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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.

These are row percentages, so the comparisons estimate the conditional probability of recorded vaccination within each NHS trust category. The percentages increase across the ordered trust categories asThe extreme-category difference is approximately 49.31 percentage points, rather than a 49.31% relative increase. The ordinal categorical variable records an order; the categories do not establish equal numerical trust increments.

Under the null hypothesis of independence, the expected count in a cell is its row total times its column total divided by 748. The Pearson chi-squared statistic compares observed and expected counts throughThere are $(5-1)(2-1)=4$ statistical degrees of freedom, giving $p\approx 6.49\times 10^{-21}$. The minimum expected count is approximately 5.015, so the usual expected-count concern is much less acute here than in the income strata. The table gives strong evidence of a statistical association. The likelihood-ratio test of independence in a contingency table agrees: its statistic is approximately 81.361 on four statistical degrees of freedom, with a very small p-value. The displayed 0.000 is rounding, not a zero probability.

For magnitude, Cramér's V isIt describes the size of the departure from independence without using the category order or giving a direction. Calling it “moderate” is only a rough contextual description; the actual percentage differences are easier to interpret.

The ordered coefficients describe a positive direction. Goodman and Kruskal's gamma uses only concordant pairs and discordant pairs:It excludes tied pairs, which are frequent when the response has only two categories. It is therefore not a claim that 54.7% of respondents were vaccinated because of their trust. The directional Somers' D for vaccination as response includes pairs tied on that response in its denominator:The smaller value reflects the retained ties; it does not contradict the positive statistical association.

The small column-dependent Goodman and Kruskal's lambda has a useful prediction interpretation. Without trust information, always predicting the more frequent vaccinated category gives 121 mistakes. Predicting the majority category separately in each trust row gives $13+15+26+34+28=116$ mistakes. HenceTrust reduces this particular modal-prediction error by only about 4.13%, despite the strong evidence against independence. Most rows retain the same majority category. Evidence of association, its ordinal direction and improvement in a chosen prediction rule are different questions. None establishes that changing trust would change an individual's vaccination status.

Treat the two row totals as fixed sample sizes, with independent multinomial observations within each row. More generally let $N_{ij}$ be an $r$ by $c$ table, row totals $n_i$, column totals $m_j$ and total $N$. Under the unrestricted alternative, row $i$ has its own category probabilities $p_{ij}$; independence means that all rows have the same probabilities $p_j$.

Ignoring the multinomial coefficients, which cancel in a likelihood ratio, the unrestricted likelihood is ${\prod }_{ij}{p}_{ij}^{N_{ij}}$. Maximizing ${\sum }_j{N}_{ij}\log p_{ij}$ under ${\sum }_j{p}_{ij}=1$ gives ${\hat{p}}_{ij}=N_{ij}/n_i$. Under the null, the log likelihood is ${\sum }_j{m}_j\log p_j$, giving ${\hat{p}}_j=m_j/N$. Thus the null fitted counts are $E_{ij}=n_i{m}_j/N$, and the generalized likelihood ratio isA zero observed count contributes zero by continuity. Small $\Lambda$, equivalently large $G^2$, is evidence against independence. The unrestricted model has $r(c-1)$ free probabilities and the null has $c-1$, a difference of $(r-1)(c-1)$. Under the usual large-sample likelihood-ratio approximation at an interior null, $G^2$ has a chi-squared distribution with that many degrees of freedom. This is the likelihood-ratio test of independence in a contingency table. Sampling the entire table multinomially instead gives the same statistic and degrees of freedom after estimating the row probabilities.

For the supplied data the fitted counts areSubstitution givesThere are $(2-1)(3-1)=2$ degrees of freedom. The supplied 1% critical value is $9.21$, and $0.9701<9.21$. Do not reject independence at the 0.01 level. The approximate chi-squared tail probability is $e^{-G^2/2}\simeq 0.616$. This lack of rejection does not establish that the classifications are independent; the test supplies no evidence against that null here. All fitted counts are at least ten, making the stated asymptotic approximation reasonable.

Treat the cell counts as independent Poisson random variables. Under independence of Month and Hospital, the independence log-linear model for a two-way contingency table iswith no Month–Hospital interaction. Compare its Poisson deviance with the appropriate upper quantile of a chi-squared distribution. This is the likelihood-ratio test of independence in a contingency table.

The approximation assumes independent counts with correctly specified Poisson means, identifiable parameters, and sufficiently large fitted cell means for the asymptotic chi-squared law to be accurate. Equivalently, one may condition on the margins and use the corresponding multinomial sampling formulation.

For the $12\times 3$ month table there are $36$ cells and $1+11+2=14$ independent model parameters, so the residual degrees of freedom areSincemodel 1 does not reject Month–Hospital independence at the $1\%$ level.

After combining months into four quarters, the table is $4\times 3$, givingdegrees of freedom. Nowso model 2 rejects Quarter–Hospital independence at the $1\%$ level.

There is no contradiction. Relabelling twelve months as four quarters aggregates the contingency table, changes the null hypothesis, and reduces the degrees of freedom. The quarter-level hospital pattern is coherent enough to cross the much lower six-degree-of-freedom threshold even though the more detailed month-level omnibus test does not. This is an instance of how aggregation can change a contingency-table independence test.

Both variables are ordinal categorical variables, and the proposed alternative is directional: higher scores should tend to occur at higher prices. Assign increasing numerical scores to low, medium, high price and to poor, good, excellent quality, and use a one-sided linear-by-linear association test. Its statistic is based on the sample covariance, or equivalently the correlation, of those scores under the fixed margins.

This test uses one degree of freedom and directs its rejection region toward positive association. The four-degree-of-freedom likelihood-ratio test of independence in a contingency table spends power on every possible departure from independence, including unordered and negative patterns, so the ordered one-sided test is more powerful when the stated alternative is credible.