Question

Consider the data set shown below. Find the 95% confidence interval for the slope of the regression line.

y 0 3 2 3 8 10 11

x -2 0 2 4 6 8 10

9) A) 0.94643 ± 0.27603 B) 0.94643 ± 0.33306 C) 0.94643 ± 0.28377 D) 0.94643 ± 0.36203

Answer 1

Answer:-

Given That:-

Consider the data set shown below. Find the 95% confidence interval for the slope of the regression line.

y 0 3 2 3 8 10 11

x -2 0 2 4 6 8 10

Given,

X	$Y_i$	x	$Y_i^2$	$X_iY_i$	$\hat{Y_i}$		$e_i=Y_i-\hat{Y_i}$
-2	0	4	0	0	-0.3927	at x = -2	0.3927
0	3	0	9	0	1.5001	at x = 0	1.4999
2	2	4	4	4	3.3929	at x = 2	-1.3929
4	3	16	9	12	5.2857	at x = 4	-2.2857
6	8	36	64	49	7.1785	at x = 6	0.8215
8	10	64	100	80	9.0713	at x = 8	0.9287
10	11	100	121	110	10.9641	at x =10	0.0359

$\bar{X}=\frac{\Sigma X_i}{n}$

= 28/7

= 4

$\bar{Y}=\frac{\Sigma Y_i}{n}$

= 37/7

= 5.2857

$\Sigma X_iY_i=254$

$\Sigma X_i^2=224$

$\hat{\beta}=\frac{\Sigma X_iY_i-n\bar{X}\bar{Y}}{\Sigma X^2_i-n\bar{X}^2}$

$\hat{\beta}=\frac{254-7*4*5.2857}{224-7*(4)^2}$

= 106.0004/112

= 0.9464

$\hat{\beta}_0=\bar{Y}-\hat{\beta}_1\bar{X}$

= 5.2857 - 0.9464 * 4

= 1.5001

Simple linear regression model

$Y=\beta_0+\beta_1 X+\varepsilon \sim N(0, \sigma^2)$

$\hat{Y}=\beta_0+\beta_1 X+\varepsilon \sim N(0, \sigma^2)$

$\hat{Y}=1.5001+0.9464X_i$

$\Sigma e^2_i=(0.3927)^2+(1.4999)^2+(-1.3929)^2+(-2.2857)^2+(0.8215)^2+(0.9287)^2+(0.0359)^2$

$\Sigma e^2_i=11.1072=SS_{Residual}$

$MS_{Residual}=\frac{SS_{Residual}}{n-2}$

= 11.1072/7-2

= 2.2213

$S_{xx}=\sum_{1}^{7}x^2_i-n\bar{x}^2$

= 224 - 7*4 *4

= 112

100(1 - $\alpha$ )% confidence interval for  $\beta_1$ is

$\beta_1\pm t_{\alpha/2}n-2\sqrt{\frac{MS_{Res}}{S_{xx}}}$

i.e,

$P(\hat{\beta_1}-t_{\alpha/2}*n-2\sqrt{\frac{MS_{Residual}}{S_{xx}}}\leq \beta_1 \leq \hat{\beta_1}+t_{\alpha/2}*n-2\sqrt{\frac{MS_{Residual}}{S_{xx}}})=1-\alpha$

$\Rightarrow P(0.9464-2.571\sqrt{\frac{2.2213}{112}}\leq \beta_1 \leq 0.9464+2.571\sqrt{\frac{2.2213}{112}})=0.95$

$\Rightarrow P(0.9464-2.571*0.1408\leq \beta_1 \leq 0.9464+2.571*0.1408)=0.95 \, \, \, \, \, \left \{ \because t_{0.05/2},7-2=2.571\, \, \alpha=0.05 \right \}$

0.9464 $\pm$ 0.3619968

0.9464 $\pm$ 0.36200

Here I take  $\hat{\beta_1}=0.9464$, Take  $\hat{\beta_1}=0.94643$ and perfrom calculation then you will get exact 0.9463 $\pm$ 0.36203 95% confidence interval.

The correct answer is option (D)

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