Stats question

So, for simple linear regression we need to find the slope $\beta_1$ and the intercept $\beta_0$ so that we can use that line to estimate torque based on diameter. There are formulas to estimate both of these (the hat means that it is estimated).

$\hat{\beta_1} = \frac{S_{X,Y}}{Var(X)}$ where $Var(X)$ is the sample variance of X and $S_{X,Y}$ is the sample covariance between X and Y.

$\hat{\beta_0} = \bar{Y} - \hat{\beta_1}\bar{X}$ where $\bar{Y}$ and $\bar{X}$ are the respective sample means of Y and X.

In order to get all the tools necessary to calculate these, we need the means and variances of X and Y, and the covariance between X and Y.

Using the familiar formula for the sample mean $\bar{X} = \frac{\sum_{i=1}^{n}X_i }{n} = \frac{1+3+7+12+13+18}{6} = \frac{54}{6} = 9$ 
Analogously $\bar{Y} = \frac{367}{60}$

Now the variance is $Var(X) = \frac{\sum (X_i - \bar{X})^2}{n-1}$ (the n-1 is there to correct for the fact that we don't know the sample mean and the variance calculation depends on the sample mean, so it is lower than it should be, this correction is always used when calculating sample variance)

So $Var(X) = \frac{(1-9)^2+(3-9)^2+(7-9)^2+(12-9)^2+(13-9)^2+(18-9)^2}{5}= 42$ and $S_X = \sqrt{42}$ since the standard deviation is the square root of the variance.
Analogously $Var(Y) = \frac{6349}{600}$ and $S_Y = \sqrt{\frac{6349}{600}}$

Okay, one more of these tedious calculations left: the covariance between X and Y
$Cov(X,Y) = \frac{\sum (X_i - \bar{X})(Y_i - \bar{Y})}{n-1} = 21.08$

Good, now we can get to answering the questions.

a) Recall the formulas for $\hat{\beta_1}$ and $\hat{\beta_0}$
$\hat{\beta_1} = \frac{21.08}{42} = 0.502$
$\hat{\beta_0} = \frac{367}{60}- 0.502*9 = 1.599$
Since the simple linear regression model is formulated as $Y = \beta_0 + \beta_1X + \epsilon$, the equation is

$T = 1.599 + 0.502D$

b) In order to estimate, we just substitute the diameter into the equation 
$\hat{T_7} = 1.599 + 0.502*7 = 5.11$

c) Here we can use the property that in simple linear regression $\hat{\beta_1} = r \frac{S_Y}{S_X}$ so that $r = \hat{\beta_1} \frac{S_X}{S_Y} = 0.5019*\frac{\sqrt{42}}{\sqrt{\frac{6349}{600}}} = 0.9999$

Hopefully that clears everything up.