3 3rd Tutorial

3.1 Recap

We want to test $H_0: \beta_1 = 0$ vs. $H_1 :\beta_1 \neq 0$ at level $\alpha$. Note that if we reject $H_0$ we are saying that the model $\hat{\beta}_0 +\hat{\beta_1}X$ has some ability to explain the variance that we are observing in Y . (i.e. There exists a linear relationship between the explanatory variables and the response variable.) Recall that ANOVA table can be used in the test for the existence of regression which is

Source of Variation	SS	df	MS	F
Regression	SSR	1	MSR	$F_0=\frac{MSR}{MSE}$
Error	SSE	n-2	MSE
Total	SSTO	n-1

Where, $$\begin{align} SSR &= \sum~(\hat{y}_{i}-\bar{y})^2 = \hat{\beta_1^2}S_{xx}\\ SSE &= \sum~(\hat{y_i}-y_i)^2=S_{yy}-\dfrac{S_{xy}^2}{S_{xx}}\\ SSTO&= \sum~(y_i-\bar{y})^2 = S_{yy}\\ MSR &= \frac{SSR}{1}\\ MSE &= \frac{SSE}{n-2} \end{align}$$

To assess how well the regression line fit the data one can use the The coefficient of determination $R^2$ which can be defined as follow:

$$R^2=\frac{SSR}{SSTO}=1-~\frac{SSE}{SSTO}$$

which measures the proportion of variability explained by the regression.

3.2 Exercieses

Exercise 3.1 A second-hand cars dealer has 10 cars for sale. He decides to investigate the relation between the cars age $X$ (in years) and the millage $Y$ (in thousands miles) by using the simple linear regression model. The dealer reported the following: The mean and standard deviation of the cars millage are given, respectively, by $40.6$ and $11.87153$. The correlation coefficient between $X$ and $Y$ is $0.9687105$. The estimated simple regression model is $\hat{Y} = 8.892 + 7.733X$.

1. Obtain $S_{YY} , S_{XX}$ and $S_{XY}$
2. Construct $90\%$ CI for the slope
3. Use ANOVA for testing the significance of the linearity.
4. What proportion of the total variation in millage is explained by age?
5. Compute the $95\%$ CI for car millage with age $7$ years.

Solution 3.1:
a)

$$S^2_{Y} = 11.87153^2 = \dfrac{S_{YY}}{n-1} ~~ \Rightarrow ~~ S_{YY} = 11.87153^2 \times 9 = 1268.399$$

Knowing $$\hat{\beta}_1 = \dfrac{S_{XY}}{S_{XX}} \Rightarrow ~ S_{XY}= \hat{\beta}_1 S_{XX}$$

Then,

$$r_{XY} = \dfrac{S_{XY}}{\sqrt{S_{XX}}~\sqrt{S_{YY}}} = \dfrac{\hat{\beta}_1 S_{XX}}{\sqrt{S_{XX}}~\sqrt{S_{YY}}}= \dfrac{\hat{\beta}_1~\sqrt{S_{XX}}}{\sqrt{1268.399}}$$
Therefore:
$$0.9687105 = \dfrac{7.733~\sqrt{S_{XX}}}{\sqrt{1268.399}}~ \Rightarrow S_{XX} = 19.9$$ b) $$\begin{align} &SSE = S_{YY}-\dfrac{S^2_{XY}}{S_{XX}} = 78.187\\ &MSE = \dfrac{SSE}{n-2} = 9.773375\\ &S(\hat{\beta}_1) = \sqrt{\dfrac{MSE}{S_{XX}}} = 0.7\\ \end{align}$$ Therefore, \ The $90\%$ CI for the slope is: $$(\hat{\beta}_1 \pm t_{(1−\alpha/2⁄ ,𝑛−2)} \times S(\hat{\beta}_{1})) = (6.432 , 9.035)$$

3. 
   

Source of Variation	SS	df	MS	F
Regression	1190.22	1	1990.22	121.787
Error	78.187	8	9.77
Total	1268.399	9

$$F_{0.95,1,8} = 5.31 < 121.787$$ Therefore $H_0$ is rejected and that implies the existence of linear relationship.
d) $$R^2 = r_{XY}^2 = 0.9384$$
e) The fitted value $\hat{Y_h} = 8.892+7.7337~\times 7 = 63.0279$
$$\hat{\beta}_0 = \bar{Y} - \hat{\beta}_1~\bar{X} ~~ \Rightarrow ~ \bar{X}= 4.1$$
$$S(\hat{Y})= MSE~\times (\dfrac{1}{n}+\dfrac{(X_h-\bar{X})^2}{S_{XX}})= 5.107$$

The $95\%$ CI is:

$$(\hat{Y}_h \pm t_{(1-\alpha/2, n-2)} S(\hat{Y}_h))=(51.251158 , 74.804642)$$

3.3 Coursework

1. Attempt Problem 3 and Problem 4 in the past exam paper here
   
2. Attempt Problem 2 and Problem 3 in the past exam paper here