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The haplotype-relative-risk design compares transmitted alleles to the corresponding untransmitted parental copies, used as internal controls for an affected offspring. If allele-1/allele-2 transmitted counts are $w,x$ and untransmitted counts are $y,z$, the allele-1 odds ratio estimate is $wz/(xy)$. An unmatched Pearson chi-squared test of independence treats this as a two-by-two table. The transmission disequilibrium test instead keeps parent-specific pairing and uses only discordant transmissions. The sampling assumptions of the unmatched calculation and any missing-parent reconstruction must be justified separately.

Read the body percentages within the populism rows. There are 50 Yes responses out of 447 Not-populist respondents and 218 out of 1182 Populist respondents. The corresponding No percentages are 76.73% and 75.38%, while Not know falls from 12.08% to 6.18%. The overall distribution therefore changes in more than just the Yes category. The row-total percentages, 27.44% and 72.56%, describe group composition in the complete-case table, not the conditional blocking rates. Their slight difference from Question 2's 28.25% and 71.75% is consistent with different missing-value patterns.

The Pearson chi-squared test of independence compares the two full response distributions. Under the null hypothesis of independence, each cell has expected frequency $E_{ij}=n_{i+}{n}_{+j}/N$, andThe minimum expected count is about 34.849, comfortably above the usual sparse-cell concern. With independent observations and a suitable sampling model, the tiny p-value rejects that omnibus null at conventional significance levels. The likelihood-ratio test of independence in a contingency table gives $G^2\approx 24.6044$, also on 2 statistical degrees of freedom, with calculated $p\approx 4.54\times 10^{-6}$. Its printed 0.000 is a rounded value, not zero.^[ref]

The generalized phi coefficient and Cramér's V both equal $\sqrt{X^2/N}\approx 0.124$ because the smaller nontrivial table dimension is one. The contingency coefficient is $\sqrt{X^2/(X^2+N)}\approx 0.123$. These are unsigned magnitude measures, not further significance tests or percentages of explained variation. A modest magnitude can be highly significant with a large sample size.

The omnibus test does not, by itself, identify which response contrast differs. To target reported Yes specifically, an explicitly derived binary table combines No and Not know as non-Yes reports, producing rows $(50,397)$ and $(218,964)$. An uncorrected two-by-two Pearson test gives $X^2\approx 12.4292$ on one statistical degree of freedom, $p\approx 0.000423$. A large-sample confidence interval for the difference of reported-Yes proportions is approximately $[0.0359,0.1092]$, or 3.59 to 10.92 percentage points. These are additional count-derived calculations under independent binomial sampling, not values printed in the exam. They address the selected Yes contrast and do not pretend that Not know and No are substantively identical in every respect. As aggregation can change a contingency-table independence test explains, the binary and three-category tests have different null hypotheses.

The printed ordinal coefficients, Goodman and Kruskal's gamma 0.295, Somers' D 0.116, Kendall's tau-b 0.117 and Stuart's tau-c 0.093, are arithmetically consistent with the displayed ordering. For example, 134660 concordant and 73239 discordant pairs give $\gamma =(134660-73239)/(134660+73239)\approx 0.295$. The full tie counts also partition all $\left(\genfrac{}{}{0ex}{}{1629}{2}\right)=1326006$ unordered pairs correctly. However, the three response labels do not establish an ordinal scale. Their arithmetic sign therefore cannot replace the direct Yes comparison or be called the proportion of variation explained. Changing the arbitrary order of Not know and No could change the ordinal summary without changing the nominal association.^[ref]

Both directional values of Goodman and Kruskal's lambda are zero. No is the modal blocking answer in both populism groups, and Populist is the modal classification within each blocking column. Knowing the other variable therefore changes no modal prediction enough to reduce the number of classification errors. This does not contradict the unequal conditional distributions or the chi-squared result. Before generalizing, missing outcomes, missing classifications, the sampling design and possible confounding must be considered; a p-value does not supply those assumptions.

