inferences on proportions

The proportion of the members of the population having the trait is:

$$p=frac{# text { members wih trait }}{text { population size }}=frac{1}{N} sum_{i=1}^{N} x_{i}$$

Sample proportion, is an unbiased logical point estimator for $p$:

$$hat{p}=overline{X}=frac{1}{n} sum_{i=1}^{n} X_{i}$$

$$frac{hat{p}-p}{sqrt{p(1-p) / n}}$$

is approximately standard-normally distributed.

Then, it follows immediately that the following is a 100(1 - $alpha$)% confidence interval for p:

$$hat{p} pm z_{alpha / 2} sqrt{p(1-p) / n}$$

But we can’t use $p$ to estimate $p$. One solution is to replace $p$ by $hat{p}$:

$$hat{p} pm z_{alpha / 2} sqrt{hat{p}(1-hat{p}) / n}$$

An important question:

“How large a sample should be selected so that $hat{p}$ lies within a specified distance $d$ of $p$ with a stated degree of confidence?”

Sample size for estimating $p$, prior estimate available

$$n=frac{z_{alpha / 2}^{2} hat{p}(1-hat{p})}{d^{2}}$$

given $d=z_{alpha / 2} sqrt{hat{p}(1-hat{p}) / n}$

Sample size for estimating $p$, no prior estimate available

$$n=frac{z_{alpha / 2}^{2}}{4 d^{2}}$$

will ensure $|p-hat{p}|<d$ with 100$(1-alpha) %$ confidence.

Let $X_{1}, ldots, X_{n}$ be a random sample of size $n$ from a Bernoulli distribution with parameter $p$ and let $hat{p}=overline{X}$ denote the sample mean. Then any test based on the statistic

$$Z=frac{hat{p}-p_{0}}{sqrt{p_{0}left(1-p_{0}right) / n}}$$

is called a large-sample test for proportion.

We reject at significance level $alpha$

$$begin{array}{l} {H_{0} : p=p_{0} text { if } |Z|&gt;z_{alpha / 2}} \ {H_{0} : p leq p_{0} text { if } Z&gt;z_{alpha}} \ {H_{0} : p geq p_{0} text { if } Z&lt;-z_{alpha}} end{array}$$

Point estimator for $p_1 - p_2$:

$$widehat{p_{1}-p_{2}}=hat{p}_{1}-hat{p}_{2}=X_1/n_1 - X_2/n_2$$

Confidence Interval on $p_1 - p_2$:

$$hat{p}_{1}-hat{p}_{2} pm z_{alpha / 2} sqrt{frac{hat{p}_{1}left(1-hat{p}_{1}right)}{n_{1}}+frac{hat{p}_{2}left(1-hat{p}_{2}right)}{n_{2}}}$$

which is valid for large sample sizes.

The test

$$Z=frac{hat{p}_{1}-hat{p}_{2}-left(p_{1}-p_{2}right)_{0}}{sqrt{frac{hat{p}_{1}left(1-hat{p}_{1}right)}{n_{1}}+frac{hat{p}_{2}left(1-hat{p}_{2}right)}{n_{2}}}}$$

is called a large-sample test for differences in proportions.

We reject at significance level $alpha$

$$begin{array}{l}{H_{0} : p_{1}-p_{2}=left(p_{1}-p_{2}right)_{0} text { if }|Z|&gt;z_{alpha / 2}} \ {H_{0} : p_{1}-p_{2} leqleft(p_{1}-p_{2}right)_{0} text { if } Z&gt;z_{alpha}} \ {H_{0} : p_{1}-p_{2} geqleft(p_{1}-p_{2}right)_{0} text { if } Z&lt;-z_{alpha}}end{array}$$

Pooled estimator for $p$ when $p_1 = p_2$:

$$hat{p}=frac{n_{1} hat{p}_{1}+n_{2} hat{p}_{2}}{n_{1}+n_{2}}$$

pooled large-sample test for equality of proportions:

$$Z=frac{hat{p}_{1}-hat{p}_{2}}{sqrt{hat{p}(1-hat{p})left(frac{1}{n_{1}}+frac{1}{n_{2}}right)}}$$

We reject at significance level $alpha$

$$begin{array}{l}{H_{0} : p_{1}=p_{2} text { if }|Z|&gt;z_{alpha / 2}} \ {H_{0} : p_{1} leq p_{2} text { if } Z&gt;z_{alpha}} \ {H_{0} : p_{1} geq p_{2} text { if } Z&lt;-z_{alpha}}end{array}$$