# Inference in Simple Linear Regression

## Inference in Simple Linear Regression

• Fact : Under normal regression model $$(b_0,b_1)$$ and $$SSE$$ are independently distributed and
$$\frac{b_0 - \beta_0}{s(b_0)} \sim t_{n-2}$$, $$\qquad \frac{b_1 - \beta_1}{s(b_1)} \sim t_{n-2}$$, $$\qquad SSE \sim \sigma^2 \chi_{n-2}^2$$.

• Confidence interval for $$\beta_0$$ and $$\beta_1$$ : $$100(1-\alpha)\%$$ (two-sided) confidence interval for $$\beta_i$$:
$$(b_i - t(1-\alpha/2;n-2) s(b_i)$$, $$b_i + t(1-\alpha/2;n-2) s(b_i))$$

for $$i=0,1$$, where $$t(1-\alpha/2;n-2)$$ is the $$1-\alpha/2$$ upper cut-off point (or $$(1-\alpha/2)$$ quantile) of $$t_{n-2}$$ distribution; i.e., $$P(t_{n-2} > t(1-\alpha/2;n-2)) = \alpha/2$$.
• Hypothesis tests for $$\beta_0$$ and $$\beta_1$$ : $$H_0 : \beta_i = \beta_{i0}$$ ($$i=0$$ or $$1$$).
Test statistic : $$T_i = \frac{b_i - \beta_{i0}}{s(b_i)}$$.
1. Alternative: $$H_1 : \beta_i > \beta_{i0}$$. Reject $$H_0$$ at level $$\alpha$$ if $$\frac{b_i - \beta_{i0}}{s(b_i)} > t(1-\alpha;n-2)$$. Or if, P-value = $$P(t_{n-2} > T_i^{observed}) < \alpha$$.
2. Alternative: $$H_1 : \beta_i < \beta_{i0}$$. Reject $$H_0$$ at level $$\alpha$$ if $$\frac{b_i - \beta_{i0}}{s(b_i)} < t(\alpha;n-2)$$. Or if, P-value = $$P(t_{n-2} < T_i^{observed}) < \alpha$$.
3. Alternative: $$H_1 : \beta_i \neq \beta_{i0}$$. Reject $$H_0$$ at level $$\alpha$$ if $$|\frac{b_i - \beta_{i0}}{s(b_i)}| > t(1-\alpha/2;n-2)$$. Or if, P-value = $$P(|t_{n-2}| > |T_i^{observed}|) < \alpha$$.

Inference for mean response at $$X = X_h$$

• Point estimate: $$\widehat Y_h = b_0 + b_1 X_h$$.

Fact: $$E(\widehat Y_h) = \beta_0 + \beta_1 X_h = E(Y_h)$$, $$Var(\widehat Y_h) = \sigma^2(\widehat Y_h) = \sigma^2\left[\frac{1}{n} + \frac{(X_h - \overline{X})^2}{\sum_i (X_i - \overline{X})^2}\right]$$. Estimated variance is $$s^2(\widehat Y_h) = MSE \left[\frac{1}{n} + \frac{(X_h - \overline{X})^2}{\sum_i (X_i - \overline{X})^2}\right]$$.

Distribution: $$\frac{\widehat Y_h - E(Y_h)}{s(\widehat Y_h)} \sim t_{n-2}$$.

Confidence interval: $$100(1-\alpha)$$% confidence interval for $$E(Y_h)$$ is $$(\widehat Y_h - t(1-\alpha/2;n-2) s(\widehat Y_h),\widehat Y_h + t(1-\alpha/2;n-2) s(\widehat Y_h))$$.

Prediction of a new observation $$Y_{h(new)}$$ at $$X = X_h$$

• Prediction : $$\widehat Y_{h(new)} = \widehat Y_h = b_0 + b_1 X_h$$.

Error in prediction : $$Y_{h(new)} - \widehat Y_{h(new)} = Y_{h(new)} - \widehat Y_h$$.

Fact : $$\sigma^2(Y_{h(new)} - \widehat Y_h) = \sigma^2(Y_{h(new)}) + \sigma^2(\widehat Y_h) = \sigma^2 + \sigma^2(\widehat Y_h) = \sigma^2\left[1+\frac{1}{n}+ \frac{(X_h - \overline{X})^2}{\sum_i (X_i - \overline{X})^2}\right]$$.