This table has 1692 observations, all respondents with an observed populism classification, because age has no missing values. The within-age Populist percentages are 66.77%, 73.18%, 66.38%, 74.03% and 77.12%, in ascending age-group order. Their Not-populist complements are 33.23%, 26.82%, 33.62%, 25.97% and 22.88%. Comparing the oldest and youngest Populist percentages gives about 10.35 percentage points, but the intermediate reversal prevents describing a uniform rise at every age transition. The row totals 310, 302, 348, 308 and 424 are the classified cases; their differences from the age totals in Figure 1 sum to the 92 missing classifications.

The Pearson chi-squared test of independence gives $X^2\approx 15.8680$ on $(5-1)(2-1)=4$ statistical degrees of freedom, $p\approx 0.003201$. The likelihood-ratio test gives $G^2\approx 15.8572$ and $p\approx 0.003217$. Both reject independence at 5% and 1% under the usual sampling assumptions. The minimum expected frequency is about 85.317, so the large-sample approximation is not threatened by sparse expected counts here. Neither test is specifically a test of a linear or monotone age trend.

The generalized phi coefficient and Cramér's V are about 0.097, and the contingency coefficient about 0.096: small unsigned association magnitudes. If the binary classification is coded Not populist before Populist and age groups are ordered younger to older, the ordinal output has a meaningful conventional direction. The 258482 concordant and 205602 discordant pairs give Goodman and Kruskal's gamma about 0.114; Somers' D for classification given age is about 0.046, Kendall's tau-b 0.065 and Stuart's tau-c 0.074. These small positive summaries describe an overall older–Populist tendency while allowing the middle-age reversal. They use different tie or table-size adjustments and are not percentages of respondents caused to change classification by ageing.

Goodman and Kruskal's lambda for predicting classification from age is zero because Populist remains the modal classification in every age row. In the reverse direction, predicting age group from classification improves modal predictions slightly: without classification the most frequent age group has 424 cases, whereas choosing a separate modal age group for each classification yields $117+327=444$ correct predictions. The error reduction is $(444-424)/(1692-424)\approx 0.0158$, matching the displayed 0.016. Direction must therefore be stated, rather than treating both lambda numbers as interchangeable. Age is a relevant control variable for the later comparison, but these cross-sectional associations do not establish an age-related causal mechanism or remove other confounders.

Greater technology distrust is associated with lower efficacy in the overall sample. The omnibus evidence against independence is strong, while the unsigned association magnitude is modest: $X^2\approx 107.727$ on 6 statistical degrees of freedom, $p\approx 6.10\times 10^{-21}$ and Cramér's V about 0.148. The observational study does not establish that changing trust would cause efficacy to change.

Rows identify the technology response and columns identify efficacy. Each body percentage is a within-row conditional probability estimate, not a column percentage. For example, $166/336=49.40\%$ of those who strongly disagree with the distrust statement have High efficacy, whereas $199/718=27.72\%$ of those who agree do. The difference is about 21.69 percentage points. High efficacy decreases across all four rows: 49.40%, 42.41%, 31.02%, 27.72%. Low efficacy is 20.54% in the strongly-disagree row, 18.00% in Disagree, 21.43% in Neutral and 34.26% in Agree. Thus the overall pattern is clear without claiming that every category changes monotonically: the low-efficacy proportion initially falls, and the Medium proportion peaks at 47.55% among neutral respondents.

