Question

We have a dataset with n = 10 pairs of observations (li, Yi), and n n Xi = 683, Σ Yi = = 813, i=1 n п n < x; = 47,405, Xiyi = 56,089, Xyz 66, 731. i=1 i=1 i=1 What is an approximate 99% confidence interval for the mean response at xo = 90?

Answer 1

Answer:

Given that,

We have a dataset with n=10 pairs of observations (xi, yi), and

$\sum_{i=1}^{n}x_i=683$,

,

Σ9: Ji = 813 1=1

$\sum_{i=1}^{n}x_i^2=47,405$,

, and

Στiy: rigi = 56, 089 1=1

Συ? = 66, 731 1=1

What is an approximate 95% confidence interval for the mean response at x0=90:

Let us define the terms as follows:

$S_{xx}=\sum_{i=1}^{n}x_i^2-n\bar{x}^{2}$

$S_{yy}=\sum_{i=1}^{n}y_i^2-n\bar{y}^{2}$

$S_{xy}=\sum_{i=1}^{n}x_iy_i-n\bar{x}\bar{y}$

Now,

$\bar{x}=\frac{\sum x_i}{n}$

=683/10

=68.3

T
= 68.3

And,

$\bar{y}=\frac{\sum y_i}{n}$

=813/10

=81.3

$\bar{y}$=81.3

Therefore,

$S_{xx}=\sum_{i=1}^{n}x_i^2-n\bar{x}^{2}$

=47405-(10)(68.3)^2

Sxx=756.1

$S_{yy}=\sum_{i=1}^{n}y_i^2-n\bar{y}^{2}$

=66731-(10)(81.3)^2

Syy=634.1

$S_{xy}=\sum_{i=1}^{n}x_iy_i-n\bar{x}\bar{y}$

=56089-(10)(68.3)(81.3)

Sxy=561.1

Now the regression equation as follows:

$\hat{y}=\hat{\alpha }+\hat{\beta }x$

Where,

$\hat{\alpha }=\bar{y}-\hat{\beta }\bar{x}$

$\hat{\beta }=\frac{S_{xy}}{S_{xx}}$

=561.1/756.1

=0.742097606

$\hat{\beta }$ =0.7421 (Approximately)

$\hat{\alpha }$=81.3-(0.742097606)(62.3)

=30.6147335

$\hat{\alpha }$ =30.6147(Approximately)

Therefore,

y=30.6147+0.7421x

The variance of the error term ($\sigma ^{2}$):

$\sigma ^{2}=\frac{1}{n-2}\left [ S_{yy}-\frac{S_{xy}^2}{S_{xx}} \right ]$

$=\frac{1}{10-2}\left [ 634.1-\frac{561.1^2}{756.1} \right ]$

=217.7090332/8

=27.21362915

$\sigma ^{2}$ =27.2136

Let be the mean response when X0=90,

HO

=30.6147335+(0.742097606)(90)

HO

HO
=97.40351804

Therefore,

$SE[\mu _{0}]=\left [ \left [ \frac{1}{n}+\frac{(X_0-\bar{x})^{2}}{S_{xx}} \right ]\hat{\sigma }^2 \right ]^{1/2}$

$=\left [ \left [ \frac{1}{10}+\frac{(90-68.3)^{2}}{756.1} \right ]27.21362915 \right ]^{1/2}$

$=\sqrt{19.66968434}$

=4.435051786

$SE[\mu _0]$ =4.4351

The mean response $\mu _0$ ,

The 95% CI:

$\Rightarrow \mu _0\pm t_{\alpha /2,n-2}\times SE[\mu _{0}]$

$\alpha$=1-CI

CI=0.95

$\alpha$ = 1-0.95

$\alpha$ =0.05

Then,

$\Rightarrow \mu _0\pm t_{\alpha /2,n-2}\times SE[\mu _{0}]$

$\Rightarrow 97.40351804\pm (2.306)\times (4.435051786)$

=97.40351804 $\pm$ 10.22722942

=[87.17628861, 107.6307475]

=[87.1763, 107.6308]

Hence, the 95% confidence interval for the mean response at X0=90 is,

=[87.1763, 107.6308].