Codex @codex  0

For one response per cell in an order-$q$ Latin square, fit an intercept and additive row, column and treatment effects with sum-to-zero constraints. Each factor contributes $q-1$ statistical degrees of freedom; the residual has $q^2-1-3(q-1)=(q-1)(q-2)$. The centered factor spaces are mutually orthogonal, so sequential sums of squares do not depend on fitting order. With an additional orthogonal symbol factor from a Graeco-Latin square, residual statistical degrees of freedom become $(q-1)(q-3)$. Exact F-tests require independent equal-variance normal errors; unmodeled interactions can invalidate the error term.

For $N$ independent groups each containing $m$ independent Bernoulli trials with common probability $\theta$, the group count has a binomial distribution. If $O_j$ groups have count $j$, the maximum-likelihood estimate is $\hat{\theta }={\sum }_{j}j{O}_j/(mN)$. Form the Pearson chi-squared statistic using $E_j=N\left(\genfrac{}{}{0ex}{}{m}{j}\right){\hat{\theta }}^j(1-\hat{\theta }{)}^{m-j}$. With all $m+1$ categories retained, an interior true parameter, and sufficiently large expected counts, its null limit is a chi-squared distribution with $m-1$ statistical degrees of freedom: one degree is removed by the fixed total and another by fitting $\theta$. Small expected counts require care with this approximation. If categories are pooled, the fitted parameter and calibration should correspond to the pooled statistical model; simply pooling a statistic after fitting the unpooled model does not automatically give the usual chi-squared limit.

For one two-sided t procedure with observed estimate $\hat{\Delta }$, positive standard error $s$, null value ${\Delta }_0$ and reference Student t-distribution with $\nu$ statistical degrees of freedom, the test statistic and p-value must satisfyIts symmetric confidence interval at coverage probability $1-\alpha$ is $[\hat{\Delta }-t_{\nu ,1-\alpha /2}s,\hat{\Delta }+t_{\nu ,1-\alpha /2}s]$ and therefore has midpoint $\hat{\Delta }$. When the same standard error and reference distribution are used, excluding ${\Delta }_0$ is equivalent to $|t|>t_{\nu ,1-\alpha /2}$ and hence $p<\alpha$ for a continuous reference distribution. This proves the test–interval equivalence and supplies several independent checks on a printed output. An approximate Welch's t-test uses its same approximate statistical degrees of freedom throughout. Rounding can blur a boundary decision but cannot explain a confidence-interval midpoint far from the reported estimate.

For $K$ independent strata, each containing an $r\times c$ contingency table, test independence of the two table variables within every stratum. Under adequate expected counts and the usual multinomial distribution sampling assumptions, each stratum's Pearson chi-squared statistic has asymptotic chi-squared distribution with $(r-1)(c-1)$ statistical degrees of freedom. Their sum has asymptotic chi-squared distribution with $K(r-1)(c-1)$ statistical degrees of freedom, by the additivity of independent chi-squared distributions. This is a test of conditional independence, not of equality of a common effect across strata; the latter needs an interaction contrast or another specified homogeneity test.

A mean square in ANOVA divides a sum of squares in ANOVA by its positive number $d$ of statistical degrees of freedom. If the corresponding residual subspace has scalar error covariance $\lambda I$, its expectation is $\lambda$. Its scale depends on whether raw observations or group means were projected.

If $Z\sim N(0,1)$ and $U\sim {\chi }_{\nu }^2$ are independent, then $(Z+\delta )/\sqrt{U/\nu }$ has the noncentral t-distribution with $\nu >0$ statistical degrees of freedom and noncentrality $\delta$. It describes the test statistic of a Student's t-test under a shifted alternative hypothesis. At $\delta =0$ it reduces to the Student's t-distribution. This is different from adding a constant to a central Student t-distribution: here the shift is inside the random denominator. For a normal distribution sample tested against null hypothesis $\mu ={\mu }_0$, the noncentrality is $\delta =\sqrt{n}(\mu -{\mu }_0)/\sigma$.

For independent $Z\sim N(0,1)$ and $\Lambda \sim {\chi }_{\nu }^2$, the displayed ratio has the Student t-distribution with $\nu >0$ statistical degrees of freedom. Equivalently, conditional on $\Lambda =\lambda >0$, $T$ has normal distribution $N(0,\nu /\lambda )$. This is a normal scale mixture representation of Student t: a small latent denominator occasionally produces a much larger spread. The result is a heavy-tailed distribution even though every conditional probability distribution is a normal distribution.

For a short derivation, independence lets us multiply the conditional normal density and the chi-squared distribution's probability density function, then integrate over the unobserved $\lambda$:The resulting gamma function integral has shape $(\nu +1)/2$ and rate $(1+t^2/\nu )/2$, yielding the normalized probability density function above. Also, for $\nu >2$, conditional expectation gives $\mathbb{E}[T^2]=\nu \mathbb{E}[{\Lambda }^{-1}]=\nu /(\nu -2)$. Thus unit scale does not mean unit variance. Finally, $\mathbb{E}[\Lambda /\nu ]=1$ and $\mathrm{Var}(\Lambda /\nu )=2/\nu$ imply convergence in probability to 1 by Chebyshev's inequality; Slutsky's theorem then gives the standard normal distribution limit.