For the Pearson chi-squared test of independence, the null hypothesis says that the two categorical variables are independent, equivalently that the efficacy distribution is the same in every technology-response row. With row totals $n_{i+}$ and column totals $n_{+j}$, the expected frequency under that hypothesis is $E_{ij}=n_{i+}{n}_{+j}/N$, andAll twelve expected frequencies exceed 79, with minimum about 79.844, so sparse expected cells are not the concern here. Under independent sampling and the usual large-sample calibration, the very small p-value rejects independence at conventional significance levels. This is an omnibus finding: it does not provide a separate adjusted significance test for every pair of row percentages.^[ref]

The likelihood-ratio test of independence in a contingency table uses $G^2=2{\sum }_{O_{ij}>0}{O}_{ij}\log (O_{ij}/E_{ij})$. The supplied $G^2\approx 105.146$ with 6 statistical degrees of freedom gives $p\approx 2.11\times 10^{-20}$. Its displayed 0.000 is rounded, not an exact zero. It corroborates the omnibus finding. The generalized phi coefficient is $\sqrt{X^2/N}\approx 0.209$, the contingency coefficient is about 0.205, and Cramér's V is $\sqrt{X^2/[N\min (3,2)]}\approx 0.148$. These are unsigned measures of association strength rather than three additional significance tests; they do not describe causal effects or percentages of variance explained. Their magnitude and the small p-value answer different questions, especially with a large sample size.

The ordinal measures use the displayed order: increasing row index means increasing distrust, while increasing column index means decreasing efficacy. There are 882033 concordant pairs and 552543 discordant pairs. Hence Goodman and Kruskal's gamma isIts positive sign therefore means that increasing distrust tends to accompany movement toward lower efficacy. Under an ascending efficacy order its sign would be negative, by sign reversal under reversal of one ordinal ordering. Goodman and Kruskal's gamma ignores tied pairs; it is not a claim that 23% of efficacy variation is explained. The displayed pair counts partition all $\left(\genfrac{}{}{0ex}{}{2466}{2}\right)=3039345$ unordered pairs correctly, including ties. Somers' D for efficacy given technology is about 0.151, Kendall's tau-b about 0.158, and Stuart's tau-c about 0.163. Their different denominators handle ties or table dimensions differently, so their differing magnitudes do not constitute disagreement about the direction.^[ref][ref]

Goodman and Kruskal's lambda concerns the proportional reduction in errors when predicting a most frequent category, not ordinal movement. Without technology information the modal efficacy category is Medium, producing $2466-972=1494$ errors. Choosing a separate modal efficacy category in each technology row yields $2466-(166+391+233+273)=1403$ errors. Thus ${\lambda }_{\mathrm{efficacy}\mid \mathrm{technology}}=91/1494\approx 0.061$. In the reverse prediction direction the reduction is $80/1544\approx 0.052$, matching the other printed lambda. The improvement in modal classification is small even though independence is decisively rejected. Both the conditional percentages and these distinct summaries support a real sample association; ethnicity and other possible confounders still require attention before any causal account.

Read the body percentages within each ethnic group. Strong disagreement with the distrust statement occurs among 24.68% of Asian respondents, 10.53% of Black respondents and 13.34% of White respondents. Agreement occurs among 27.27%, 21.05% and 29.37%, respectively. Neutral responses are 20.78%, 26.32% and 19.68%. Thus Asian respondents are especially likely to be in the strongly-disagree category, while Black respondents have the largest neutral share. The mode for Black and White respondents is Disagree; Asian respondents have a tie between Disagree and Agree at 21 observations each. The small group totals, 77 and 57, mean that a few observations can materially change the minority-group percentages.

The Pearson chi-squared test of independence has $X^2\approx 12.60265$, $(3-1)(4-1)=6$ statistical degrees of freedom and $p\approx 0.049798$. Under its sampling assumptions, this is a nominal rejection at a prespecified 5% significance level, but not at 1%. The minimum expected frequency is about 7.766, so this particular table has no expected cell below five. The alternative likelihood-ratio test of independence in a contingency table gives $G^2\approx 11.52267$ and $p\approx 0.0735$, printed as 0.074. These are different test statistics, not contradictory versions of the same calculation. Their different threshold decisions show that the evidence is close enough to the conventional boundary to warrant qualification rather than selection of the more favourable label.

