*simple linear regresssion model*. One simple way to think of it
is via scatter plots. Below are heights of mothers and daughters collected
by Karl Pearson in the late 19th century.

In [2]:

```
heights_fig = plt.figure(figsize=(10,6))
axes = heights_fig.gca()
axes.scatter(M, D, c='red')
axes.set_xlabel("Mother's height (inches)", size=20)
axes.set_ylabel("Daughter's height (inches)", size=20)
```

Out[2]:

In [3]:

```
X = 66
from matplotlib import mlab
xf, yf = mlab.poly_between([X-.5,X+.5], [50,50], [75, 75])
selected_points = (M <= X+.5) * (M >= X-.5)
mean_within_slice = D[selected_points].mean()
scatterplot_slice = axes.fill(xf, yf, facecolor='blue', alpha=0.1, hatch='/')[0]
axes.scatter([X],[mean_within_slice], s=130, c='yellow', marker='^')
heights_fig
```

Out[3]:

In [4]:

```
print mean_within_slice
```

In [5]:

```
X = 60
selected_points = (M <= X+.5) * (M >= X-.5)
mean_within_slice = D[selected_points].mean()
print mean_within_slice
```

The regression model puts a line through this scatter plot in an *optimal* fashion.

In [6]:

```
%%R -o slope,intercept
parameters = lm(D ~ M)$coef
print(parameters)
intercept = parameters[1]
slope = parameters[2]
```

In [7]:

```
axes.plot([M.min(), M.max()], [intercept + slope * M.min(), intercept + slope * M.max()],
linewidth=3)
heights_fig
```

Out[7]:

A

*linear*regression model says that the function $f$ is a sum (linear combination) of functions of ${\tt Mother}$.Simple linear regression model: $$f({\tt Mother}) = \beta_0 + \beta_1 \cdot {\tt Mother}$$ for some unknown parameter vector $(\beta_0, \beta_1)$.

Could also be a sum (linear combination) of

*fixed*functions of`Mother`

: $$f({\tt Mother}) = \beta_0 + \beta_1 \cdot {\tt Mother} + \beta_2 \cdot {\tt Mother}^2 $$

*Simple linear*regression is the case when there is only one predictor: $$ f({\tt Mother}) = \beta_0 + \beta_1 \cdot {\tt Mother}.$$

Let $Y_i$ be the height of the $i$-th daughter in the sample, $X_i$ be the height of the $i$-th mother.

Model: $$ Y_i = \underbrace{\beta_0 + \beta_1 X_i}_{\text{regression equation}} + \underbrace{\varepsilon_i}_{\text{error}}$$ where $\varepsilon_i \sim N(0, \sigma^2)$ are independent.

This specifies a

*distribution*for the $Y$'s given the $X$'s, i.e. it is a*statistical model*.

We will be using

*least squares*regression. This measures the*goodness of fit*of a line by the sum of squared errors, $SSE$.Least squares regression chooses the line that minimizes $$ SSE(\beta_0, \beta_1) = \sum_{i=1}^n (Y_i - \beta_0 - \beta_1 \cdot X_i)^2.$$

In principle, we might measure goodness of fit differently: $$ SAD(\beta_0, \beta_1) = \sum_{i=1}^n |Y_i - \beta_0 - \beta_1 \cdot X_i|.$$

For some

*loss function*$L$ we might try to minimize $$ L(\beta_0,\beta_1) = \sum_{i=1}^n L(Y_i-\beta_0-\beta_1X_i) $$

With least squares, the minimizers have explicit formulae -- not so important with today's computer power -- especially when $L$ is convex.

Resulting formulae are

*linear*in the outcome $Y$. This is important for inferential reasons. For only predictive power, this is also not so important.If assumptions are correct, then this is

*maximum likelihood estimation*.Statistical theory tells us the

*maximum likelihood estimators (MLEs)*are generally good estimators.

The choice of the function we use to measure goodness of fit, or the *loss* function, has an outcome on what
sort of estimates we get out of our procedure. For instance, if, instead of fitting a line to a scatterplot, we were
estimating a *center* of a distribution, which we denote by $\mu$, then we might consider minimizing several loss functions.

If we choose the sum of squared errors: $$ SSE(\mu) = \sum_{i=1}^n (Y_i - \mu)^2. $$ Then, we know that the minimizer of $SSE(\mu)$ is the sample mean.