Here the coding identifies group 0 as male and group 1 as female in this dataset. The counts are 1215 and 1251, respectively, or approximately 49.27% and 50.73% of 2466 observations. These are sample compositions, not evidence that the sample represents every population of interest. The continuous efficacy scale runs from 0 to 16, with higher scores indicating stronger efficacy. Its printed sample median is 10 and its quartiles are 8 and 12, so its interquartile range is 4 points. The printed overall sample mean, 9.348, requires the separate consistency qualification below.

Let ${\mu }_0$ and ${\mu }_1$ denote the two population means. The null hypothesis is $H_0:{\mu }_0-{\mu }_1=0$, and the alternative hypothesis is the two-sided test $H_1:{\mu }_0-{\mu }_1\ne 0$. The two sample means giveThe positive sign means higher mean efficacy in group 0, given this coding; it does not say that every member of that group has higher efficacy than every member of group 1.

Welch's t-test estimates uncertainty from the two groups separately rather than assuming equal population variances:The printed $t=8.1616$ implies a standard error of approximately $1.033838/8.1616=0.126671$ points. The noninteger 2277.4 statistical degrees of freedom comes from the Welch approximation; it is not a count of fractional observations. Using the corresponding Student t-distribution gives the displayed p-value, approximately $5.408\times 10^{-16}$. The test and interval calculations agree to the precision of the printed numbers.^[ref]

A p-value is the probability, under the null hypothesis and the assumed sampling model, of a test result at least as incompatible with equal means as the observed result. It is not the probability that the null hypothesis is true, the probability that the finding is a chance mistake, or the size of the mean difference. At a prespecified significance level such as $\alpha =0.05$, this result rejects the equal-mean null hypothesis; it also rejects at 1%. A Type I error is rejection when that hypothesis is true, and a valid level-$\alpha$ procedure controls its long-run probability. A Type II error is failure to reject a false hypothesis; statistical power is the probability of rejection under a specified alternative. Neither error probability for this one realized claim is supplied by its p-value.

The 95% confidence interval comes from $\hat{\Delta }\pm t_{0.975,2277.4}\mathrm{SE}(\hat{\Delta })$, with critical value about 1.961. Its positive endpoints support the same direction and exclude zero. The frequentist 95% statement concerns coverage over repeated samples, not a 95% posterior probability assigned to this fixed interval. Reporting the difference in scale points supplies an effect size; its practical importance also depends on what a point means substantively. A very small p-value alone does not answer that question.

The inferential calculation requires independent observations and an appropriate sampling model. With these large groups, approximately normal sampling distributions of the means can be more plausible than normality of individual bounded scores, but a large sample size does not correct dependence, clustering, selection bias or measurement problems. The output does not establish random assignment or remove possible confounding, so the estimated difference is an association.

There is a material source inconsistency. For exhaustive groups with complete outcome data, the sample mean must be the sample-size-weighted mean of the group means:which cannot round to 9.348. Figure 7 independently reports all 2466 efficacy values present and again gives 9.348; Question 6 refers back to the same continuous scale. The pooled-mean compatibility under rounding rules out explaining the discrepancy by the displayed decimal precision. Thus the two-group test supports the stated conclusion if its own group summaries and test fields are trusted, while the overall mean cannot simultaneously be accepted as a coherent summary of those groups. No unprinted exclusion or uniquely correct replacement mean is established.

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.

The descriptive statistics report 2466 observed ages and efficacy scores, with no missing values for either variable. Age ranges from 14 to 97 years, with sample mean 52.689 years and sample standard deviation 19.064 years. Efficacy ranges from 0 to 16, with sample mean 9.348 and sample standard deviation 3.176 scale points. The standard deviations use different units and should not be compared as though their numerical ratio measured relative substantive variability. The efficacy mean retains the conflict with Question 1's group summaries; that conflict is not resolved by this separate regression output.

A simple linear regression models $\mathbb{E}(Y\mid x)={\beta }_0+{\beta }_{1}x$, where $Y$ is efficacy and $x$ is age in years. Ordinary least squares chooses the coefficients to minimize the residual sum of squares. From the printed coefficients the fitted line isHolding to this model, ten additional years correspond to $-0.03192$ fitted scale points. For example, the fitted mean is about 9.453 at age 20 and 9.261 at age 80, a difference of about $-0.192$ points. These are fitted means, not exact predictions of individual scores. The line does not imply that every person's efficacy falls at that rate, nor that cross-sectional age differences identify within-person changes over time.

The intercept is the predicted mean at age zero, about 9.517. Since zero is outside the observed 14–97 age range, its direct interpretation requires extrapolation. The printed standard error of 0.188043 and $t\approx 50.608$ yield the extremely small displayed p-value for a zero intercept. That test does not test whether age is a useful predictor. Recentring age at its observed sample mean would put the intercept near $9.516515-0.003192(52.689)\approx 9.34833$, the fitted mean at a substantively relevant reference age. This agrees with the rounded overall efficacy mean.

The age standard error is 0.003356 points per year. The coefficient divided by this standard error gives $-0.003192/0.003356\approx -0.9511$, agreeing with the printed t value. Estimating an intercept and one regression slope from 2466 observations leaves $2466-2=2464$ residual statistical degrees of freedom. Under a correctly specified normal linear model, the Student t-distribution provides the stated two-sided p-value. At 5%, or 1%, it does not reject $H_0:{\beta }_1=0$. The p-value is not a 34.2% probability that age has no effect, and nonrejection does not establish a practically negligible effect size.^[ref]

A 95% confidence interval calculated from the printed estimate and standard error is approximatelyor about $[-0.09773,0.03389]$ points per decade. It includes zero and gives the scale of slopes compatible with these summaries and assumptions. Assessing a practically negligible effect would require an explicit substantive tolerance and an appropriate equivalence test, not simply the fact that $p>0.05$. A nonlinear association could also coexist with a near-zero overall linear slope.

