Confidence Interval for the Slope

t Interval and t Test for the Slope

By the end of this lesson, you will be able to construct and interpret a confidence interval for the slope, and conduct and interpret a significance test for the slope.

Confidence Interval for the Slope β

$$b \pm t^* \cdot SE_b$$

where $t^*$ uses df = n − 2 (not n − 1, because we estimated two parameters: the intercept and the slope).

SE_b comes directly from the computer output ("Std Error" or "SE Coef" column next to the slope).

Interpretation: "We are C% confident that for each 1-unit increase in [x], the true mean [y] changes by between [lower] and [upper] [units]."

Significance Test for the Slope

Hypotheses:

• $H_0: \beta = 0$ (no linear relationship between x and y in the population)
• $H_a: \beta \neq 0$ (or one-sided: $\beta > 0$ or $\beta < 0$)

Test statistic: $$t = \frac{b - 0}{SE_b} = \frac{b}{SE_b}, \quad df = n - 2$$

The p-value is given directly in computer output. Compare to $\alpha = 0.05$.

If p < α: Reject $H_0$. There is convincing evidence of a linear relationship between x and y in the population.

If p ≥ α: Fail to reject $H_0$. We don't have convincing evidence of a linear relationship.

Reading Computer Output (Full Example)

Predictor   Coef    SE Coef    T      P
Constant   87.3     5.62     15.5   0.000
Age         0.72    0.09      8.0   0.000

S = 12.1   R-Sq = 56.8%

A study of age (x, in years) and blood pressure (y, in mmHg) for 50 adults:

From output:

• b = 0.72 (slope); SE_b = 0.09; t = 8.0; p < 0.0001
• df = 50 − 2 = 48

95% CI for β: $0.72 \pm t^(0.09)$, with $t^ \approx 2.011$ (df = 48)

$= 0.72 \pm 0.181 = (0.539, , 0.901)$ mmHg per year

Interpretation: We are 95% confident that each additional year of age is associated with a true mean increase in blood pressure of between 0.539 and 0.901 mmHg.

Test conclusion: Since p < 0.0001 < 0.05, we reject $H_0$. There is convincing evidence of a positive linear relationship between age and blood pressure in adults.