The generalized phi coefficient is about 0.071, the contingency coefficient about 0.071 and Cramér's V about 0.051. They indicate a small overall unsigned association. This does not erase the particular descriptive percentage differences, and a small p-value is not a measure of their substantive size. Conversely, failure of the likelihood-ratio test to reject at 5% is not evidence that every ethnic group has exactly the same population response distribution.

Ethnicity is a nominal categorical variable, with no natural rank Asian–Black–White. Therefore the printed Goodman and Kruskal's gamma of 0.079, Somers' D of 0.058, Kendall's tau-b of 0.022 and Stuart's tau-c of 0.009 have no invariant substantive ordinal interpretation here. Permuting the ethnicity labels can change the ordinal pair comparisons without changing the underlying observations or their nominal association. These values are not evidence for an ethnic hierarchy; Cramér's V and the actual within-group percentages are appropriate summaries for this question.^[ref]

Both directional values of Goodman and Kruskal's lambda are zero. Knowing ethnicity does not improve the number of correct modal predictions of technology response: Disagree remains a modal category within every ethnic group, including the Asian tie. In the reverse direction, White is the most frequent ethnicity within every technology-response column, so knowing technology does not improve modal ethnicity predictions either. Zero lambda therefore coexists with unequal conditional distributions and a nonzero Pearson chi-squared statistic; modal prediction discards changes that do not change the most frequent category. The supported conclusion is a small, weakly evidenced nominal association, with its descriptive pattern and uncertainty stated explicitly.

Each table fixes ethnicity and compares the efficacy conditional distributions across technology responses. The body percentages use the technology-response row total within that stratum. The same labels, coding direction and efficacy order apply as before. The strata contain 77, 57 and 2332 observations; they sum to 2466 and their cells reproduce the overall table exactly. Consequently the comparison is genuinely a stratification of the supplied overall counts, rather than a comparison of unrelated samples.

For Asian respondents, High efficacy is 78.95% among those who strongly disagree with the distrust statement, falling to 23.81% among those who disagree, 18.75% among neutral respondents and 19.05% among those who agree. The strong-disagreement versus agreement contrast is about 59.90 percentage points, based on 15 of 19 versus 4 of 21 respondents. This is a large descriptive contrast, but the rest of the table is not a simple monotone progression: Low efficacy peaks at 47.62% in Disagree, versus 10.53% in Strongly disagree, 18.75% in Neutral and 23.81% in Agree; Medium efficacy exceeds 57% in the last two rows. The striking first row is part of a more irregular categorical pattern.

The Asian Pearson chi-squared test of independence gives $X^2\approx 28.41883$, 6 statistical degrees of freedom, $p\approx 7.83\times 10^{-5}$. The likelihood-ratio test gives $G^2\approx 27.20973$, $p\approx 0.0001323$; its displayed 0.000 is a rounding convention. Cramér's V is about 0.430, compared with 0.148 overall. The generalized phi coefficient is 0.608 and the contingency coefficient 0.519. Two of twelve expected frequencies are below five, with minimum 4.156. This is a modest sparse-cell warning rather than a reason to delete the observed cells, but asymptotic calibration should still be qualified and could be checked by an exact or fixed-margin Monte Carlo test.

The Asian Goodman and Kruskal's gamma is about 0.323, from 1025 concordances and 524 discordances, so its displayed-order direction agrees with the overall association. Somers' D for efficacy given technology is about 0.226, Kendall's tau-b 0.241 and Stuart's tau-c 0.253. These stronger ordinal summaries do not imply that every individual row transition is monotone. Goodman and Kruskal's lambda is about 0.362 for predicting efficacy from technology and 0.286 in the reverse direction: the technology row changes the modal efficacy prediction sufficiently to reduce its errors by about 36.2%. This is a reduction in modal-classification errors, not a fraction of efficacy variance explained.