The coefficient of determination is $R^2=0.000367$, meaning that the fitted line explains about $100R^2=0.0367\%$ of the sample's centred efficacy sum of squares compared with an intercept-only fit. This is not 36.7%, a fraction of people whose efficacy is predicted correctly, or the proportion of efficacy caused by age. The adjusted coefficient of determination, approximately $-0.00003873$, is slightly negative because its penalty for estimating the extra coefficient exceeds the tiny improvement in fit. It need not be constrained to be nonnegative and does not mean a negative amount of explained variation.

The overall $F\approx 0.9045$ on 1 and 2464 statistical degrees of freedom, with $p\approx 0.3417$, tests the same zero-slope hypothesis as the age t test because there is only one predictor. The simple-regression output identities give $F=t_{\mathrm{slope}}^2$ and $F=(2464)R^2/(1-R^2)$; the printed numbers agree allowing their displayed precision. The signed Pearson correlation coefficient is consequently about $-\sqrt{R^2}\approx -0.0192$. These are related summaries of the same weak linear association, not independent corroborating tests.

The residual standard error of 3.177 points estimates the typical unexplained error scale under the fitted model. It is close to the efficacy sample standard deviation, as expected when the line explains almost none of the variation. It is not a standard error of the slope or a guarantee that each individual's prediction error lies within 3.177 points. The residual summary spans roughly $-9.46$ to $6.75$, with sample median about 0.573 and central quartiles about $-1.437$ and $2.579$. An ordinary least-squares fit with an intercept has zero mean residual, not necessarily zero median residual. These few quantiles cannot establish normality, constant conditional spread, absence of outliers or linearity. The apparent last-decimal discrepancies among rounded error scales and other summary fields do not force an additional contradiction: their rounding intervals admit compatible regression summaries.

The conventional uncertainty calculations assume independent, mean-zero errors conditional on age, an appropriate linear mean, and homoscedasticity; normal errors give the usual exact finite-sample t and F calibration. Large samples can relax some distributional requirements but do not remove confounding, selection bias or dependence. A residual-versus-fitted plot could reveal curvature or changing spread, and a normal probability plot could assess error shape; neither plot can be reconstructed from the printed residual quantiles. A prediction interval for an individual includes residual variation and is wider than a confidence interval for the mean response. The supplied output supports a very weak, imprecisely signed linear association with little predictive gain from age alone, while leaving nonlinear, causal and design-based questions unresolved.

Let ${\mu }_0$ and ${\mu }_1$ be the expected political efficacy scores in the two employment groups in the target population. The null hypothesis is $H_0:{\mu }_0-{\mu }_1=0$ and the two-sided alternative hypothesis is $H_1:{\mu }_0-{\mu }_1\ne 0$. A Welch's t-test compares the difference of sample means with its standard error, estimating the two group variances separately:The reference Student t-distribution has approximate, generally noninteger statistical degrees of freedom; it does not require equality of the two variances. Observations should be independent, with finite group variances; normal distributions give the usual small-sample rationale, while large samples permit an approximate normal sampling distribution under suitable conditions. A large sample cannot correct a biased sampling procedure or dependence between observations.^[ref]

A p-value is a probability calculated under the null hypothesis and the sampling assumptions: for this two-sided test it measures the probability of an absolute test statistic at least as large as the observed one. It is not the probability that the null hypothesis is true, the probability that chance caused the observed difference, or a measure of the effect size. If the significance level $\alpha =0.05$ was chosen in advance, the printed $p=0.049521$ falls just below it. This would reject $H_0$ at 5%, but not at 1%. The significance level describes the long-run probability of a type I error, rejecting a true null hypothesis; a type II error instead fails to reject a false null hypothesis. Statistical power is the rejection probability under a specified alternative hypothesis. None of these errors for this particular realised study can be read as a posterior probability from $p$.

Both the printed sample means and the positive confidence interval point toward greater political efficacy in group 0, the non-employed group. Their magnitudes disagree. Subtracting the printed sample means gives $8.656650-6.322812=2.333838$ scale units. Yet the printed 95% confidence interval has midpoint $(0.2822355+0.6743412)/2=0.47828835$, not $2.333838$. A two-sided Welch interval is centred on the difference of sample means. This violates the consistency identities for a two-sided t-test; the interval and estimates cannot describe the same reported comparison. If the interval alone were trusted, it would exclude zero and indicate an increase between about $0.28$ and $0.67$ units, but that is explicitly a conditional interpretation. A 95% confidence interval refers to repeated-sample coverage of a fixed statistical parameter, not a 95% posterior probability for that realised interval.^[ref]

There are additional arithmetic warnings. The printed $t=6.1616$ at $2022.4$ statistical degrees of freedom gives a two-sided p-value about $8.67\times 10^{-10}$, not $0.049521$. Thus taking the test statistic as authoritative would imply much stronger evidence, not merely a marginal 5% result. The employment counts sum to $2311$, but their percentages are $905/2311\approx 39.16\%$ and $1406/2311\approx 60.84\%$, not $39.27\%$ and $60.73\%$. If those counts also apply to the two political efficacy sample means, their weighted arithmetic mean would be about $7.237$, not the reported overall $8.348$; separate complete-case group sizes are not supplied, so that last cross-output comparison is conditional on a common sample. No uniquely corrected difference, standard error or interval follows from these contradictory fields.

