Inference for Means & Proportions

One-Sample z-Test/CI for Proportion

• Test: $z = \frac{\hat{p} - p_0}{\sqrt{p_0(1-p_0)/n}}$
• CI: $\hat{p} \pm z^*\sqrt{\hat{p}(1-\hat{p})/n}$
• Conditions: random, Large Counts ($np \geq 10$, $n(1-p) \geq 10$), 10% rule.

One-Sample t-Test/CI for Mean

• Test: $t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}$, $df = n-1$.
• CI: $\bar{x} \pm t^* \cdot s/\sqrt{n}$
• Conditions: random, Normal/Large Sample ($n \geq 30$ or no strong skew), 10% rule.

Two-Sample z-Test for Proportions

$z = \frac{(\hat{p}_1 - \hat{p}_2) - 0}{\sqrt{\hat{p}_c(1-\hat{p}_c)\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}$ where $\hat{p}_c = (X_1+X_2)/(n_1+n_2)$ is the pooled proportion.

Two-Sample t-Test for Means

$t = \frac{(\bar{x}_1 - \bar{x}_2) - 0}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$ Use the conservative $df = \min(n_1-1, n_2-1)$ or Welch's approximation.

Paired t-Test

When observations are naturally paired (before/after, matched subjects):

• Compute differences $d_i = x_{1i} - x_{2i}$.
• Apply one-sample t-test to the differences. $t = \frac{\bar{d} - 0}{s_d/\sqrt{n}}$

Which Test to Use?

Problem: Men ($n_1 = 100$, $\hat{p}_1 = 0.45$) and women ($n_2 = 120$, $\hat{p}_2 = 0.55$). Test if proportions differ ($\alpha = 0.05$).

Solution:

$H_0: p_1 = p_2$, $H_a: p_1 \neq p_2$.

$\hat{p}_c = (45+66)/220 = 111/220 = 0.5045$.

$SE = \sqrt{0.5045(0.4955)(1/100+1/120)} = \sqrt{0.5045(0.4955)(0.01833)} = \sqrt{0.00458} = 0.0677$.

$z = (0.45-0.55)/0.0677 = -1.477$.

p-value $= 2(0.0698) = 0.1396 > 0.05$. Fail to reject $H_0$.

• Choose the correct procedure based on data type and design.
• Always state hypotheses, check conditions, compute test statistic/p-value, conclude in context.
• Confidence intervals and hypothesis tests are two sides of the same coin.
• Use pooled proportion for two-prop z-tests; paired t for matched data.