For Black respondents, High efficacy is 0 of 6, or 0%, in Strongly disagree; 8 of 24, or 33.33%, in Disagree; 2 of 15, or 13.33%, in Neutral; and 7 of 12, or 58.33%, in Agree. Low efficacy is 16.67%, 25.00%, 20.00% and 8.33%, respectively. The strongest-disagreement versus agreement High-efficacy contrast is therefore $-58.33$ percentage points, opposite to the Asian and White contrasts. The zero observed High count in the first row does not prove zero population probability, especially with only six respondents in that row. These percentages are compatible with a reversed tendency, but considerable sampling uncertainty is unavoidable.

The Black Pearson chi-squared test of independence gives $X^2\approx 10.80719$, 6 statistical degrees of freedom and $p\approx 0.09452$. The likelihood-ratio test gives $G^2\approx 12.26209$, $p\approx 0.05637$, also above 5%. Thus neither nominal test rejects at 5%; this does not establish exact independence. Eight of twelve expected frequencies are below five, with minimum about 1.158, making the usual large-sample calibration particularly questionable. An exact conditional analysis or Monte Carlo test would be more defensible for a calibrated inferential claim; no uncomputed simulated p-value is asserted.^[ref]

The Black Cramér's V is about 0.308, the generalized phi coefficient 0.435 and the contingency coefficient 0.399. An unsigned magnitude of this size can coexist with a nonsignificant test in a small sample. There are 254 concordances and 457 discordances, giving Goodman and Kruskal's gamma about $-0.286$, Somers' D about $-0.179$, Kendall's tau-b about $-0.191$ and Stuart's tau-c about $-0.187$. Their negative signs agree with the reversed displayed-order direction. Goodman and Kruskal's lambda is about 0.107 for efficacy given technology and zero for technology given efficacy. In the latter direction, Disagree remains a modal technology category within every efficacy column, so modal prediction gains nothing despite the changed conditional percentages.

For White respondents, High efficacy decreases from 48.55% in Strongly disagree through 43.10% and 32.03% to 27.45% in Agree. The extreme-row contrast is about 21.11 percentage points, from 151 of 311 versus 188 of 685. Low efficacy is 21.22%, 17.10%, 21.57% and 35.04%, respectively. This closely resembles the overall distribution, including the fact that the Low category is not monotone at every step. The White Pearson chi-squared test of independence gives $X^2\approx 107.70835$, 6 statistical degrees of freedom and $p\approx 6.15\times 10^{-21}$; the likelihood-ratio test gives $G^2\approx 105.54201$, $p\approx 1.75\times 10^{-20}$. All expected frequencies exceed 74, so sparse cells are not the problem for these nominal tests.

The White Cramér's V is about 0.152, the generalized phi coefficient 0.215 and the contingency coefficient 0.210. Goodman and Kruskal's gamma is about 0.236, Somers' D 0.155, Kendall's tau-b 0.162 and Stuart's tau-c 0.167: close to the overall ordinal summaries and in the same direction. The two values of Goodman and Kruskal's lambda are about 0.061 for efficacy given technology and 0.062 for technology given efficacy. They again indicate only a small improvement in modal prediction. The large sample size helps explain why the White test has a much smaller p-value than the smaller strata, without implying that its association magnitude is the largest.

The original figure below plots selected empirical conditional probabilities from all the cells. It gives the opposite small-stratum pattern and the dominant White pattern a common scale; it is not a reproduction of a source figure, an estimated continuous trend or a display of uncertainty intervals.

The elaboration model asks how the original association changes when a third variable is held fixed. If the association disappeared throughout the strata, an explanation by the control variable might be plausible, subject to a credible causal ordering. Here it persists in the White group, which supplies 94.57% of the sample, and is stronger on several descriptive measures in the Asian group. Therefore ethnicity does not simply explain away the relationship. The White result is a replication of the broad pattern; the reversed Black direction and the Asian shape suggest descriptive specification or possible effect modification. The output supplies no temporal or causal design that would distinguish an actual causal mechanism from other confounding, measurement differences or sampling variation. It also does not demonstrate a Simpson's paradox in which the association reverses in every stratum: that is not what the displayed tables show.

