Inference for \(\beta_1\)

Since \(\beta_1\) captures the relationship between X and Y , we would like to be able to do inference on this parameter.

Confidence interval or test about \(\beta_1\), (not \(\hat\beta_1\))

Result: If \(Y | X = x\) has a normal distribution and \(S_{\hat \beta_1}\) is the estimated standard error of \(\hat \beta _1\), then

\[ {\hat \beta_1-\beta_1\over S_{\hat \beta_1}}\sim T_{n-2} \]

\(S_{\hat \beta_1}\) is the standard error of \(\hat \beta_1\). depend on \(\sigma_\epsilon^2\). You will not required to calculate the value by hand in this course.

This holds approximately if \(Y | X = x\) is not normal, but \(n \geq 30\).

Confidence Interval for \(\beta_1\)

A \((1-\alpha)100\)% confidence interval for \(\beta_1\) is

\[ \hat \beta_1\pm t_{n-1,\alpha/2}S_{\hat \beta_1} \]

In R:

confint(model, level =0.95)
##                  2.5 %     97.5 %
## (Intercept) 26.7346495 33.1002241
## Mheight      0.4908201  0.5926739

The intercept is \(\alpha_1\), Mheight is \(\beta_1\)

Hypothesis Test for \(\beta_1\)

  1. State the hypotheses

\[ H_0:\beta_1=\beta_{1,0},H_0:\beta_1=\beta_{1,0},H_0:\beta_1=\beta_{1,0}\\ H_a:\beta_1>\beta_{1,0},H_a:\beta_1<\beta_{1,0},H_a:\beta_1\neq\beta_{1,0} \]

\(H_a:\beta_1\neq\beta_{1,0}\) means there is no linear relationship between Y and X.

\(H_a:\beta_1>\beta_{1,0}\) means Y and X are positively correlated, or Y will increase as X increases.

\(H_a:\beta_1<\beta_{1,0}\) means Y and X are negatively correlated, or Y will decrease as X increases.

  1. Compute the test statistic

\[ T_{H_0}={\hat\beta_1-\beta_{1,0}\over S_{\hat\beta_1}} \]

  1. Reach a conclusion

  • If \(H_0\) is true, \(T_{H_0}\sim T_{n-2}\)

  • We can find rejection rules and p-values using the same method as our previous T tests.

  1. State the conclusion in the context of the problem.

Example: Test whether mothers’ height and daughters’ height have a positive linear relationship using \(\alpha = 0.05\).

summary(model)
## 
## Call:
## lm(formula = Dheight ~ Mheight, data = heights)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -7.397 -1.529  0.036  1.492  9.053 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 29.91744    1.62247   18.44   <2e-16 ***
## Mheight      0.54175    0.02596   20.87   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 2.266 on 1373 degrees of freedom
## Multiple R-squared:  0.2408, Adjusted R-squared:  0.2402 
## F-statistic: 435.5 on 1 and 1373 DF,  p-value: < 2.2e-16

\(\hat\beta_1=0.54175>0\), standard error of \(\hat \beta_1\) is 0.02596

\(T_{H_0}=20.87\), p-value is realllly small.

The p-value is two-sided by default in R.

\(H_0:\beta_1=0\), \(H_a:\beta_1>0\)

$T_{H_0}={ 0.54175 }=20.87 $

\(p-value=2\div (2\times 10^{-16})<0.05\)

Reject \(H_0\) p-value is smaller than \(\alpha\)

Prediction

The estimated regression line is

\[ \hat Y=\hat \alpha_1+\hat \beta_1 x \]

We can use this regression line to predict values of Y for specific values of x. We will denote this predicted value with \(\hat \mu_{Y|X}(x)\).

Example: What is the predicted height for a woman whose mother is 69 inches tall?

\(\hat y=\hat \alpha_1+\hat\beta_1x=29.917+0.542\times 69=67.315\)

\(\hat \mu_{Y|X=69}=\hat E(Y|X=69)=67.315\) inches.

This value represents two types of prediction.

  1. The average value of Y for the sub-population with \(X=x\)

Average height of all women whose mothers are 69 inches

Smaller error due to averaging

  1. An individual value of Y when \(X=x\)

The height of an individual (future observation) women whose mother is 69 inches.

Higher error for individual prediction

The errors and intervals associated with these two cases are different.