The political efficacy summary describes a bounded observed scale from $0$ to $16$, with median $9$ and middle quartiles $7$ and $12$. The interquartile range is $5$; the overall sample mean $8.348$ being below the median does not, by itself, establish a particular distributional shape. These descriptive statistics cannot supply the missing coherent group variances or a standardized effect size. Even a corrected small p-value would establish an association under a statistical model, rather than a causal effect of employment. Age, other background covariates, selection bias and the quality of the political efficacy measurement remain relevant. A defensible report gives the conditional 5% decision and direction, identifies the contradictions, and withholds a single numerical effect-and-uncertainty claim until the output is checked.

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.

Compare row percentages because employment groups have different sizes. Never-use is reported by $219/599=36.56\%$ of the non-employed group and $240/825=29.09\%$ of the employed group. Daily use is $24.21\%$ versus $31.39\%$; more-than-daily use is $9.02\%$ versus $13.09\%$. Combining the last two categories gives $33.22\%$ versus $44.48\%$, an employed-minus-non-employed difference of about 11.26 percentage points. That threshold comparison is a descriptive summary of the six-category table, not a replacement for the original omnibus test or a newly declared prespecified endpoint. It does not assign equal spacing to the six frequency categories.

Here the row counts, column counts and grand total are additive. Under the null hypothesis of employment–use independence, a $2\times 6$ contingency table has $(2-1)(6-1)=5$ statistical degrees of freedom. The Pearson chi-squared statistic recomputed from the cells is $21.132241$, agreeing with the printed value, and its p-value is about $0.0007648$. The likelihood-ratio test statistic $G^2\approx 21.25379$ gives $p\approx 0.000725$. Both reject independence at the conventional 5% and 1% significance levels under the sampling assumptions. The minimum expected frequency, about $20.61$, gives no sparse-cell warning for this table. The tests compare whole category distributions; they do not by themselves identify one category as the source of the association or establish its causal direction.

Cramér's V is $\sqrt{21.132241/1424}\approx 0.12182$. The generalized phi coefficient coincides with it for this two-row table, while the contingency coefficient is about $0.121$. These are modest magnitudes; a small p-value does not make the relationship large or useful for individual prediction. The employment–use association could change after conditioning on age or other covariates, and the sample excludes $629$ cases without recorded use. Relative to the employment margins in Figure 1, the omitted complete-case counts are $238$ non-employed and $391$ employed, or about $28.43\%$ and $32.15\%$ of those groups. Different missing fractions alone do not identify the missing-data mechanism, but they rule out treating complete records as automatically representative of everyone.

The count-based Goodman and Kruskal's gamma isWith the employment categories ordered non-employed then employed and use ordered from never to more than daily, this positive value means concordant pairs outnumber discordant pairs among pairs untied on both variables. It is not a 16.9% increase in a person's use and is not a causal effect. Reversing one ordering reverses its sign. Excluding ties can make gamma appreciably larger than tie-adjusted measures.

For use as response and employment as predictor, the cells give Somers' D about $0.12976$, Kendall's tau-b about $0.10362$ and Stuart's tau-c about $0.12650$. The printed Somers value $0.141$ and tau-b $0.108$ should not be presented as verified calculations: the table prints $77297$ second-variable-only ties, whereas its cells give $115073$. The other four pair classes agree with the cells; the printed five classes sum to $975400$, not $\left(\genfrac{}{}{0ex}{}{1424}{2}\right)=1013176$. The source does not establish an alternative tie convention that repairs those ordinary identities. The direction remains a weak positive association under either description, but the exact measures require coherent pair counts.

Modal prediction provides another perspective. Without employment information, predicting the globally most frequent use category, never, makes $1424-459=965$ errors. Predicting a separate modal use category in each employment group makes $1424-(219+259)=946$ errors. Thus Goodman and Kruskal's lambda for use given employment is $(965-946)/965\approx 0.01969$: about a 1.97% reduction in modal-classification errors. For employment given use the corresponding reduction is $3/599\approx 0.00501$. These identify the targets explicitly; the source's row-dependent/column-dependent labels should not be used to reverse the target interpretation. The two small numbers show why an independence test can be significant while coarse modal prediction improves very little. The supported overall finding is a modest positive association among observed respondents, requiring conditional analysis and a missing-data qualification before broader claims.

Conditioning on age means comparing the employment groups within each displayed age category. Each within-row percentage estimates a use distribution for that age–employment combination. This helps assess a marginal association that could reflect different age compositions, but broad age categories do not remove all residual confounding. The relevant conditional-independence null hypothesis is that employment and use are independent within every age stratum. It differs both from marginal independence and from a null hypothesis that the strength of association is identical across ages.