On the other hand, if we choose the sum of the absolute errors $$ SAD(\mu) = \sum_{i=1}^n |Y_i - \mu|.$$ Then, the resulting minimizer is the sample median.

Both of these minimization problems also have

*population*versions as well. For instance, the population mean minimizes, as a function of $\mu$ $$ \mathbb{E}((Y-\mu)^2) $$ while the population median minimizes $$ \mathbb{E}(|Y-\mu|). $$

Let's take some a random scatter plot and view the loss function.

In [8]:

```
X = np.random.standard_normal(50)
Y = np.random.standard_normal(50) * 2 + 1.5 + 0.1 * X
plt.scatter(X, Y)
```

Out[8]:

`R`

. Let's take a look at how we might
fit it in `python`

directly.

*loss* as a function of the parameters. Note that the
*true* intercept is 1.5 while the *true* slope is 0.1.

In [11]:

```
squared_error_loss
```

Out[11]:

Let's contrast this with the sum of absolute errors.

In [13]:

```
absolute_error_loss
```

Out[13]:

The following picture will be with us, in various guises, throughout much of the course. It depicts the geometric picture involved in least squares regression.

It requires some imagination but the picture should be thought as representing vectors in $n$-dimensional space, l where $n$ is the number of points in the scatterplot. In our height data, $n=1375$. The bottom two axes should be thought of as 2-dimensional, while the axis marked "$\perp$" should be thought of as $(n-2)$ dimensional, or, 1373 in this case.

The (squared) lengths of the above vectors are important quantities in what follows.

There are three to note: $$ \begin{aligned} SSE &= \sum_{i=1}^n(Y_i - \widehat{Y}_i)^2 = \sum_{i=1}^n (Y_i - \widehat{\beta}_0 - \widehat{\beta}_1 X_i)^2 \\ SSR &= \sum_{i=1}^n(\overline{Y} - \widehat{Y}_i)^2 = \sum_{i=1}^n (\overline{Y} - \widehat{\beta}_0 - \widehat{\beta}_1 X_i)^2 \\ SST &= \sum_{i=1}^n(Y_i - \overline{Y})^2 = SSE + SSR \\ R^2 &= \frac{SSR}{SST} = 1 - \frac{SSE}{SST} = \widehat{Cor}(\pmb{X},\pmb{Y})^2. \end{aligned} $$

An important summary of the fit is the ratio
$$
R^2 = \frac{SSR}{SST} = 1 - \frac{SSE}{SST}
$$
which measures *how much variability in $Y$* is explained by $X$.

*lm* for the wage
data and verify that some of the equations we present for the
least squares solutions agree with the output.
The data was compiled from a study in econometrics Learning about Heterogeneity in Returns to Schooling.

In [14]:

```
%%R
url = 'http://stats191.stanford.edu/data/wage.csv'
wages = read.table(url, sep=',', header=T)
print(head(wages))
mean(logwage)
```

`wages`

we `attach`

it so that the variables
are in the toplevel namespace.

In [15]:

```
%%R
attach(wages)
mean(logwage)
```

Let's fit the linear regression model.

In [16]:

```
%%R
wages.lm = lm(logwage ~ education)
print(wages.lm)
```

As in the mother-daughter data, we might want to plot the data and add the regression line.

In [17]:

```
%%R -h 800 -w 800
plot(education, logwage, pch=23, bg='red', cex=2, cex.lab=3)
abline(wages.lm, lwd=4, col='black')
```

There are explicit formulae for the least squares estimators, i.e. the minimizers of the error sum of squares.

For the slope, $\hat{\beta}_1$, it can be shown that $$ \widehat{\beta}_1 = \frac{\sum_{i=1}^n(X_i - \overline{X})(Y_i - \overline{Y} )}{\sum_{i=1}^n (X_i-\overline{X})^2} = \frac{\widehat{Cov}(X,Y)}{\widehat{Var}( X)}.$$

Knowing the slope estimate, the intercept estimate can be found easily: $$ \widehat{\beta}_0 = \overline{Y} - \widehat{\beta}_1 \cdot \overline{ X}.$$

In [18]:

```
%%R
beta.1.hat = cov(education, logwage) / var(education)
beta.0.hat = mean(logwage) - beta.1.hat * mean(education)
print(c(beta.0.hat, beta.1.hat))
print(coef(wages.lm))
```

There is one final quantity needed to estimate all of our parameters in our (statistical) model for the scatterplot. This is $\sigma^2$, the variance of the random variation within each slice (the regression model assumes this variance is constant within each slice...).