The predominance of White respondents explains why the pooled pattern resembles theirs. By the law of total probability, each technology row's efficacy distribution is a mixture over ethnicity, using the ethnic proportions within that technology row. Those weights need not equal the overall ethnic proportions. As an optional descriptive comparison, standardization with common weights using the overall ethnic shares gives High-efficacy proportions of about 48.38% in Strongly disagree and 27.90% in Agree, a 20.48-percentage point contrast, compared with the crude 21.69-point contrast. This explicitly defined adjustment leaves the broad contrast intact; it is not a causal estimate and does not turn the three different stratum patterns into one common effect.

Finally, a significant test in one group and a nonsignificant test in another is not itself a test that the groups' associations differ. A formal statistical interaction requires a directly specified heterogeneity test or model on a chosen effect scale. The omnibus tests also address whole categorical distributions, whereas gamma and the High-efficacy contrast describe particular ordered features. Multiple stratum tests create a multiple testing issue if used to select a discovery. The defensible result is an association that survives stratification in the dominant group, with important descriptive heterogeneity that merits a direct, sparse-cell-aware investigation rather than an assertion that ethnicity causes the difference.

Each row percentage estimates an employment proportion within that age group. For example, $105/182=57.69\%$ of the youngest displayed group are employed. The employed proportions in the next four rows are $61.57\%$, $62.26\%$, $56.69\%$ and $57.23\%$. The largest displayed proportion is in the 45–54 group and the smallest in the 55–64 group, a difference of about $5.57$ percentage points. These are conditional percentages, not the age distribution of employed people. The sequence rises and then falls; there is no simple increasing or decreasing pattern.

The null hypothesis for a Pearson chi-squared test of independence is that the joint category probabilities factor into their age and employment marginals. Its expected frequencies are $E_{ij}=n_{i+}{n}_{+j}/n$, and its test statistic adds the standardized squared count discrepancies, $X^2={\sum }_{ij}(O_{ij}-E_{ij}{)}^2/E_{ij}$. For a complete $r\times c$ contingency table with positive margins, the large-sample reference chi-squared distribution has $(r-1)(c-1)$ statistical degrees of freedom. Independent observations and an appropriate sampling scheme are essential; adequate expected counts make the approximation more plausible.^[ref]

Reading only the printed test block, $X^2=4.83463$, $5$ statistical degrees of freedom and $p\approx 0.4364$ are mutually compatible. The printed likelihood-ratio test gives nearly the same p-value, about $0.435$. At a prechosen 5% significance level neither rejects independence. This is a failure to find evidence, not proof of equal employment probabilities at every age. The printed Cramér's V around $0.049$ suggests a small sample association; the contingency coefficient around $0.048$ has the same broad message. The generalized phi coefficient is an unsigned chi-squared magnitude here, not evidence for a positive age trend. A small association magnitude and a large p-value answer different questions.

But only a $5\times 2$ table is actually visible. Those rows total $1730$, with employment column sums $698$ and $1032$, rather than the printed $2053$ with columns $837$ and $1216$. Computing solely from the five visible rows givesIts likelihood-ratio test statistic is about $4.03243$, again with $4$ statistical degrees of freedom. The smallest count-based expected frequency is about $73.43$, comfortably large for this ordinary approximation, but different from the printed minimum $74.20068$. These figures are a labelled analysis of the visible subset, not a repaired full dataset. An unprinted sixth age row might explain some intended metadata, but neither its label nor a uniquely justified reconstruction is supplied. The oldest printed label also conflicts with Figures 1 and 8.