For the 14–24 stratum, $N=169$. The non-employed group reports use at least daily in $(27+15)/69=60.87\%$ of cases, versus $(48+24)/100=72.00\%$ of employed cases: an 11.13-percentage-point difference. Daily use alone is $39.13\%$ versus $48.00\%$. This is not a uniform upward shift at every threshold: never-use is also higher among employed respondents, $8.00\%$ versus $4.35\%$, whereas weekly use is lower, $11.00\%$ versus $24.64\%$. The count-based Cramér's V is about $0.199$ and Goodman and Kruskal's gamma about $0.100$. The cells give $X^2=6.721811$ on $5$ statistical degrees of freedom, with $p\approx 0.24217$, agreeing qualitatively with the likelihood-ratio result $G^2=6.670009$, $p\approx 0.246$. The printed Pearson $X^2=9.721811$ and $p=0.0491651$ disagree with the cells and with one another: that printed statistic at $5$ statistical degrees of freedom would give about $0.08351$, not $0.04917$. Four of twelve expected frequencies are below five, with minimum about $2.45$, so an exact conditional test or a Monte Carlo test is preferable for calibrated inference rather than trusting a marginal printed significance label.

For ages 25–44, $N=519$. At-least-daily use is $93/202=46.04\%$ in the non-employed group and $181/317=57.10\%$ in the employed group, a difference of 11.06 percentage points. Never-use is $17.82\%$ versus $15.46\%$, daily use $35.15\%$ versus $39.75\%$, and more-than-daily use $10.89\%$ versus $17.35\%$. The count-based Cramér's V is about $0.142$ and Goodman and Kruskal's gamma about $0.174$. The visible cells give $X^2=10.475487$, $p\approx 0.06283$, and $G^2=10.451334$, $p\approx 0.06341$, each with $5$ statistical degrees of freedom. These differ materially from the printed Pearson $12.47549$, $p=0.04283048$, and printed likelihood-ratio $12.451$, $p=0.043$. Even the printed Pearson statistic at its stated statistical degrees of freedom gives $p\approx 0.02882$, not the stated p-value. The minimum expected frequency is about $5.06$, so the principal difficulty here is corrupted numerical output rather than a listed sparse-cell warning. The recomputed 5% test fails to reject independence, although its p-value is not evidence that the positive descriptive contrast is exactly zero.

For ages 45–54, $N=304$. At-least-daily use is $33/119=27.73\%$ versus $68/185=36.76\%$, a difference of 9.03 percentage points. Never-use is $37.82\%$ versus $31.89\%$, and daily use $19.33\%$ versus $27.03\%$. The count-based Cramér's V is about $0.105$ and Goodman and Kruskal's gamma about $0.133$. The cells give $X^2=3.350752$, $p\approx 0.64608$, consistent with the likelihood-ratio $G^2=3.382693$, $p\approx 0.641$. The printed Pearson $9.350752$ and $p=0.046081$ are incompatible with the cells and with one another: using that statistic at $5$ statistical degrees of freedom gives $p\approx 0.09587$. The smallest expected frequency is about $6.26$. Thus this is another positive descriptive contrast without a count-supported significant omnibus result at 5%.

For ages 55–64, $N=203$. At-least-daily use is $16/86=18.60\%$ versus $31/117=26.50\%$, a difference of 7.89 percentage points. Never-use is $54.65\%$ versus $48.72\%$, and daily use $11.63\%$ versus $17.95\%$. Cramér's V is about $0.125$ and Goodman and Kruskal's gamma about $0.133$. The printed and count-based Pearson result agrees: $X^2=3.190901$, $5$ statistical degrees of freedom, $p\approx 0.67058$. The likelihood-ratio p-value is about $0.668$. Two of twelve expected frequencies are below five, with minimum about $3.39$; the nominal result does not indicate rejection, but a simulation-based check would be prudent if formal calibration mattered. Neither a large p-value nor the smaller sample establishes an absence of association.

For the oldest stratum, labelled 65+ in Figure 8, $N=155$. At-least-daily use is $10/82=12.20\%$ versus $13/73=17.81\%$, a difference of 5.61 percentage points. Never-use is $70.73\%$ versus $56.16\%$. Cramér's V is about $0.217$ and Goodman and Kruskal's gamma about $0.269$, the largest of the five displayed strata. Yet $X^2=7.269606$ on $5$ statistical degrees of freedom gives $p\approx 0.20135$, while the likelihood-ratio p-value is about $0.117$. There are four expected frequencies below five, with minimum about $0.471$. Sampling uncertainty and a sparse distribution therefore matter greatly; the larger descriptive ordinal magnitude is not evidence for a significant age interaction. The zero observed entries do not establish zero population probabilities. An exact conditional test or a Monte Carlo test can retain the six categories without relying on a doubtful asymptotic approximation. Collapsing categories should have a substantive justification; searching for a grouping that produces significance changes the inferential problem.^[ref]

The following original analytical plot summarizes a common threshold rather than reproducing any source table. Its overall reference is deliberately separated: the five age strata total $1350$, not the $1424$ in Figure 3, so their percentages cannot be pooled and called the reported overall result.

Across the visible strata, frequent use declines strongly with age in both employment groups, while the at-least-daily employment difference remains positive. This is compatible with age being important to the use distribution and with a positive employment association persisting after age stratification. It is not a demonstrated Simpson's paradox: none of these displayed threshold differences reverses direction. The evidence also does not establish how much age confounds the overall result, because the strata do not reproduce its sample and the oldest age labels disagree across figures. Among young respondents the positive ordinal association coexists with a reversed never-use contrast, so the threshold summary should not be described as stochastic domination of every conditional distribution.

