线性回归分析-HW4

HW.

Prob-1

1-a

$$\begin{align} H_0 & : \beta_{1}=3 \\ H_a & : \beta_1\neq 3 \end{align}$$

the statistic is

$$t=\frac{\hat{\beta}_1-3}{s(\hat{\beta}_{1})}\overset{H_0}{\sim}t_{22}$$

we know from the table that

$$\hat{\beta}_{1}=3.5702,\ s(\hat{\beta}_{1})=0.347$$

so $t=1.643228$.

The decision rule is

$$|t|>|t_{0.975,23}|=2.068658$$

So the conclusion is that we don't have enough evidence to say that the increase in the expected number of work hours in the Toluca Company does not equal this standard.

1-b

$$\mathrm{power}(3.5)=P(\left| \frac{\hat{\beta}_1-3}{s(\hat{\beta}_{1})} \right|>t_{0.975,22} \mid \beta_{1}=3.5)$$

Because when $\beta_{1}=3.5$,

$$\frac{\hat{\beta}_1-3}{s(\hat{\beta}_{1})} \sim t\left( 23, \frac{0.5}{\sigma(\hat{\beta}_{1})} \right)$$

so

$$\mathrm{power}(3.5)=0.277781$$

1-c

$F^{*}$ tests

$$H_0':\beta_{1}=0$$

the null hypothesis is different.

Prob-2

2-a

2-b

1. 1.19
2. 84.94683
3. 4.4432518
4. 8.173

Prob-3

3-a

$$s^{2}=MSE\cdot\left[ \frac{1}{n}+ \frac{(X_{h}-\bar{X})^{2}}{S_{x x}} \right]=8.173\cdot\left[ \frac{1}{60}+\frac{(60-59.98333)^{2}}{8210.983} \right]=0.1362194$$

CI:

$$84.94683\pm s\cdot t_{0.975,58} = [84.81061,85.08305]$$

3-b

$$s^{2}=MSE\cdot\left[ {\color{red}1+}\frac{1}{n}+ \frac{(X_{h}-\bar{X})^{2}}{S_{x x}} \right]=8.173\cdot\left[ 1+\frac{1}{60}+\frac{(60-59.98333)^{2}}{8210.983} \right]=8.309219$$

CI:

$$84.94683\pm s\cdot t_{0.975,58} = [76.63761,93.25605]$$

3-c

Yes it is wider and it should be wider.

Prob-4

4-a

SSR seems to be larger. This implies that $R^{2}>0.5$.

4-b

4-c

$$\begin{align} H_0 & : \beta_{1}=0 \\ H_a & : \beta_{1}\ne 0 \end{align}$$

Decision rule:

$$F>F_{0.95,1,58}$$

Conclusion: Reject.

4-d

$R^{2}=0.7501$. Unexplained proportion: $\approx 25\%$. It is relatively small.

4-e

$R^{2}= 0.750,\quad r = −0.866$.

Prob-5

5-a

$$\begin{align} \mathbb{E}[MSR] & =\sigma^{2}+\beta_{1}^{2}S_{x x} = 1026.36\\ \mathbb{E}[MSE] & =\sigma^{2}=0.36 \end{align}$$

5-b

$S_{x x}$ would be smaller, leading to less significance. So the answer is it would be worse for determining whether or not a regression relation exists.

The mean response for $X=8$, the variances of the estimates based on the two sets of observations are respectively

$$0.36\cdot\left[ \frac{1}{5}+ \frac{(8-8)^{2}}{114} \right],\ 0.36\cdot \left[ \frac{1}{5}+ \frac{(8-8)^{2}}{10} \right]$$

they are the same. So both sets of observations are OK. But if it is not for $X=8$, I would say the former set is better.

Prob-6

$$\begin{align} \mathbb{E}(Y_{i}-\bar{Y})^{2} & = \mathbb{E}(\beta_{1}(X_{i}-\bar{X})+\varepsilon_{i}-\bar{\varepsilon})^{2} \\ & = \beta_{1}^{2}(X_i-\bar{X})^{2}+ \frac{n-1}{n}\sigma^{2} \end{align}$$

so

$$\mathbb{E}\left( \sum (\hat{Y}_{i}-\bar{Y})^{2} \right)=\beta_{1}^{2}S_{x x}+(n-1)\sigma^{2}$$

instead of $(n-1)\sigma^{2}$.