Question

1. Suppose we have the following data on a dependent variable (Y) and an explanatory variable (X):

X          Y

0          140

1          140

4             0

0          180

3           80

2          120

4             0

1          200

2          120

            3            40

1. Calculate the simple linear regression equation by hand. Show all your work. Using your result, predict the value of Y when X = 3.5. (15 points)
2. Calculate the R²and the adjusted R²measures (by hand) and provide an interpretation of what they tell us. (12 points)
3. Test (by hand) to see whether this is a good or bad regression model using all three tests discussed in lecture. Make sure to show the null and alternative hypotheses and interpret results. (18 points)

Answer 1

(a) The simple linear regression equation is:
y = 188 - 43*x

The value of Y when X = 3.5 is 37.5.

(b) R2 = 0.834

Adjusted R2 = 0.813

83.4% of the variation in the model is explained.

(c) The hypothesis being tested is:

H0: β1 = 0

H1: β1 ≠ 0

The p-value is 0.0002.

Since the p-value (0.0002) is less than the significance level (0.05), we can reject the null hypothesis.

Therefore, we can conclude that the model is significant.

X	Y
0	140
1	140
4	0
0	180
3	80
2	120
4	0
1	200
2	120
3	40
	r²	0.834
	r	-0.913
	Std. Error	30.373
	n	10
	k	1
	Dep. Var.	Y
ANOVA table
Source	SS	df	MS	F	p-value
Regression	36,980.000	1	36,980.000	40.09	.0002
Residual	7,380.000	8	922.5000
Total	44,360.000	9
Regression output				confidence interval
variables	coefficients	std. error	t (df=8)	p-value	95% lower	95% upper
Intercept	188.0000
X	-43.0000	6.7915	-6.331	.0002	-58.6613	-27.3387
Predicted values for: Y
		95% Confidence Interval	95% Prediction Interval
X	Predicted	lower	upper	lower	upper	Leverage
3.5	37.500	5.213	69.787	-39.623	114.623	0.213