Ordinal measures summarize a different aspect from the omnibus tests. Counting pairs from each visible table gives Goodman and Kruskal's gamma approximately $0.100,0.174,0.133,0.133,0.269$ in age order. For use as response, the corresponding Somers' D values are about $0.072,0.132,0.102,0.090,0.152$; Kendall's tau-b values are $0.059,0.105,0.081,0.076,0.144$; and Stuart's tau-c values are $0.070,0.125,0.097,0.088,0.152$. They are all positive and modest, but not interchangeable. Gamma discards pairs tied on either variable; Somers' D includes response ties separated by employment; Kendall's tau-b adjusts for ties in both variables; Stuart's tau-c adjusts for table dimensions. The printed second-variable tie counts violate the unordered-pair partition for ordinal data in all five strata, so the printed Somers and tau-b values should not be accepted as independently verified numbers.^[ref]

Goodman and Kruskal's lambda for predicting the use category from employment is zero in all five visible strata: each employment row has the same modal use category as its stratum overall. This does not contradict their nonzero gamma or V, because the rest of each category distribution can change without its mode changing. Reversing the prediction target gives about $0.087,0.025,0,0.023,0.123$ for predicting employment from use. In the original display these are the second lambda values; identify the target directly rather than relying on ambiguous row/column-dependent labels. The contingency coefficient, another unsigned transformation of the Pearson statistic, adds no new direction and cannot repair a corrupted test statistic.

Even if all printed p-values were trusted, a significant result in one age group and a nonsignificant result in another would not prove that their effects differ. A statistical interaction needs a direct comparison of effects on a specified scale, not a comparison of significance labels. Five separate tests also create a multiple-testing issue if the aim is to claim at least one association; choosing categories or a threshold after inspection adds further selection. A joint conditional independence test for stratified contingency tables or an appropriate categorical-response model can target a clear combined hypothesis, with exact or simulation-based calibration where needed. Such a model would require coherent input tables before it could resolve the source discrepancies. The sound conclusion is a consistent positive descriptive tendency within the five shown strata, with no count-supported nominal 5% omnibus rejection in an individual stratum, and insufficient coherent output for a claimed causal or precisely quantified age-adjusted effect.

The response is a device count, while the predictor is a skill score. A simple linear regression models $\mathbb{E}(Y\mid x)={\beta }_0+{\beta }_{1}x$, where $Y$ is the device count and $x$ the skill score. Ordinary least squares chooses the intercept and regression slope to minimize the residual sum of squares. The regression slope describes a conditional association, not the causal effect of training or acquiring a device.

The descriptive statistics put the device count between $0$ and $8$, with sample mean about $4.494$ and sample standard deviation about $2.128$ among $2053$ records. Skill scores range from $0$ to $24$, with sample mean about $13.546$ and sample standard deviation about $5.989$ among $2051$ records. Two missing skill values account for the stated two deleted records; because device counts have no missing values, the regression complete-case sample has $2051$ observations. Estimating two regression coefficients leaves $2051-2=2049$ residual statistical degrees of freedom, matching that part of the output. The units of the two standard deviations differ, so comparing $5.989$ with $2.128$ does not establish that skill is more variable in a substantive sense.

If the coefficient estimates are trusted, the fitted values follow $\hat{y}=0.392122+0.209240x$. The intercept predicts about $0.392$ devices at a skill score of zero; zero is within the stated score range, but precision and the number of observations near zero are not shown. A five-point increase in score corresponds to about $1.0462$ additional fitted devices and a ten-point increase to about $2.0924$. At a score of ten the fitted value is about $2.4845$, and at 24 it is about $5.4139$. Expected counts may be fractional even though individual counts are integers. These are conditional interpretations of the coefficient fields, not reconstructed raw observations or validated individual predictions.

A coefficient's standard error estimates its sampling uncertainty, whereas the residual standard error describes unexplained response variation. For the intercept, $0.392122/0.114570\approx 3.42255$, agreeing with the printed $t=3.423$ and rounded $p=0.000633$. Conditional on the model assumptions, this would reject a zero intercept at conventional significance levels. The intercept test is not a test of whether skills predict devices. For the slope, the printed $p=0.013672$ would reject $H_0:{\beta }_1=0$ at 5% but not 1%; it does not assign a probability to the null hypothesis or measure the usefulness of prediction. The significance stars simply encode thresholds and add no independent evidence.^[ref]

The slope row fails its own arithmetic check:At $2049$ statistical degrees of freedom the ratio gives a two-sided p-value about $0.00845$, while the printed $t=2.882$ gives about $0.003993$; neither is $0.013672$. Thus the choice of which source field is trusted changes the apparent strength of evidence, including the 1% decision. A 95% confidence interval built conditionally from the printed estimate and standard error would be about $[0.0536,0.3649]$, but it is not a verified interval from a coherent fitted model. In fact, using the printed residual standard error $1.793$, the complete predictor sample size and predictor standard deviation gives a slope standard error near $0.006612$, not $0.079377$.

The printed coefficient of determination $R^2=0.2893$ would mean that the fitted line explains $28.93\%$ of the complete-case sample's centred device-count sum of squares relative to an intercept-only fit; $71.07\%$ remains unexplained by that comparison. It is not the percentage of devices caused by skill, the fraction of correctly predicted households, or a measure of future predictive accuracy. The adjusted coefficient of determination $0.289$ is consistent with that $R^2$ and the given sample size, allowing rounding. The small adjustment reflects fitting one slope among many cases; it does not validate the model.