The ordinal measures look for an ordered tendency. Goodman and Kruskal's gamma compares concordant pairs with discordant pairs among pairs untied on both variables. The printed negative gamma, Somers' D, Kendall's tau-b and Stuart's tau-c therefore suggest, at most, a very weak tendency toward lower employment at higher age categories under the specified ordering. Their precise magnitudes are not reliable here. The ordinary unordered-pair partition for ordinal data fails for the printed pair counts, even when compared with the printed $2053$ cases. On the five visible rows the respective count-based values are about $-0.0356$, $-0.0172$ for employment conditional on age, $-0.0218$ and $-0.0266$. None gives a substantial monotone pattern; a near-zero ordinal measure would not exclude a nonmonotone relationship in general.

Goodman and Kruskal's lambda is zero in both predictive directions on the visible table: knowing age does not improve the number of correct predictions made by always choosing the most common employment category, and knowing employment does not improve modal age-category prediction. Zero modal-prediction improvement does not mean every category distribution is identical. Both the reported test block and the separately recomputed visible-table test fail to reject independence; neither justifies asserting no association, a causal age effect, or a coherent full-sample calculation from the contradictory printout.

Arrange rows by marker genotype, with cases and controls as the two columns. Both column totals are 100, and the three row totals are 100, 60 and 40. Under the null of no genetic association, estimate each common row probability from its pooled proportion. The expected count is row total times column total divided by the grand total. Thus the expected contingency table isThe Pearson chi-squared test of independence usesUnder independent sampled individuals and the no-association null, its reference chi-squared distribution has $(3-1)(2-1)=2$ degrees of freedom. All expected cells are well above five. The null is overwhelmingly rejected: the approximate p-value is $e^{-93.44/2}\simeq 5.13\times 10^{-21}$. This establishes association in the sampling model, not by itself a causal role for the marker.

Each diploid individual contributes two marker alleles. Count $A$ copies as twice the $AA$ count plus the $Aa$ count, and count $a$ copies analogously. The observed chromosome-level contingency table, followed by its null expected table, isThere are 200 chromosomes in each group; pooled totals are 140 $A$ and 260 $a$. The usual allelic Pearson chi-squared test of independence givesWith the ordinary independent-chromosome null model, use one degree of freedom; its p-value is about $1.04\times 10^{-25}$, providing very strong evidence of genetic association.

The two alleles from the same person are not automatically two independent observations. The usual one-degree-of-freedom calibration is justified, for example, for unrelated individuals with Hardy-Weinberg equilibrium under the null. If that assumption is not suitable, retain individuals as the sampling units and test their allele dosages using an appropriate variance or permutation of case-control labels, or use the genotype-level test from part (iii). No Hardy-Weinberg equilibrium assumption was needed for that genotype-level test.

For the first sampling design, treat the fixed groups of 60 and 90 people as two independent, representative samples from the respective populations. Within each sample, assume independent individuals with common eye-colour probabilities and mutually exclusive, exhaustive categories. This gives two independent multinomial distributions. The null hypothesis is that the two populations have the same eye-colour probabilities; its estimated common probabilities are $40/150$, $70/150$ and $40/150$.

The expected counts under the null hypothesis areThe Pearson chi-squared test of homogeneity usesThere are $2(3-1)=4$ unrestricted category parameters and $3-1=2$ under the null hypothesis, so the asymptotic chi-squared distribution has two degrees of freedom. All expected counts exceed five, making the usual large-sample approximation reasonable. Reject equality at the 5% level, but not at the 1% level: $5.99<T<9.21$. The approximate p-value is $\mathbb{P}({\chi }_2^2\ge T)=e^{-25/7}\simeq 0.0281$. Independence and representative sampling, rather than merely the count sizes, are essential to population inference; the result does not establish a causal explanation.