The estimate most commonly used is $$ \hat{\sigma}^2 = \frac{1}{n-2} \sum_{i=1}^n (Y_i - \hat{\beta}_0 - \hat{\beta}_1 X_i)^2 = \frac{SSE}{n-2} = MSE $$

Above, note the practice of replacing the quantity $SSE(\hat{\beta}_0,\hat{\beta}_1)$, i.e. the minimum of this function, with just $SSE$.

The term *MSE* above refers to mean squared error: a sum of squares divided by what we call its *degrees of freedom*. The degrees of freedom
of *SSE*, the *error sum of squares* is therefore $n-2$. Remember this $n-2$ corresponded to $\perp$ in the picture above...

Using some statistical calculations that we will not dwell on, if our simple linear regression model is correct, then we can see that
$$
\frac{\hat{\sigma}^2}{\sigma^2} \sim \frac{\chi^2_{n-2}}{n-2}
$$
where the right hand side denotes a *chi-squared* distribution with $n-2$ degrees of freedom.

In [19]:

```
%%R
sigma.hat = sqrt(sum(resid(wages.lm)^2) / wages.lm$df.resid)
sigma.hat
```

The summary from *R* also contains this estimate of $\sigma$:

In [20]:

```
%%R
summary(wages.lm)
```

Generally, by inference, we mean "learning something about the relationship between the sample $(X_1, \dots, X_n)$ and $(Y_1, \dots, Y_n)$."

In the simple linear regression model, this often means learning about $\beta_0, \beta_1$. Particular forms of inference are

*confidence intervals*or*hypothesis tests*. More on these later.Most of the questions of

*inference*in this course can be answered in terms of $t$-statistics or $F$-statistics.First we will talk about $t$-statistics, later $F$-statistics.

One sample problem: given an independent sample $\pmb{X}=(X_1, \dots, X_n)$ where $X_i\sim N(\mu,\sigma^2)$, the

*null hypothesis $H_0:\mu=\mu_0$*says that in fact the population mean is some specified value $\mu_0$.Two sample problem: given two independent samples $\pmb{Z}=(Z_1, \dots, Z_n)$, $\pmb{W}=(W_1, \dots, W_m)$ where $Z_i\sim N(\mu_1,\sigma^2)$ and $W_i \sim N(\mu_2, \sigma^2)$, the

*null hypothesis $H_0:\mu_1=\mu_2$*says that in fact the population means from which the two samples are drawn are identical.

We test a null hypothesis, $H_0$ based on some test statistic $T$ whose distribution is fully known when $H_0$ is true.

For example, in the one-sample problem, if $\bar{X}$ is the sample mean of our sample $(X_1, \dots, X_n)$ and
$$
S^2 = \frac{1}{n-1} \sum_{i=1}^n (X_i-\bar{X})^2
$$
is the sample variance. Then
$$
T = \frac{\bar{X}-\mu_0}{S/\sqrt{n}}
$$
has what is called a Student's t distribution with $n-1$ degrees of freedom *when $H_0:\mu=\mu_0$ is true.* When the null
hypothesis is not true, it does not have this distribution!

A $t$ statistic with $k$ degrees of freedom, has a form that becomes easy to recognize after seeing it several times.

It has two main parts: a numerator and a denominator. The numerator $Z \sim N(0,1)$ while $D \sim \sqrt{\chi^2_k/k}$ that is assumed

*independent*of $Z$.The $t$-statistic has the form $$ T = \frac{Z}{D}. $$

Another form of the $t$-statistic is $$ T = \frac{\text{estimate of parameter} - \text{true parameter}}{\text{accuracy of the estimate}}. $$

In more formal terms, we write this as $$ T = \frac{\hat{\theta} - \theta}{SE(\hat{\theta})}. $$ Note that the denominator is the accuracy of the

*estimate*and not the true parameter (which is usually assumed fixed, at least for now). The term $SE$ or*standard error*will, in this course, usually refer to an estimate of the accuracy of estimator. Therefore, it is often the square root of an estimate of the variance of an estimator.In our simple linear regression model, a natural $t$-statistic is $$ \frac{\hat{\beta}_1 - \beta_1}{SE(\hat{\beta}_1)}. $$ We've seen how to compute $\hat{\beta}_1$, we never get to see the true $\beta_1$, so the only quantity we have anything left to say about is the standard error $SE(\hat{\beta}_1)$.