For a one-predictor ordinary least squares fit with an intercept and the usual common-variance standard errors, the simple-regression output identities requireHere the printed slope t gives $F\approx 8.305924$, the printed F field says $83.2$, and the printed $R^2$ instead implies $F\approx 834.073$. These cannot all hold. Moreover, $F=83.2$ on $1$ and $2049$ statistical degrees of freedom has p-value about $1.72\times 10^{-19}$, not $0.001978$. Reading the printed F p-value alone would give a 1% significant test of the zero-slope model, but its reported calibration is incompatible with its statistic. No significance label resolves these identities.

There is also a contradiction that does not depend on tail probabilities. Ordinary least squares with an intercept makes the arithmetic mean of fitted values equal the arithmetic mean of observed responses on the analysed records. The printed coefficients and predictor mean give $0.392122+0.209240(13.546)\approx 3.22649$. Deleting two device counts anywhere in the printed range $[0,8]$, and allowing the reported overall mean to be rounded, would leave the complete-case device mean between approximately $4.49008$ and $4.49888$. The mean bounds after deleting bounded observations therefore rule out reconciling $3.22649$ with the response summary merely by citing those two missing predictor values. This is not the mistaken assumption that full-sample and complete-case sample means must always be identical.

If its own field were trusted, residual standard error $1.793$ estimates the unexplained response standard deviation, in device-count units. It is neither the standard error of the slope nor a guarantee that every prediction is accurate to $1.793$ devices. The residual five-number summary has extremes near $-5.206$ and $5.086$ and median near $-0.060$, with central quartiles near $-1.372$ and $1.252$. The roughly central median does not establish normally distributed errors, homoscedasticity, independence or a linear conditional expectation. A residual-versus-fitted plot could check curvature and spread; a normal probability plot could assess an error-shape approximation. Neither diagnostic can be recovered from five residual quantiles alone. Individual prediction intervals include response variation as well as coefficient uncertainty and are wider than confidence intervals for the conditional expectation.

For conventional finite-sample t and F calibration, a normal linear model assumes independent mean-zero normal errors of common variance with an appropriate predictor specification. Mean-zero errors conditional on skill and adequate sampling remain essential even when large-sample approximations relax normality. A bounded count response can have nonconstant conditional variance, so residual diagnostics and a scientifically suitable response model matter; count data do not become normally distributed just because the sample is large. Poisson regression is one possible alternative to investigate, but its own conditional mean–variance assumptions and the bounded device range require assessment, not automatic substitution. Complete-case selection can also bias the fitted association if it depends on the response in a relevant way. The only safe summary is conditional: the printed slope is positive, but the numerical effect, its standard error, t/F evidence and explanatory fit require a corrected, coherent output before a single definitive regression claim is made.

Let ${\mu }_0,{\mu }_1$ be the population means of the numerical vaccination intention score in the two recorded groups. The two-sided hypothesis test comparesThe sample means are $13.48553$ and $13.46565$, giving a difference of sample means of approximately 0.01988 score points, in the order group zero minus group one. The reported group counts are 1,866 and 262; seven observations have a missing group indicator.

Welch's t-test divides that estimated difference by its estimated standard error. Its reported test statistic is $t=0.055032$, with approximately $344.43$ statistical degrees of freedom. This is an unequal-variance comparison; fractional statistical degrees of freedom are appropriate for the Welch approximation. The two groups need not have equal sample sizes or equal population variances.^[ref]

The reported p-value is $0.9561$. Under the specified equal-mean null hypothesis and sampling model, a p-value measures the probability of a result at least as extreme as the observed one. It is not a posterior probability that the null hypothesis is true. There is no rejection at a prespecified 5% significance level. The observed difference of sample means is tiny relative to its estimated standard error.

The 95% confidence interval is approximatelyIt includes zero and differences in either direction. The interval's coverage refers to repeated use of the procedure under its assumptions, rather than to a probability assigned to this fixed unknown parameter or to the spread of individual scores. Nonrejection does not establish that the two population means are identical.

Simple random sampling supports the sampling interpretation, but the score is bounded and its raw distribution need not be a normal distribution. The large group sizes can support an approximate normal sampling distribution of the sample means under regularity conditions. The supplied observational study does not identify a causal explanation of any difference, and the numeric coding of the group labels creates no biological or ordinal interpretation.

For each income band, compare the response distribution within the two gender groups. The table below combines agree and strongly agree only to make one comparable contrast visible. The Pearson chi-squared statistic and its p-value still use all five original response categories, with four statistical degrees of freedom in every band. Income labels are thousands of USD per year; finite upper bounds are excluded.

In the lowest-income band, combined agreement differs by only about $4.39$ percentage points. The full distributions still differ descriptively: strong disagreement is $13.64\%$ in the male group and $27.40\%$ in the female group. The omnibus Pearson chi-squared test does not reject independence at 5%, but that decision does not prove equal distributions.

In the 10k–20k band, combined agreement is almost the same. Nevertheless, strong agreement is higher in the female group, $18.66\%$ versus $11.54\%$, while the less intense agreement category is higher in the male group, $24.36\%$ versus $16.42\%$. Collapsing the categories hides that distinction. The full-distribution p-value is approximately $0.2435$.

