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CI
Confidence Intervals — Mean, Proportion, and Difference
Three CI formulas: 95% CI for mean [123.1, 133.7], proportion [64.7%, 77.3%], difference between means [1.04, 10.96].
📊 95% CI for mean: [123.1, 133.7]
Step-by-step solution
📏 17 · Confidence Intervals
Three Essential Confidence Interval Formulas
CI
① 95% CI for a Population Mean
Scenario: n=30 blood pressure readings, M=128.4 mmHg, SD=14.2. Construct a 95% CI.
$$CI=\bar{X}\pm t^*\frac{SD}{\sqrt{n}},\quad SE=\frac{14.2}{\sqrt{30}}=\frac{14.2}{5.477}=2.593,\quad t^*_{(29,\,0.025)}=2.045$$
$$CI=128.4\pm2.045\times2.593=128.4\pm5.30=\mathbf{[123.1,\;133.7]\text{ mmHg}}$$
128.4
Sample Mean
±5.3
Margin of Error
[123.1, 133.7]
95% CI
② 95% CI for a Proportion
Scenario: 142 out of 200 patients are satisfied with care (p̂=71%). Construct a 95% CI.
$$CI=\hat{p}\pm z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}},\quad SE=\sqrt{\frac{0.71\times0.29}{200}}=\sqrt{0.001030}=0.0321,\quad z^*=1.96$$
$$CI=0.71\pm1.96\times0.0321=0.71\pm0.063=\mathbf{[0.647,\;0.773]}$$
71.0%
Proportion
±6.3%
Margin
[64.7%, 77.3%]
95% CI
③ 95% CI for a Difference in Means
Scenario: Drug group (n=20, M=85.2, SD=7.4) vs. Placebo (n=20, M=79.6, SD=8.1). Construct a 95% CI for the mean difference.
$$SD_p=\sqrt{\frac{19(7.4)^2+19(8.1)^2}{38}}=\sqrt{\frac{1038+1246}{38}}=\sqrt{60.1}=7.75$$
$$SE_{diff}=SD_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}=7.75\sqrt{\frac{2}{20}}=7.75\times0.316=2.45$$
$$CI=(85.2-79.6)\pm t^*_{(38)}\times2.45=6.0-2.024\times2.45\;\Rightarrow\;\mathbf{[1.04,\;10.96\text{ points}]}$$
6.0
Mean Diff.
±4.96
Margin
[1.04, 10.96]
95% CI
✅ CI does not include 0 → the difference is statistically significant at α=.05. The drug improves scores by 1 to 11 points.
APA-7
A 95% confidence interval for the mean difference indicated that the drug group scored 6.0 points higher than the placebo group, 95% CI [1.04, 10.96]. Because the interval excludes zero, the difference is statistically significant at α = .05.