Question

Student	Math-proficiency   X1	SAT-Math X2	Calculus final (college 1st year)  Y
1	72	462	71
2	96	545	92
3	68	585	72
4	86	580	82
5	70	592	74
6	73	516	71
7	91	638	100
8	75	615	87
9	76	596	81

The least-square equation is Y=-26.6+0.777X₁+0.082 X₂, that is :

Calculus final=-26.6+0.777 *math-proficiency +0.082* Sat-Math

1. Find the approximate 95% confidence interval for the mean calculus final for all the students having X₁=70 and X₂=592. (who scored 70 in Math-proficiency and 592 on Sat-Math).
2. Compute the TSS and ESS from data and regression equation. Please take a screen shot of you lists in TI-84 and submit the screen shot showing the data, the predicted value, the residuals and the squared residuals.  This is a proof that you did the problem.
3. Compute the coefficient of multiple determination and interpret its value in the context of the problem.  You must write down TSS, ESS and how r² is computed from them.

Conduct ANOVA F-test for the entirety of the regression. You must write down the ANOVA table with all the details.  

Answer 1

a)

Find the approximate 95% confidence interval for the mean calculus final for all the students having X1=70 and X2=592. (who scored 70 in Math-proficiency and 592 on Sat-Math).

		95% Confidence Interval
X1	X2	lower	upper
70	592	71.359	81.206

b)

Compute the TSS and ESS from data and regression equation. Please take a screen shot of you lists in TI-84 and submit the screen shot showing the data, the predicted value, the residuals and the squared residuals.  This is a proof that you did the problem.

	SS
Residual	97.3230
Total	848.8889

The regression equation is:

y = -26.6155 + 0.7763*x1 + 0.0820*x2

c)

Compute the coefficient of multiple determination and interpret its value in the context of the problem.  You must write down TSS, ESS and how r2 is computed from them.

R2 = 0.885

88.5% of the variability in the model is explained.

1. Conduct ANOVA F-test for the entirety of the regression. You must write down the ANOVA table with all the details.  (30 points)

Source	SS	df	MS	F	p-value
Regression	751.5659	2	375.7829	23.17	.0015
Residual	97.3230	6	16.2205
Total	848.8889	8

The hypothesis being tested is:

H0: β1 = β2 = 0

H1: At least one βi ≠ 0

The p-value is 0.0015.

Since the p-value (0.0015) is less than the significance level (0.05), we can reject the null hypothesis.

Therefore, we can conclude that the model is significant.

The calculations are:

	R²	0.885
	Adjusted R²	0.847
	R	0.941
	Std. Error	4.027
	n	9
	k	2
	Dep. Var.	Y
ANOVA table
Source	SS	df	MS	F	p-value
Regression	751.5659	2	375.7829	23.17	.0015
Residual	97.3230	6	16.2205
Total	848.8889	8
Regression output				confidence interval
variables	coefficients	std. error	t (df=6)	p-value	95% lower	95% upper
Intercept	-26.6155
X1	0.7763	0.1465	5.300	.0018	0.4179	1.1347
X2	0.0820	0.0270	3.039	.0228	0.0160	0.1481
Predicted values for: Y
		95% Confidence Interval		95% Prediction Interval
X1	X2	lower	upper		lower	upper
70	592	71.359	81.206		65.266	87.299