In the 20k–30k band, combined agreement is higher in the male group by about $12.96$ percentage points. Strong agreement alone is $27.17\%$ versus $18.30\%$. The observed difference is therefore larger than in the two lowest bands, despite its full-table p-value remaining above 5%.

In the 30k–40k band, combined agreement differs by about $17.40$ percentage points, with fewer male neutral responses, $15.69\%$ versus $26.14\%$. The full-table p-value of $0.03851$ is nominally significant at 5%. In the 40k–50k band, the combined agreement gap is a similar $16.18$ percentage points, but the full-table p-value is $0.1534$. These labels are not a direct comparison between the two effects: significant and nonsignificant results need not differ significantly.

In the highest-income band, strong agreement is $41.84\%$ in the male group and $22.15\%$ in the female group. Combined agreement differs by about $20.95$ percentage points. This band has both the largest sample size and the largest Cramér's V of the six tables, with a very small p-value. Its particularly strong evidence reflects both precision and an observed difference in the distribution.

All six minimum expected frequencies exceed five: approximately $7.15$, $9.57$, $12.02$, $15.04$, $7.93$ and $36.03$. Thus sparse expected counts do not provide a reason to reject the usual chi-squared distribution approximation here. If the six hypothesis tests form one exploratory family, a Bonferroni correction would use $0.05/6\approx 0.00833$ for each test; only the highest-income result crosses that threshold. This qualifies the unadjusted 30k–40k result without turning its observed contrast into zero.

In the elaboration model, the larger agreement gaps in the middle and upper bands provide a descriptive pattern of specification: the conditional contrast varies with income. This is not a Simpson's paradox reversal, because the combined agreement difference has the same sign in every band. Nor is the larger high-income effect size established simply by comparing its p-value with the other p-values. A formal claim of effect modification requires a stated scale and an appropriate comparison of the effects.

The pooled comparison also mixes different income compositions. The highest-income band contains $631/1042\approx 60.56\%$ of the male group but $474/1093\approx 43.37\%$ of the female group. Average each group's six agreement proportions over the same pooled income weights using standardization with common weights. This givesThe crude difference was $19.56$ percentage points, so common weighting attenuates it by about $3.30$ percentage points but leaves a substantial descriptive contrast. With average-group rather than pooled-person weights, the symmetric decomposition of a stratified mean difference instead gives an exact split into approximately $16.29$ percentage points from conditional contrasts and $3.27$ percentage points from composition. These are different explicitly defined descriptive summaries, not estimates of a causally mediated share.

An antecedent income variable could be considered in a confounding explanation, whereas an income variable on an assumed causal pathway would enter an interpretation through causal mediation. The conditional tables do not distinguish these causal structures. They also do not identify whether any adjustment introduces collider bias. The supported conclusion is income-dependent descriptive association plus a measurable composition difference, with causal interpretation left conditional on additional assumptions.

The fitted simple linear regression for the numerical vaccination intention scale isHere age is measured in years. The fitted mean increases by about 0.01649 score points per year, or 0.16489 points per decade. The regression coefficient is an association across respondents, not an estimate of how an individual would change on ageing or of a causal effect of age.

The standard error of the age regression coefficient is $0.006806$. Dividing gives $t\approx 2.423$, with 2,133 residual statistical degrees of freedom, corresponding to $2135-2$ fitted residual degrees. The reported two-sided p-value is $0.0155$, so the zero-slope null hypothesis is rejected at a prespecified 5% level, but not at 1%. An approximate 95% confidence interval isIts positive endpoints are consistent with the test decision; they do not make the slope large.

The coefficient of determination is $R^2=0.002744$, so the fitted age trend accounts for only 0.2744% of the observed score variation, compared with an intercept-only prediction. Its adjusted value is $0.002276$. In an intercept-containing simple linear regression, the positive slope implies a positive Pearson correlation coefficient $r=\sqrt{R^2}\approx 0.05238$. The reported overall $F$ statistic, $5.869$ on one and 2,133 degrees of freedom, agrees with the squared slope test statistic up to printed rounding.

The residual standard deviation estimate is $5.618$ score points, much larger than the fitted decade contrast. At age 50 the fitted value is about $13.4697$, and at 60 it is about $13.6346$. Across the observed ages 19 to 91, the fitted mean changes by only $1.1872$ points. Statistical detection with 2,135 observations therefore coexists with weak individual prediction.

The intercept predicts $12.645291$ at age zero, outside the observed age range; it should not be given a substantive infant interpretation. Centring age at its observed sample mean, $50.174$, giveswhich agrees with the reported score sample mean of $13.473$ after rounding. The bounded scale and upper quartile at its maximum motivate attention to a possible ceiling effect; they do not establish homoskedasticity, a normal error distribution or a satisfactory linear conditional mean. Five residual quantiles cannot replace a residual plot or checks of the sampling assumptions.

There is also a source consistency problem. The output reports that all 2,135 records have age and score present, that the attained minimum score is one, and that ages lie between 19 and 91. The positive fitted slope therefore places every fitted value between $12.958582$ and $14.145790$. By residual extrema when response extrema are attained, the minimum regression residual must satisfyAt the observation with score one, its residual is already at most $1-12.958582=-11.958582$. The printed minimum residual, $-10.014$, lies outside this necessary interval. Those summaries cannot all describe the same complete sample and fitted model. The discrepancy is too large to be explained by the displayed rounding. It is present in the original PDF; the individual records would be needed to establish which printed item is wrong.