How many degrees of freedom would this $T$ have?

In [22]:

```
density_fig
```

Out[22]:

*rejection rule* for hypothesis tests based on the $T_k$ distribution.
For instance, for the standard normal, the 5% rejection rule is to reject if the so-called $Z$-score is larger than about 2 in absolute value.

In [24]:

```
density_fig
```

Out[24]:

For the $T_{10}$ distribution, however, this rule must be modified.

In [26]:

```
density_fig
```

Out[26]:

Above, we used the one sample problem as an example of a $t$-statistic. Let's be a little more specific.

Given an independent sample $\pmb{X}=(X_1, \dots, X_n)$ where $X_i\sim N(\mu,\sigma^2)$ we can test $H_0:\mu=0$ using a $T$-statistic.

We can prove that the random variables $$\overline{X} \sim N(\mu, \sigma^2/n), \qquad \frac{S^2_X}{\sigma^2} \sim \frac{\chi^2_{n-1}}{n-1}$$ are independent.

Therefore, whatever the true $\mu$ is $$ \frac{\overline{X} - \mu}{S_X / \sqrt{n}} = \frac{ (\overline{X}-\mu) / (\sigma/\sqrt{n})}{S_X / \sigma} \sim t_{n-1}.$$

Our null hypothesis specifies a particular value for $\mu$, i.e. 0. Therefore, under $H_0:\mu=0$ (i.e. assuming that $H_0$ is true), $$\overline{X}/(S_X/\sqrt{n}) \sim t_{n-1}.$$

The following are examples of confidence intervals you may have already seen.

One sample problem: instead of deciding whether $\mu=0$, we might want to come up with an (random) interval $[L,U]$ based on the sample $\pmb{X}$ such that the probability the true (nonrandom) $\mu$ is contained in $[L,U]$ equal to $1-\alpha$, i.e. 95%.

Two sample problem: find a (random) interval $[L,U]$ based on the sampl es $\pmb{Z}$ and $\pmb{W}$ such that the probability the true (nonrandom) $\mu_1-\mu_2$ is contained in $[L,U]$ is equal to $1-\alpha$, i.e. 95%.

In the one sample problem, we might be interested in a confidence interval for the unknown $\mu$.

Given an independent sample $(X_1, \dots, X_n)$ where $X_i\sim N(\mu,\sigma^2)$ we can test construct a $(1-\alpha)*100\%$ using the numerator and denominator of the $t$-statistic.

Let $q=t_{n-1,(1-\alpha/2)}$

$$ \begin{aligned} 1 - \alpha &= P\left(-q \leq \frac{\mu - \overline{X}} {S_X / \sqrt{n}} \leq q \right) \\ &= P\left(-q \cdot {S_X / \sqrt{n}} \leq {\mu - \overline{X}} \leq q \cdot {S_X / \sqrt{n}} \right) \\ &= P\left(\overline{X} - q \cdot {S_X / \sqrt{n}} \leq {\mu} \leq \overline{X} + q \cdot {S_X / \sqrt{n}} \right) \\ \end{aligned} $$

Therefore, the interval $\overline{X} \pm q \cdot {S_X / \sqrt{n}}$ is a $(1-\alpha)*100\%$ confidence interval for $\mu$.

Recall our model $$ Y_i = \beta_0 + \beta_1 X_i + \varepsilon_i,$$ errors $\varepsilon_i$ are independent $N(0, \sigma^2)$.

In our heights example, we might want to now if there really is a linear association between ${\tt Daughter}=Y$ and ${\tt Mother}=X$. This can be answered with a

*hypothesis test*of the null hypothesis $H_0:\beta_1=0$. This assumes the model above is correct, but that $\beta_1=0$.Alternatively, we might want to have a range of values that we can be fairly certain $\beta_1$ lies within. This is a

*confidence interval*for $\beta_1$.

The second model is the *null model* in which we have set $\beta_1=0$.
This model says that
$$
Y_i = \beta_0 + \varepsilon_i.
$$
This model says that the mean of the $Y$'s is unrelated to that of $X$.

