AP Statistics Unit 3 Review: Proportions

A point estimator uses a single sample statistic to estimate a population parameter. The sample proportion p-hat is the point estimator for the population proportion p. An estimator is unbiased if its sampling distribution is centered at the true parameter value, meaning it does not systematically overestimate or underestimate. Bias and variability are separate concerns: a large sample reduces variability but does not fix a biased estimator.

When you take a random sample of size n from a population with proportion p, the sample proportion p-hat varies from sample to sample. Its sampling distribution has mean mu-p-hat = p and standard deviation sigma-p-hat = sqrt(p(1-p)/n). The distribution is approximately normal when the Large Counts condition is satisfied. Two conditions support independence of observations: the randomization condition and the 10% condition.

A one-sample z-interval for a population proportion estimates p using the formula p-hat plus or minus z-star times sqrt(p-hat(1-p-hat)/n). The standard error SE = sqrt(p-hat(1-p-hat)/n) replaces the true standard deviation because p is unknown. The margin of error equals z-star times SE. To find a required sample size, use n = (z-star)^2 times p-hat(1-p-hat) divided by (MOE)^2; if p-hat is unknown, substitute 0.5 for the most conservative estimate. Interpretation: we are C% confident the interval (a, b) contains the true population proportion. Increasing confidence level widens the interval; increasing sample size narrows it proportionally to 1/sqrt(n).

A one-sample z-test for a population proportion tests H0: p = p0 against a one-sided or two-sided alternative. The test statistic is z = (p-hat minus p0) divided by sqrt(p0(1-p0)/n). Notice that the denominator uses p0, the null value, not p-hat, because you assume H0 is true when computing the statistic. The p-value is the probability of getting a z as extreme or more extreme than observed, assuming H0 is true. Compare the p-value to the significance level alpha: if p-value is less than or equal to alpha, reject H0 and conclude there is convincing evidence for Ha; otherwise, fail to reject H0. A test can never prove H0 true. The normality condition for the test uses np0 and n(1-p0), not p-hat.

Every hypothesis test risks two kinds of errors. A Type I error is rejecting H0 when it is actually true; its probability equals alpha. A Type II error is failing to reject H0 when Ha is actually true; its probability equals 1 minus power. Power is the probability of correctly rejecting a false H0. Power increases when sample size increases, when the true parameter is farther from the null value, when alpha increases, or when standard error decreases. Before a study, consider the real-world consequences of each error type to guide choices about alpha and sample size.

When comparing two independent groups, the parameter of interest is p1 minus p2. The sampling distribution of p-hat1 minus p-hat2 has mean p1 minus p2 and standard deviation sqrt(p1(1-p1)/n1 + p2(1-p2)/n2). A two-sample z-interval estimates p1 minus p2 using (p-hat1 minus p-hat2) plus or minus z-star times sqrt(p-hat1(1-p-hat1)/n1 + p-hat2(1-p-hat2)/n2). Conditions mirror the one-proportion case but apply to both samples. Key interpretation: if the interval contains 0, there is insufficient evidence of a difference between the two population proportions; if the interval is entirely above or below 0, the data support a directional claim.

A two-sample z-test for a difference of proportions tests H0: p1 = p2 (equivalently, p1 minus p2 = 0). Because H0 assumes the proportions are equal, the normality check uses the pooled proportion p-hat-c = (n1p-hat1 + n2p-hat2) / (n1 + n2) rather than separate sample proportions. The test statistic is z = (p-hat1 minus p-hat2) divided by sqrt(p-hat-c(1-p-hat-c)(1/n1 + 1/n2)). The p-value comes from the standard normal distribution. Conclusions follow the same reject or fail-to-reject logic as the one-proportion test, stated in context with non-definitive language.

Chi-square procedures handle two-way tables of categorical counts. A chi-square test for homogeneity compares the distribution of one categorical variable across two or more populations or treatment groups. A chi-square test for independence checks whether two categorical variables are associated within a single population. Both use the same mechanics: compute expected counts as (row total times column total) divided by table total, then calculate chi-squared = sum of (Observed minus Expected) squared divided by Expected. Degrees of freedom = (rows minus 1)(columns minus 1). The p-value comes from a chi-square distribution, which is right-skewed and has only positive values. The expected counts condition requires all expected counts to be greater than 5.