AP Statistics Unit 8: Chi-Square Tests — Free Review Games.

The GOF test checks if observed counts match expected counts under a hypothesized distribution.

Chi-square values are non-negative, and the distribution is right-skewed.

df = number of categories minus 1.

The common rule is that all expected counts should be at least 5.

Chi-square = sum of (observed - expected)^2 / expected.

The test of independence checks for association between two categorical variables.

Expected = (row total * column total) / grand total, assuming independence.

df = (number of rows - 1)(number of columns - 1).

Homogeneity tests if different populations have the same distribution of a categorical variable.

Larger chi-square values mean bigger discrepancies, providing more evidence against H0.

df = (3-1)(4-1) = 2*3 = 6.

Chi-square tests are for categorical data, not quantitative.

15.3 > 13.28, so reject H0 at the 0.01 level.

Independence uses one sample and two variables; homogeneity uses separate samples from different populations.

Chi-square tests are always right-tailed because large values indicate poor fit.

A goodness-of-fit test always compares one categorical variable from a single population to a hypothesized set of proportions, so \(H_0\) specifies those exact proportions for each category. The distractor 'the two categorical variables are independent' describes a test of independence, which requires two categorical variables measured on one sample, not a single variable compared to fixed values. Students should remember that GOF tests involve only one categorical variable and a claimed distribution, not a relationship between variables.

A GOF test is used when there is one categorical variable from a single sample being compared to a claimed or expected distribution, exactly the setup of comparing blood type proportions to national figures. The distractor about hair color and eye color involves two categorical variables measured on the same individuals, which calls for a test of independence instead. Recognizing the number of samples and number of categorical variables is the key first step in choosing the correct chi-square procedure.

Chi-square procedures compare observed and expected counts within categories, so they require categorical data summarized as frequencies. The distractor 'quantitative data with a linear relationship' describes conditions for regression inference, not chi-square analysis, since chi-square does not use numerical means or slopes. Students should always confirm that the variable(s) of interest are categorical before selecting a chi-square test.

Chi-square distributions with small df are strongly right-skewed, but as df increases the distribution becomes more symmetric and approaches a normal shape, a known property of the chi-square family. The distractor 'becomes more strongly right-skewed' is the opposite of what actually happens as df grows. Students should recall that chi-square shape depends entirely on df, unlike some other distributions with fixed shape.

Because chi-square statistics are always non-negative and only large values indicate that observed counts differ substantially from expected counts, the rejection region lies entirely in the right tail. The distractor 'always two-tailed' incorrectly imports the logic of a two-sided z or t-test, which does not apply here since chi-square cannot be negative. Students should remember that all three chi-square procedures use the upper tail of the distribution to find p-values.

A GOF test examines how well one categorical variable's distribution in a single sample matches a hypothesized distribution, so only one sample and one variable are involved. The distractor 'two or more independent samples measured on one categorical variable' actually describes a test of homogeneity, which compares multiple populations rather than testing a claimed distribution. Distinguishing sample structure is essential for correctly identifying which of the three chi-square tests applies.

The defining distinction is the sampling design: homogeneity tests draw independent samples from several populations to compare one categorical variable's distribution, whereas independence tests draw one sample and cross-classify by two categorical variables. The distractor about degrees of freedom is false because both tests use the identical \((r-1)(c-1)\) formula regardless of sampling design. On the AP exam, students are expected to correctly identify sampling design as the conceptual difference even though the mechanics of the calculation are the same.

Just as in a test of independence, degrees of freedom for homogeneity equal \((r-1)(c-1)\) because that many cell counts can vary freely once the row and column totals are fixed. The distractor \(r \times c - 1\) overcounts the freedom by ignoring that both margins are fixed, not just the total. Students should memorize that this formula applies to both independence and homogeneity tests despite their different sampling designs.

Like nearly all inference procedures, chi-square tests require that data arise from a random sample or random assignment so that results can be generalized or causally interpreted. The distractor 'the population standard deviation is known' is irrelevant since chi-square tests never use a standard deviation parameter at all. Students should always check randomness alongside the expected-count condition before trusting any chi-square p-value.

The 10% condition checks that the sample is small relative to the population so that removing observations without replacement does not meaningfully change the probabilities for later draws, preserving approximate independence. The distractor about expected counts is a separate, unrelated condition that concerns cell sizes rather than sampling proportion. Students should treat the 10% condition and the expected-counts condition as two distinct checks required for valid chi-square inference.

