Question

B =X	Y
1	17
3	15
5	10
7	7
9	1

HW 1 For the date above

1-1. Construct an Analysis of Regression table, below (Below is the regression table I made)

Source of Variation	Degrees of Freedom, d.f.	Sums of Squares, SS	Mean Squares, MS	F, FR, MSR/MSE
Total	4	164
Regression	1	160	160	120.3
Residual	3	4	1.33

(I need a help with this) 1-2. Test B= 0 with a = 1%

Answer 1

Hypotheses are:

$H_{0}:\beta_{1}=0,H_{a}:\beta_{1}\neq 0$

Following table shows the calculations;

	X	Y	X^2	Y^2	XY
	1	17	1	289	17
	3	15	9	225	45
	5	10	25	100	50
	7	7	49	49	49
	8	1	64	1	8
Total	24	50	148	664	169

Sample size: n =5

Now,

$S_{yy}=\sum y^{2}-\frac{\left (\sum y \right )^{2}}{n}=164$

$S_{xx}=\sum x^{2}-\frac{\left (\sum x \right )^{2}}{n}=32.8$

$S_{xy}=\sum xy-\frac{\left (\sum x \right )\left (\sum y \right )}{n}=-71$

Slope of the regression equation is

$b_{1}=\frac{S_{xy}}{S_{xx}}=-2.1646$

and intercept of the equation will be

$b_{0}=\frac{1}{n}(\sum y - b_{1} \sum x)=20.3902$

Let us find SSE first :

$SSE=\sum y^{2}-b_{0}\sum y-b_{1}\sum xy=10.31097561$

So standard error of estimate will be

$S_{e}=\sqrt{\frac{SSE}{n-2}}=1.853912584$

The standard error of slope is

$s_{b_{1}}=\frac{S_{e}}{\sqrt{S_{xx}}}=0.3237$

T-statistics is

$t=\frac{b_{1}-0}{s_{b_{1}}}=-6.687$

Degree of freedom of test is df=n-2=5-2=3

The p-value using excel function "=TINV(0.01,3)" are +/- 5.841.

Since t does not lie between the critical values so we reject the null hypothesis.