Question

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.909785963
R Square	0.827710499
Adjusted R Square	0.826591736
Standard Error	7.177298036
Observations	156
ANOVA
	df	SS	MS	F	Significance F
Regression	1	38112.05194	38112.05194	739.8443652	1.09619E-60
Residual	154	7933.095493	51.5136071
Total	155	46045.14744
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	8.67422449	2.447697434	3.543830365	0.000522385	3.838827439	13.50962154	3.838827439	13.50962154
X Variable 1	0.801382837	0.029462517	27.20008024	1.09619E-60	0.743179986	0.859585688	0.743179986	0.859585688

(d)	How much of the variation in the sample values of total point earned does the model you estimated in part (b) explain?
	If required, round your answer to two decimal places.
	%
(e)	Mark Sweeney spent 80 hours studying. Use the regression model you estimated in part (b) to predict the total points Mark earned.
	If required, round your answer to the nearest whole number. Do not round intermediate calculations.
	$

Answer 1

(D) Amount of variation is equal to the value of coefficient of determination or R squared

Using the given output data table

R square = 0.8277 or 82.77

So, 82.77% is the variation in the sample values of total point earned does the model you estimated in part (b) explain

(E) Regression equation using the given data output table is

Total points earned = 8.6742 + 0.80138*(number of hours)

we have to find total points earned when number of hours = 80

this gives us

= 8.6742 + 0.80138*(80)

= 72.78

= 73 (rounded to nearest whole number)