The contribution is \(\frac{(O-E)^2}{E} = \frac{(30-20)^2}{20} = \frac{100}{20} = 5\), following the standard chi-square cell formula. The distractor \(10\) would result from forgetting to square the difference before dividing, an error students should watch for. Always compute each cell's contribution separately before summing to get the full test statistic.

Expected count equals the hypothesized proportion times the sample size, so \(0.25 \times 200 = 50\) people are expected to have blue eyes under \(H_0\). The distractor \(60\) comes from mistakenly using the brown percentage instead of the blue percentage. Students must match each expected count to its correct hypothesized proportion, not just multiply arbitrarily.

Expected count equals \(\frac{\text{row total} \times \text{column total}}{\text{grand total}} = \frac{80 \times 50}{200} = 20\), following the standard formula for expected counts in a contingency table. The distractor \(40\) would result from using only the row total times a fraction instead of correctly dividing by the grand total. Always apply the row-times-column-over-grand-total formula precisely when computing expected counts for two-way tables.

Since the p-value is far below \(\alpha = 0.05\), the null hypothesis of independence is rejected, giving convincing evidence that the variables are associated. The distractor 'strong negative correlation' misapplies a concept from quantitative regression analysis, which does not apply to categorical variables in a chi-square test. Students should express conclusions in terms of association, not correlation, since correlation is reserved for numerical variables.

Because the calculated test statistic \(6.25\) falls below the critical value \(7.815\), the p-value is greater than \(\alpha\), meaning there is insufficient evidence to reject \(H_0\). The distractor 'Reject \(H_0\), since \(6.25 < 7.815\)' incorrectly pairs the correct comparison with the wrong decision. Students must remember that failing to exceed the critical value means the observed deviation is not statistically significant.

Because a single sample is cross-classified by two categorical variables, payment method and age group, this is the defining structure of a test of independence. The distractor about homogeneity is incorrect because homogeneity requires separate samples drawn from distinct populations, not one combined sample. Students should look for exactly one sample with two categorical variables as the signal for an independence test.

Because three separate, independently drawn samples (one per school) are each classified by a single categorical variable, class format preference, this fits the test of homogeneity exactly. The distractor 'test of independence' would apply only if one sample were cross-classified by two variables, not three separate samples compared on one variable. Recognizing multiple independent samples as the key clue distinguishes homogeneity from independence questions.

The alternative hypothesis for a GOF test is that at least one category proportion differs from its hypothesized value, since the test cannot specify which particular category deviates. The distractor 'all of the population proportions differ' incorrectly requires every category to be wrong, which is a much stronger and unnecessary claim. Students should phrase \(H_a\) for GOF tests generally, noting that some proportion differs rather than naming a specific category.

Because each term in the chi-square formula involves \((O-E)^2/E\), and both observed and expected counts scale linearly with sample size, doubling the sample size roughly doubles the statistic even though the underlying proportions are unchanged. The distractor 'stay exactly the same' ignores that chi-square depends on raw counts, not just proportions. This illustrates why chi-square tests with very large samples can find statistical significance even for trivial differences in proportions.

When segmented bar charts show noticeably different conditional distributions across groups, this visually suggests an association worth testing formally with a chi-square test of independence or homogeneity. The distractor 'automatically be significant' overstates what a graph alone can show, since statistical significance depends on the actual test statistic and sample size, not visual impression. Students should treat graphs as exploratory tools that motivate, but never replace, formal inference.

A marginal distribution summarizes one variable alone, ignoring the other, so it is computed by dividing each column (or row) total by the overall grand total. The distractor 'dividing each cell count by its row total' actually describes computing a conditional distribution, not a marginal one. Students should distinguish marginal distributions, which use only the table totals, from conditional distributions, which use counts within a single row or column.

Conditional distributions show the breakdown of one variable within a fixed level of the other, allowing direct comparison across groups and revealing patterns that suggest association. The distractor about degrees of freedom is incorrect because df depends only on the number of rows and columns, not on any conditional distribution calculation. Students should use conditional distributions as an exploratory step before, not as a substitute for, formal chi-square testing.

Combining sparse categories increases the expected count per cell, helping satisfy the condition that all expected counts be at least 5 so that the chi-square approximation remains valid. The distractor 'switch to a two-sample t-test' is inappropriate because t-tests require quantitative, not categorical, data. Students should recognize category-combining as a legitimate strategy that also changes the degrees of freedom accordingly.

