# MSE and MSR in regression question

In a small-scale regression study, five observations on $Y$ were obtained corresponding to $X = 1,4,10, 11$, and $14$. Assume that $\sigma=0.6,B_0=5,B_1=3$

a. What are the expected values off MSR and MSE here?

b. For determining whether or not a regression relation exists, would it has been better or worse to have made the five observations at $X = 6,7, 8, 9$, and $10$? Why? Would the same answer apply if the principal purpose were to estimate the mean response for $X = 8$? Discuss.

$$Y_i=B_0+B_1X_i+\epsilon_i$$ $$\hat{Y_i}=\hat{B_0}+\hat{B_1}X_i$$

$$MSR=\sum(\hat{Y_i}-\overline{Y})^2$$

$$MSE=\frac{\sum (Y_i-\hat{Y_i)}^2}{n-2}=\frac{\sum(B_0+B_1X_i+\epsilon_i-\hat{B_0}-\hat{B_1}X_i)^2}{n-2}$$

I'm still doesn't understand what they want, they want $$E(MSE);E(MSR)$$?

What do they mean by expected values?

• What is your understanding of the "expected values" requested in (a)? – whuber Oct 4 '15 at 21:28
• The question does not concern calculated values: it explicitly asks for expected values. That is intended in the sense of expectations of random variables. MSR and MSE are both random because they depend on the random variables $\epsilon_i$. Thus, you can answer this question only by making assumptions about the expectations (and, it will turn out, variances and covariances) of the $\epsilon_i$ and knowing the formulas for MSE and MSR in terms of the $\epsilon_i$. Since $\epsilon_i$ can be computed as $Y_i - (B_0+B_1X_i)$, start with formulas in terms of the $X_i$ and $Y_i$. – whuber Oct 4 '15 at 21:41
• Often, unnecessary extra information is provided in questions either to make them easier for those with limited algebra skills or to distract those who do not understand the concepts at all. – whuber Oct 5 '15 at 13:27
• You don't seem to be asking a question any more after your latest edits. If you wish to post an answer to your original question, then please do so as an answer, not by changing the question. BTW, what calculation supports your answer to (b)? – whuber Oct 6 '15 at 3:28
• That conclusion is incorrect, however. You need to look at the formula for the standard error of $B_1$. – whuber Oct 7 '15 at 0:04

$$Y_i=B_0+B_1X_i+\epsilon_i$$ $$\hat{Y_i}=\hat{B_0}+\hat{B_1}X_i$$
a) $$E[MSE]=E[\frac{\sum(Y_i-\hat{Y_i})^2}{n-2}]=\sigma^2=0.6^2$$ $$E[MSR]=E[\sum(\hat{Y_i}-\overline{Y})^2]=\sigma^2+B_1\sum(X_i-\overline{X})^2=1026.36$$
b) $$\sigma(\hat{B_1})=\sqrt{\frac{\sigma^2}{\sum(X_i-\overline{X})^2}}=\frac{0.6}{\sqrt{\sum(X_i-\overline{X})^2}}$$ for the case where $X=(1,4,10,11,14)$ we have that $\sigma(\hat{B_1})=0.05619515$ and for the case where $X=(6,7,8,9,10)$ $\sigma(B_1)=0.1897367$, then the first set is better I think.
But why I need to look $\sigma(\hat{B_1})$?