the t-statistics in the table to test the hypotheses.

t = Coefficient/Standard Error

Ho: B1 = 0

t = 2.7011

p-value = 0.03059

As the p-value < 0.05, we can reject the null hypothesis.

Ho: B2 = 0

t = 4.4545

p-value = 0.003

As the p-value < 0.05, we can reject the null hypothesis.