Estimate of $$\sigma^2(Y_{h(new)} - \widehat Y_h)$$ is $$s^2(Y_{h(new)} - \widehat Y_h) = MSE \left[1+\frac{1}{n}+ \frac{(X_h - \overline{X})^2}{\sum_i (X_i - \overline{X})^2}\right]$$.

Distribution : $$\frac{Y_{h(new)} - \widehat Y_h}{s(Y_{h(new)} -\widehat Y_h)} \sim t_{n-2}$$.

Prediction interval : $$100(1-\alpha)$$% prediction interval for $$Y_{h(new)}$$ is $$(\widehat Y_h - t(1-\alpha/2;n-2) s(Y_{h(new)}-\widehat Y_h),\widehat Y_h + t(1-\alpha/2;n-2) s(Y_{h(new)}-\widehat Y_h))$$.
• Confidence band for the regression line : At $$X=X_h$$ the $$100(1-\alpha)$$% confidence band for the regression line is given by $$\widehat Y_h \pm w_\alpha s(\widehat Y_h), \qquad \mbox{where} \sim w_\alpha = \sqrt{2F(1-\alpha; 2, n-2)}$$.

Here $$F(1-\alpha;2,n-2)$$ is the $$1-\alpha$$ upper cut-off point (or, $$(1-\alpha)$$ quantile) for the $$F_{2,n-2}$$ distribution ($$F$$ distribution with d.f. $$(2,n-2)$$).

Example $$\PageIndex{1}$$: Simple linear regression

We consider a data set on housing price. Here$$Y=$$ selling price of houses (in $1000), and $$X=$$ size of house (100 square feet). The summary statistics are given below: $$n = 19$$, $$\overline{X} = 15.719$$, $$\overline{Y} = 75.211$$ $$\sum_i(X_i - \overline{X})^2 = 40.805$$, $$\sum_i (Y_i - \overline{Y})^2 = 556.078$$, $$\sum_i (X_i - \overline{X})(Y_i - \overline{Y}) = 120.001$$. Estimates of $$\beta_1$$ and $$\beta_0$$ : $b_1 = \frac{\sum_i (X_i - \overline{X})(Y_i - \overline{Y})}{\sum_i(X_i - \overline{X})^2} = \frac{120.001}{40.805} = 2.941$ and $b_0 = \overline{Y} - b_1 \overline{X} = 75.211 - (2.941)(15.719) = 28.981.$ • Fit and Prediction: The fitted regression line : $$Y = 28.981 + 2.941 X$$. When $$X = 18.5 = X_h$$, the predicted value, that is an estimate of the mean selling price (in$1000) when size of the house is 1850 sq. ft., is $$\widehat Y_h = 28.981 + (2.941) (18.5) = 83.39$$.
• MSE: The degrees of freedom (df) $$= n-2 = 17$$. $$SSE = \sum_i(Y_i - \overline{Y})^2 - b_1^2\sum_i(X_i - \overline{X})^2 = 203.17$$. So, $$MSE = \frac{SSE}{n-2} = \frac{203.17}{17} = 11.95$$.
• Standard Error Estimates: $$s^2(b_0) = MSE \left[\frac{1}{n} + \frac{\overline{X}^2}{\sum_i(X_i - \overline{X})^2} \right] = 73.00$$, $$\qquad s(b_0) = \sqrt{s^2(b_0)} = 8.544$$.
$$s^2(b_1) = \frac{MSE}{\sum_i(X_i - \overline{X})^2} = 0.2929$$, $$\qquad s(b_1) = \sqrt{s^2(b_1)} = 0.5412$$.
• Confidence Intervals: We assume that the errors are normal to ﬁnd conﬁdence intervals for the parameters $$\beta_0$$ and $$\beta_1$$. We use the fact that $$\frac{b_0 - \beta_0}{s(b_0)} \sim t_{n-2}$$ and $$\frac{b_1 - \beta_1}{s(b_1)} \sim t_{n-2}$$ where $$t_{n-2}$$ denotes the $$t$$-distribution with $$n-2$$ degrees of freedom. Since $$t(0.975;17) = 2.1098$$, it follows that 95% two-sided confidence interval for $$\beta_1$$ is $$2.941 \pm (2.1098)(0.5412) = (1.80, 4.08)$$. Since $$t(0.95;17) = 1.740$$, the 90% two-sided confidence interval for $$\beta_0$$ is $$28.981\pm (1.740)(8.544) = (14.12,43.84)$$.

## Contributors

• Agnes Oshiro
(Source: Spring 2012 STA108 Handout 4)