Intervals for a difference of proportions or means

Three quick questions from earlier lessons.

A 95% interval for $p_1 - p_2$ is (0.02, 0.10). What does it suggest about the two proportions?

Compare two groups with an interval for the difference.

(\hat{p}_1 - \hat{p}_2) \pm z^*\sqrt{\dfrac{\hat{p}_1(1 - \hat{p}_1)}{n_1} + \dfrac{\hat{p}_2(1 - \hat{p}_2)}{n_2}}

; for means

(\bar{x}_1 - \bar{x}_2) \pm t^*\sqrt{\dfrac{s_1^2}{n_1} + \dfrac{s_2^2}{n_2}}

. If 0 is in the interval, "no difference" is plausible.

$\hat{p}_1 = 0.6$ , $\hat{p}_2 = 0.5$ , $n_1 = n_2 = 100$ .

SE = \sqrt{0.0049} = 0.07

0.0024 + 0.0025.

ME \approx 0.137

1.96 × 0.07.

(-0.037, 0.237)

: no convincing difference

0.1 ± 0.137 contains 0.

Your turn: variances of the two estimates are 0.0009 and 0.0016. The SE of the difference.

0.05

√0.0025.

Independent samples of 100 give p̂₁ = 1/10 and p̂₂ = 6/10. Using z* = 1.96, find the margin of error for p₁ − p₂.
Independent samples of 100 give p̂₁ = 3/10 and p̂₂ = 8/10. Using z* = 1.96, find the margin of error for p₁ − p₂.

A 95% interval for p₁ − p₂ is (−0.07, 0.08). What does it suggest?
1. No convincing difference: 0 is a plausible value of p₁ − p₂
2. p₁ is larger, since the upper end is positive
3. It proves p₁ = p₂

Three questions to check you have got it.

Independent samples of 100 give p̂₁ = 1/10 and p̂₂ = 6/10. Using z* = 1.96, find the margin of error for p₁ − p₂.
Independent samples of 100 give p̂₁ = 3/10 and p̂₂ = 8/10. Using z* = 1.96, find the margin of error for p₁ − p₂.
Independent samples of 100 give p̂₁ = 7/10 and p̂₂ = 8/10. Using z* = 1.96, find the margin of error for p₁ − p₂.

Intervals for differences

Estimate difference ± critical value × √(sum of variances).

Watch out: does the interval contain 0?

About this lesson