Question

mework 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. 9--1.4053 +.02352, +.004973 where z = 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 Significance F Regression 1.7621 Residual Total 1.8 P-value Intercept Coefficients -1.4053 0.0235 0.0049 Standard Error 0.4848 0.0087 0.0011 b. Using a = .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. ! d. Use the t test and a .05 to test H : Bi=0 and Ho: B =0. Use t table. The input in the box below will not be graded, but may be reviewed and considered by your instruction.

Answer 1

a) The table has been filled with explanations in each cell given. Some key formulas are:

MS = SS/df

df(Regression) = p

df(Residual) = n - p - 1

F = MS(Reg)/MS(Res)

t-stat = Coefficient/SE

	df	SS	MS	F	Significance F
Regression	2	1.7621	1.7621/2 = 0.8810	0.88105/0.0168 = 52.4434	6.12516E-05
Residual	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) The F-statistic is significant having a p-value which is less than 0.01. Hence, there exists a significant relationship

c) The goodness of fit of data can be explained by R-square which can be computed by:

R-square = SSR/SST = 1.7621/1.88 = 0.9372
93.72% variation can be explained by the independent variables in the regression equation. It is a good fit to the data.

d) The t-statistics are already computed in the table above:

t = Coefficient/Standard Error

Ho: B1 = 0

t = 2.7011, p-value = 0.03059 (From the t-table)

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

Ho: B2 = 0

t = 4.4545, p-value = 0.003

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