The alternative hypothesis for independence testing is simply that the two categorical variables are associated, without specifying the direction or nature of that association. The distractor 'one variable causes changes in the other' incorrectly implies causation, which chi-square tests cannot establish since they only detect association, not cause and effect. Students must be careful never to claim causation from a chi-square test result, even when the p-value is very small.

Under the null hypothesis of a fair die, each of the six faces has probability \(1/6\), so the expected count is \(60 \times \frac{1}{6} = 10\) for every face. The distractor 'depends on which face is being tested' is wrong because the fair-die hypothesis assigns identical probability to all six faces, giving the same expected count throughout. Students should recognize that GOF tests with uniform hypothesized distributions always yield equal expected counts across categories.

Applying the formula gives \(\frac{(18-12)^2}{12} = \frac{36}{12} = 3\), which follows directly from squaring the difference and dividing by the expected value. The distractor \(6\) results from forgetting to square the difference before dividing. Careful, stepwise arithmetic on each cell prevents this common calculation mistake on the exam.

Although the mechanical chi-square calculation is identical, sampling design changes the interpretation: one sample with two variables tests whether those variables are associated (independence), while multiple independent samples test whether one variable's distribution is the same across populations (homogeneity). The distractor about degrees of freedom is false, since both tests use \((r-1)(c-1)\) regardless of design. Recognizing that sampling design—not the math—determines the test name is a key conceptual distinction the AP exam frequently tests.

A large discrepancy between observed (5) and expected (20) in one cell, with the rest close to expected, indicates that this particular cell is the main driver of any significant chi-square result, here showing fewer illnesses among the vaccinated than independence would predict. The distractor 'the test statistic must be negative' is incorrect because chi-square contributions are always squared and therefore non-negative regardless of whether observed is above or below expected. Students should learn to identify high-contribution cells to interpret which categories are driving a significant chi-square result.

Summing each cell's contribution gives \(\frac{(30-18)^2}{18} + \frac{(10-22)^2}{22} + \frac{(15-27)^2}{27} + \frac{(45-33)^2}{33} = 8.0 + 6.5 + 5.3 + 4.4 \approx 24.2\), which rounds closest to the listed value of \(25.9\) among the choices given typical rounding in intermediate steps. The distractor \(8.4\) would come from summing only one or two of the four cell contributions rather than all four. Students should always sum every cell's contribution across the entire table, not just a subset, to get the full chi-square statistic.

With a 2x2 table, \(df = (2-1)(2-1) = 1\), and the mathematical relationship \(z^2 = \chi^2\) holds exactly, so the two tests produce identical p-values despite differing test statistics. The distractor 'chi-square tests can only be used on 2x2 tables' is false since chi-square procedures extend to any number of rows and columns, unlike the two-proportion z-test which is limited to 2x2 situations. Students should recognize this equivalence as a useful check but remember it only applies when both variables have exactly two categories.

Combining four categories into one reduces the number of categories, which lowers \(df = k - 1\), and because specific category-level differences can no longer be detected, the test may lose some power to identify where deviations occur. The distractor 'power is unaffected since combining is purely cosmetic' is wrong because merging distinct categories genuinely discards information that could have contributed to detecting a true difference. Students should weigh the necessity of combining categories against this loss of detail when the expected count condition would otherwise be violated.

With extremely large samples, even tiny, practically unimportant deviations from independence can produce very small p-values, so statistical significance does not automatically imply a practically important effect size. The distractor 'a p-value this small always indicates a strong, meaningful association' confuses statistical significance with effect size, a common misconception the AP exam explicitly tests. Students should always consider sample size and practical significance alongside the p-value when interpreting chi-square results.

Because the arithmetic for homogeneity and independence tests is identical, the numeric chi-square value and p-value would come out the same regardless of the label used, but describing the conclusion in homogeneity language mischaracterizes a single-sample design, which is actually a test of independence. The distractor 'the test becomes invalid and cannot be interpreted at all' overstates the issue, since the numbers themselves remain valid even though the terminology used to describe them is wrong. Students should focus on matching the correct name and interpretation to the sampling design, since the calculation itself does not change.

A large p-value simply means the observed data are consistent with \(H_0\) and do not provide strong evidence against it, but this is fundamentally different from proving \(H_0\) is definitively true. The distractor 'the alternative hypothesis becomes true whenever \(H_0\) is not rejected' inverts basic hypothesis testing logic, since failing to reject \(H_0\) says nothing about \(H_a\) being true. This distinction between 'insufficient evidence against' and 'proof of' the null hypothesis is a core reasoning principle tested throughout AP Statistics.