*Strictly speaking,
we should write $Y_i|X$ on the left hand side as this is a model of the $Y_i$'s given the entire set of $X$ observations.
If the pairs of mothers and daughters are drawn from some population independently than we may write $Y_i | X_i$.*

Let $L$ be the subspace of $\mathbb{R}^n$ spanned $\pmb{1}=(1, \dots, 1)$ and ${X}=(X_1, \dots, X\ _n)$.

Then, $${Y} = P_L{Y} + ({Y} - P_L{Y}) = \widehat{{Y}} + (Y - \widehat{{Y}}) = \widehat{{Y}} + e$$

In our model $\mu=\beta_0 \pmb{1} + \beta_1 {X} \in L$ so that $$ \widehat{{Y}} = \mu + P_L{\varepsilon}, \qquad {e} = P_{L^{\perp}}{{Y}} = P_{L^{\perp}}{\varepsilon}$$

Our assumption that $\varepsilon_i$'s are independent $N(0,\sigma^2)$ tells us that: ${e}$ and $\widehat{{Y}}$ are independent; $\widehat{\sigma}^2 = \|{e}\|^2 / (n-2) \sim \sigma^2 \cdot \chi^2_{n-2} / (n-2)$.

In turn, this implies $$ \widehat{\beta}_1 \sim N\left(\beta_1, \frac{\sigma^2}{\sum_{i=1}^n(X_i-\overline{X})^2}\right).$$

Therefore, $$\frac{\widehat{\beta}_1 - \beta_1}{\sigma \sqrt{\frac{1}{\sum_{i=1}^n(X_i-\overline{X})^2}}} \sim N(\ 0,1).$$

The other quantity we need is the

*standard error*or SE of $\hat{\beta}_1$. This is obtained from estimating the variance of $\widehat{\beta}_1$, which, in this case means simply plugging in our estimate of $\sigma$, yielding $$ SE(\widehat{\beta}_1) = \widehat{\sigma} \sqrt{\frac{1}{\sum_{i=1}^n(X_i-\overline{X})^2}} \qquad \text{independent of $\widehat{\beta}_1$}$$

Suppose we want to test that $\beta_1$ is some pre-specified value, $\beta_1^0$ (this is often 0: i.e. is there a linear association)

Under $H_0:\beta_1=\beta_1^0$ $$\frac{\widehat{\beta}_1 - \beta^0_1}{\widehat{\sigma} \sqrt{\frac{1}{\sum_{i=1}^n(X_i-\overline{X})^2}}} = \frac{\widehat{\beta}_1 - \beta^0_1}{ \frac{\widehat{\sigma}}{\sigma}\cdot \sigma \sqrt{\frac{1}{ \sum_{i=1}^n(X_i-\overline{X})^2}}} \sim t_{n-2}.$$

- Reject $H_0:\beta_1=\beta_1^0$ if $|T| > t_{n-2, 1-\alpha/2}$.

Let's perform this test for the wage data.

In [27]:

```
%%R
SE.beta.1.hat = (sigma.hat * sqrt(1 / sum((education - mean(education))^2)))
Tstat = beta.1.hat / SE.beta.1.hat
data.frame(beta.1.hat, SE.beta.1.hat, Tstat)
```

Let's look at the output of the `lm`

function again.

In [28]:

```
%%R
summary(wages.lm)
```

*R* performs this test in the second row of the `Coefficients`

table. It is clear that
wages are correlated with education.

Observing a large $|T|$ is unlikely if $\beta_1 = \beta_1^0$: reasonable to conclude that $H_0$ \ is false.

Common to report $p$-value: $$\mathbb{P}(|T_{n-2}| > |T|) = 2 \mathbb{P} (T_{n-2} > |T|)$$

In [29]:

```
%%R
2*(1 - pt(Tstat, wages.lm$df.resid))
```

In [30]:

```
%%R
detach(wages)
```

Suppose we have a parameter estimate $\widehat{\theta} \sim N(\theta, {\sigma}_{\theta}^2)$, and standard error $SE(\widehat{\theta})$ such that $$ \frac{\widehat{\theta}-\theta}{SE(\widehat{\theta})} \sim t_{\nu}.$$

We can find a $(1-\alpha) \cdot 100 \%$ confidence interval by: $$ \widehat{\theta} \pm SE(\widehat{\theta}) \cdot t_{\nu, 1-\alpha/2}.$$

