# Predicting using linear regression

(1 pt) In a large class, there were 230 students who wrote both the midterm and the final exam. The standard deviation of the midterm grades was 11, and that of the final exam was 16. The correlation between the grades on the midterm and the final was 0.66. Based on the least squares regression line fitted to the data of the 230 students, if a student scored 24 points below the mean on the midterm, then how many points below the mean on the final would you predict her final exam grade to be?

So I have:
SD for midterm $= 11$
SD for final $= 16$
$r = 0.66$

I tried approaching this problem but I have no clue how to do it. I used the linear regression model, but since the means aren't given, I can't use it to find an intercept; I can only get a slope.

• Since you only care about how far a score is from the mean score, just make up a mean score as its value won't affect the answer. – Harvey Motulsky Feb 1 '14 at 2:11
• Hint: write down the prediction equation using deviations from means (which is what the question asks). Then remember how the constant term in a simple regression is calculated. – Alecos Papadopoulos Feb 1 '14 at 2:15

Since the OP... got it, let's not leave the question in the unanswered list: Let $\hat y$ be the estimated final score and $x$ the actual midterm score. The question asks to calculate

$$d = \bar y - \hat y$$

And we are also given that $x = \bar x -24$. The prediction equation is

$$\hat y = \hat a + \hat b x \Rightarrow \bar y - d = \hat a + \hat b(\bar x -24)$$ $$\Rightarrow \bar y -\hat b\bar x -\hat a+24\hat b= d$$

But $\bar y -\hat b\bar x =\hat a$ so $$d= 24\hat b$$

Further

$$\hat b = \frac {\operatorname{\hat Cov}(y,x)}{\operatorname{\hat Var}(x)} = r\frac {SD(y)}{SD(x)} \Rightarrow d= 24r\frac {SD(y)}{SD(x)}$$

All magnitudes are now known.