In the second design, the total sample size is fixed but both row and column classifications are observed for each sampled person. Assume an independent, representative sample with a common joint distribution of handedness and eye colour. The six cell counts then have one multinomial distribution. The null hypothesis is independence of the two classifications, so joint probabilities factor into row and column probabilities. Their maximum-likelihood estimates are the observed marginal proportions, again giving exactly $E$ above.

The unrestricted joint distribution has five free parameters; independence leaves $(2-1)+(3-1)=3$. The Pearson chi-squared test of independence again has two degrees of freedom, the same statistic, and the same significance conclusion. The numerical analysis is unchanged, but the sampling model changes from two fixed-row samples to one joint sample. If handedness groups had instead been deliberately sampled with fixed row sizes, the first, homogeneity formulation would apply to those groups too.

The unmatched haplotype-relative-risk counts satisfy $w=a+b$, $x=c+d$, $y=a+c$ and $z=b+d$. For the raw table this givesThe Pearson chi-squared test of independence in this two-by-two table has statisticThus the unmatched calculation gives nominal evidence at 5%, with $p\simeq 0.030$. It differs from the transmission disequilibrium test because it uses an unmatched table, and it also inherits the missing-parent selection issue from the raw counts.

For the ten complete trios, $(w,x,y,z)=(12,8,6,14)$, giving odds ratio $7/2$ and the unmatched Pearson chi-squared test of independence statistic $40/11\simeq 3.636$, with $p\simeq 0.057$. That calculation does not reject at 5%. The nominal significance from the raw table therefore should not be presented as a robust association conclusion without a justified model for incomplete families and the unmatched-table assumptions.

Under the null hypothesis that treatment and recovery are independent, estimate each expected cell count byand form the Pearson chi-squared test of independence statistic ${\chi }^2=\sum (O-E{)}^2/E$. For a $2\times 2$ table its asymptotic null law is a chi-squared distribution with one degree of freedom. A second-order Taylor expansion of the generalized likelihood-ratio test statistic $-2\log \Lambda$ around the fitted counts gives this Pearson statistic, which explains its asymptotic null distribution.

Writing the four residuals over their fitted counts and simplifying givesFor $(a,b,c,d)=(50,10,15,5)$,whose one-degree-of-freedom p-value is about $0.41$. There is no evidence of an effect. One fitted count is only $3.75$, however, so the chi-squared approximation is questionable; Fisher's exact test would be preferable.

Under the null hypothesis of independence between treatment and outcome, the expected counts in each treatment row are one half of the column totals:The Pearson chi-squared test of independence statistic isThe number of degrees of freedom isAt the 5% level the critical value is $7.81$. Since $8.30>7.81$, we reject the null hypothesis and find statistically significant evidence that the drug and placebo have different effects.

For observed counts $O_{ij}$ and positive expected counts $E_{ij}$ in a contingency table, the statistic is $X^2={\sum }_{ij}(O_{ij}-E_{ij}{)}^2/E_{ij}$. Under common category probabilities or statistical independence, estimate $E_{ij}$ by the product of the row and column totals divided by the grand total. With large expected counts, this gives the test statistic for the Pearson chi-squared test of homogeneity or the Pearson chi-squared test of independence, with asymptotic chi-squared distribution on $(r-1)(c-1)$ degrees of freedom.

The Pearson chi-squared test of homogeneity tests whether several independent categorical samples have the same category probabilities. Under the null, expected cell counts equal the product of the corresponding row and column totals divided by the total sample size. The Pearson chi-squared statistic for contingency tables $\sum (O-E{)}^2/E$ has an asymptotic chi-squared distribution with $(r-1)(c-1)$ degrees of freedom, when expected counts are sufficiently large. The sampling design differs from a Pearson chi-squared test of independence, though the statistic has the same form.

The Pearson chi-squared test compares observed categorical counts with null expected counts using the displayed statistic. Its applications include Pearson chi-squared goodness-of-fit test for specified probabilities and Pearson chi-squared test of independence for contingency tables; the fitted restrictions determine the limiting degrees of freedom.