Confidence interval for the slope

Three quick questions from earlier lessons.

b = 0.5, SE_b = 0.1, t* = 2. Which interval estimates β₁?

An interval for the population slope.

b \pm t^* \cdot SE_b

, with

t^*

from the t distribution with

n - 2

df. If the interval does not contain 0, there is convincing evidence of a linear relationship; interpret as the change in mean

y

per unit increase in

x

.

$n = 30$ , $b = 0.42$ , $SE_b = 0.10$ , $t^* = 2.048$ .

ME = 0.2048

2.048 × 0.10.

(0.215, 0.625)

0.42 ± 0.2048.

Evidence of a positive slope

0 excluded.

Your turn: ME = 0.15. The width of the interval.

0.3

2 × 0.15.

With 30 pairs, b = 2/10 and SE_b = 7/100. Using t* = 2.048 (28 df, 95%), find the upper end of the interval for β₁.
With 30 pairs, b = 4/10 and SE_b = 1/100. Using t* = 2.048 (28 df, 95%), find the upper end of the interval for β₁.

SE_b = 1/100 and t* = 2.048. Find the width of the 95% confidence interval for β₁.

A 95% interval for the slope β₁ is (0.4, 0.8). What does it suggest?
1. Convincing evidence of a positive linear relationship: 0 is not in the interval
2. No evidence of a relationship
3. x causes y to increase

Three questions to check you have got it.

With 30 pairs, b = 2/10 and SE_b = 7/100. Using t* = 2.048 (28 df, 95%), find the upper end of the interval for β₁.
With 30 pairs, b = 4/10 and SE_b = 1/100. Using t* = 2.048 (28 df, 95%), find the upper end of the interval for β₁.
With 30 pairs, b = 3/10 and SE_b = 6/100. Using t* = 2.048 (28 df, 95%), find the upper end of the interval for β₁.

Interval for β₁

$b \pm t^* SE_b$ (df $n - 2$ ); 0 outside → evidence of a relationship.

Watch out: association, not causation.

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