To prove this, expand the absolute value as we did for the one-sample CI $$ 1 - \alpha = \mathbb{P}\left(\left|\frac{\widehat{\theta} - \theta}{SE(\widehat{\theta})} \right| < t_{\nu, 1-\alpha/2}\right).$$

Applying the above to the parameter $\beta_1$ yields a confidence interval of the form $$ \hat{\beta}_1 \pm SE(\hat{\beta}_1) \cdot t_{n-2, 1-\alpha/2}.$$

We will need to compute $SE(\hat{\beta}_1)$. This can be computed using this formula $$ SE(a_0\hat{\beta}_0 + a_1\hat{\beta}_1) = \hat{\sigma} \sqrt{\frac{a_0^2}{n} + \frac{(a_0\overline{X} - a_1)^2}{\sum_{i=1}^n \left(X_i-\overline{X}\right)^2}}.$$

*R*, this is computed by the function `qt`

.

In [31]:

```
%%R
alpha = 0.05
n = length(M)
qt(1-0.5*alpha,n-2)
```

In [32]:

```
%%R
qnorm(1-0.5*alpha)
```

*R* has some built in functions
to compute confidence intervals.

In [33]:

```
%%R
L = beta.1.hat - qt(0.975, wages.lm$df.resid) * SE.beta.1.hat
U = beta.1.hat + qt(0.975, wages.lm$df.resid) * SE.beta.1.hat
data.frame(L, U)
```

In [34]:

```
%%R
confint(wages.lm)
```

Once we have estimated a slope $(\hat{\beta}_1)$ and an intercept $(\hat{\beta}_0)$, we can predict the height
of the daughter born to a mother of any particular height by the plugging-in the height of the new mother, $M_{new}$ into
our regression equation:
$$
\hat{D}_{new} = \hat{\beta}_0 +\hat{\beta}_1 M_{new}.
$$
This equation says that our best guess at the height of the new daughter born to a mother of height $M_{new}$ is $\hat{D}_{new}$.
Does this say that the height will be *exactly* this value? No, there is some random variation in each slice, and we would expect the same random variation for this new daughter's height as well.

We might also want a confidence interval for the average height of daughters born to a mother of height $M_{new}=66$ inches: $$ \hat{\beta}_0 + 66 \cdot \hat{\beta}_1 \pm SE(\hat{\beta}_0 + 66 \cdot \hat{\beta}_1) \cdot t_{n-2, 1-\alpha/2}. $$

Recall that the parameter of interest is the average within the slice. Let's look at our picture again:

In [35]:

```
heights_fig
```

Out[35]:

In [36]:

```
%%R
height.lm = lm(D~M)
predict(height.lm, list(M=c(66,60)), interval='confidence', level=0.90)
```

There is yet another type of interval we might consider: can we find an interval that covers the height of a particular daughter knowing only that her mother's height as 66 inches?

This interval has to cover the variability of the new random variation with our slice at 66 inches. So, it must be at least as wide as $\sigma$, and we estimate its width to be at least as wide as $\hat{\sigma}$.

In [37]:

```
%%R
predict(height.lm, list(M=66), interval='prediction', level=0.90)
```

In [38]:

```
%%R
(69.41-61.94)
```

`2 * qnorm(0.95) * sigma.hat.height`

.

In [39]:

```
%%R
sigma.hat.height = sqrt(sum(resid(height.lm)^2) / height.lm$df.resid)
2 * qnorm(0.95) * sigma.hat.height
```

The actual width will depend on how accurately we have estimated $(\beta_0, \beta_1)$ as well as $\hat{\sigma}$. Here is the full formula. Again it is based on the $t$ distribution, the only thing we need to change is what we use for the SE.

$$ SE(\widehat{\beta}_0 + \widehat{\beta}_1 X_{\text{new}} + \varepsilon_{\text{new}}) = \widehat{\sigma} \sqrt{1 + \frac{1}{n} + \frac{(\overline{X} - X_{\text{new}})^2}{\sum_{i=1}^n \left(X_i-\overline{X}\right)^2}}. $$

The final interval is $$ \hat{\beta}_0 + \hat{\beta}_1 X_{\text{new}} \pm t_{n-2, 1-\alpha/2} \cdot SE(\hat{\beta}_0 + \hat{\beta}_1 X_{\text{new}} + \varepsilon_{\text{new}}). $$

In [40]:

```
```