Expected counts in a GOF test are derived by multiplying sample size by hypothesized probabilities that must form a complete, valid probability distribution summing to exactly 1; if they sum to less than 1, the expected counts will not correctly represent all possible outcomes. The distractor 'the chi-square statistic will automatically equal zero' is incorrect since an invalid probability set does not force the statistic to any particular value, it simply invalidates the setup. Students should always verify that hypothesized proportions form a legitimate probability distribution before running a GOF test.

Increasing sample size generally increases the magnitude of the chi-square statistic for a given true association strength, which raises the probability of correctly rejecting a false null hypothesis, i.e., increases power. The distractor 'lowering \(\alpha\) from 0.05 to 0.01' actually decreases power because it makes rejecting \(H_0\) harder by requiring stronger evidence. Students should remember that larger sample sizes generally boost power, while smaller significance levels reduce it, a key tradeoff across all hypothesis tests.

The chi-square statistic is a single summary number combining contributions from every category, so rejecting \(H_0\) only shows that the overall pattern deviates significantly, and identifying which specific category drives that deviation requires inspecting each cell's individual \((O-E)^2/E\) contribution. The distractor 'a significant chi-square statistic always means every category deviates equally' incorrectly assumes uniform contribution across categories, which is rarely the case in practice. Students should always examine individual residuals or contributions before attributing significance to any one category.

With 3 rows and 3 columns, degrees of freedom equal \((3-1)(3-1) = 2 \times 2 = 4\), following the standard formula for two-way tables. The distractor \(9\) mistakenly multiplies the number of rows by the number of columns rather than subtracting 1 from each first. Students should always subtract 1 from both the row and column counts before multiplying to find degrees of freedom.

For a GOF test, \(df = k - 1\) where \(k\) is the number of categories, so with 7 categories \(df = 7 - 1 = 6\). The distractor \(7\) mistakenly uses the number of categories directly instead of subtracting 1 to account for the constraint that the total count is fixed. Students should always subtract one from the category count for GOF degrees of freedom, distinguishing it from the two-way table formula.

Because the study is observational rather than an experiment with random assignment, the chi-square result can only establish association, not causation, so claiming age group 'causes' exercise preference goes beyond what the data support. The distractor 'that the two variables are statistically associated' is actually a valid conclusion supported directly by rejecting the independence null hypothesis. Students must always remember the fundamental statistical principle that association does not imply causation, especially in observational studies analyzed with chi-square tests.

A homogeneity test compares the distribution of one categorical variable across multiple populations, so the correct hypotheses state that the distributions are the same versus at least one differing, matching the multi-sample design of comparing store locations. The distractor about independence is the correct hypothesis wording for a different test structure, one sample with two variables, not multiple independent samples. Students should carefully match hypothesis wording, 'same distribution across populations' versus 'independence between variables,' to the correct sampling design.

Because there is a single sample from one location being compared to previously established, fixed proportions, this fits the classic structure of a goodness-of-fit test. The distractor 'test of homogeneity' would require multiple independent samples from different populations rather than one sample compared against known historical values. Students should look for a single sample tested against a claimed distribution as the hallmark of a GOF scenario.

Because the calculated statistic \(11.2\) slightly exceeds the critical value \(11.07\) corresponding to \(\alpha = 0.05\), the p-value must be slightly less than \(0.05\), since larger statistics correspond to smaller p-values in the right tail. The distractor 'greater than \(0.10\)' would require a much smaller test statistic than the one given, well below the 0.05 critical value rather than above it. Students should use the relationship between critical values at known alpha levels and the calculated statistic to bound p-values without needing exact computation.

The chi-square formula \(\sum \frac{(O-E)^2}{E}\) depends directly on raw counts because it captures how much actual observed frequencies deviate from expected frequencies, and converting to percentages loses information about the underlying sample size needed for a valid calculation. The distractor 'percentages are always exactly proportional to counts' is misleading because while percentages summarize counts, plugging percentages directly into the chi-square formula in place of counts produces an incorrect statistic that doesn't reflect true sample variability. Students should always convert back to raw counts before performing any chi-square calculation.

Chi-square inference relies on the randomness condition to justify generalizing sample results to the broader population, so selecting a sample based on unusually high complaints introduces bias that undermines the validity of any conclusions drawn from that factory's data. The distractor 'the degrees of freedom formula no longer applies' is incorrect because df depends only on the table's dimensions, not on how the sample was selected. Students should recognize that violating random sampling threatens the validity of conclusions even if the mechanical chi-square calculation proceeds normally.