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An alpha-spending function assigns the cumulative Type I error budget to successive information fractions in a group sequential design. Its total at the final analysis is the overall significance level. The rejection boundaries must be calibrated using the joint distribution of the sequential test statistics; independent fixed-level tests do not automatically spend the assigned budget correctly.

Consider independent binomial counts $X_i\sim \mathrm{Bin}(n_i,p)$ under the common-probability null hypothesis, with fixed positive row sizes $n_i$. Let $N={\sum }_i{n}_i$, $K={\sum }_i{X}_i$ and assume $0<K<N$. Conditional on $K$, the probability mass function is the displayed multivariate hypergeometric distribution: in the joint likelihood, the factor $p^K(1-p{)}^{N-K}$ is common to all admissible tables and cancels. The Vandermonde identity gives the denominator, so the conditional distribution is free of the unknown nuisance parameter $p$.

For a table $x$, its Pearson chi-squared statistic isAn exact conditional p-value sums the displayed probabilities over all tables with the same margins and $Q(x)\ge Q(x_{\mathrm{obs}})$. The ordering is by the Pearson chi-squared statistic, rather than by the probability of an individual table; different exact-test orderings can give different answers. Enumeration is practical when $K$ and the number of rows are small. Monte Carlo tests can approximate the same tail by sampling tables with fixed margins.^[ref]

An exact arithmetic comparison is easy too. Let $M$ be the least common multiple of the row sizes. Compare the integersince every table has the same positive denominator $MK(N-K)$. This includes boundary ties without a floating-point tolerance. Conditional p-values are discrete and can be conservative; “exact” concerns their null calibration, not equality of rejection probability with every chosen significance level.

The expected rate of detections exceeding a stated threshold under a specified background-only model. A time rate and a per-trial significance level require different normalizations.

The one-sample Kolmogorov-Smirnov test checks goodness of fit by comparing an empirical distribution function $F_n$ with a specified cumulative distribution function $F_0$. Its Kolmogorov-Smirnov statistic $D_n$ is the largest vertical gap anywhere between the two curves. Every threshold asks the same concrete question: what proportion of the observations are at or below it, and what probability does the proposed probability distribution assign there? A large discrepancy supplies evidence against the null hypothesis.

For example, four stipulated observations $(1/10,2/5,4/5,9/10)$ compared with the uniform distribution on $[0,1]$ give $D_4=3/10$. Just before $4/5$, only two observations have been counted, so the empirical distribution function approaches $1/2$, while $F_0$ approaches $4/5$. The Kolmogorov-Smirnov statistic section calculates every jump-side gap. These are exact illustrative numbers, not experimental measurements or a claim of statistical significance.

The ordinary one-sample calibration assumes independent and identically distributed real observations and a simple hypothesis $H_0:F=F_0$, with $F_0$ a continuous function fixed before inspecting these observations. All statistical parameters of $F_0$ must be specified.^[ref] Under that null hypothesis, the probability integral transform gives $U_i=F_0(X_i)$ with the uniform distribution on $(0,1)$, retaining independence. Consequently the sampling distribution of $D_n$ depends on the sample size $n$, but not on the particular continuous $F_0$. This is the useful sense in which the calibration is distribution-free.

For the exponential distribution of rate one, $F_0(x)=1-e^{-x}$ for $x\ge 0$. Take $x_i=-\log (1-u_i)$ for the four stipulated $u_i$ above. The probability integral transform changes the horizontal coordinates but preserves both the order statistics and their cumulative heights, so the same Kolmogorov-Smirnov statistic $3/10$ appears on both scales. More generally, for continuous $F_0$, the jump-side formulas depend only on $F_0(X_{(i)})$. Strict increase everywhere is unnecessary: intervals on which $F_0$ is constant have probability zero under the null hypothesis. This explains the invariance without assuming a strictly positive probability density function.

The observed Kolmogorov-Smirnov statistic is a discrepancy, whereas its p-value is a tail probability under the null hypothesis:Here $D_n^{\mathrm{new}}$ is computed from a fresh sample of the same sample size. Choose a significance level $\alpha$ before the test and reject when the calibrated p-value is at most $\alpha$. The value $D_4=3/10$ alone supplies neither a p-value nor a rejection decision. Failure to reject does not establish that the null hypothesis is true. A finite-sample calibration and the asymptotic Kolmogorov-Smirnov theorem are distinct; a cutoff for $D_n$ must be rescaled before use with $\sqrt{n}{D}_n$.^[ref]

Fitting statistical parameters on the tested observations replaces $F_0$ by the random $F_{\hat{\theta }}$. The resulting transformed observations need not be independent and identically distributed uniforms, so the ordinary reference law fails. The Lilliefors test addresses a particular fitted normal distribution. More general parametric bootstrap calibration must simulate from the fitted statistical model and repeat the fitting in every replicate; holding the fitted statistical parameters fixed during those comparisons calibrates a different procedure. Such a parametric bootstrap requires its own validity justification rather than inheriting exactness from the simple-null test.

For a discrete probability distribution, $F_0(X)$ is generally not uniform. Equal observations also arise from rounding, so tied data require a calibration that accounts for the discrete probability distribution or the rounding rule. The discrepancy remains meaningful, but continuous-null p-values are not automatically valid. Statistical dependence likewise changes the reference law. The two-sample Kolmogorov-Smirnov test compares two empirical distribution functions and has a different calibration.^[ref]

For the same normal distribution model, a two-sided test compares the null hypothesis $H_0:\mu ={\mu }_0$ with the alternative hypothesis $H_1:\mu \ne {\mu }_0$. Under the null hypothesis, the standardized test statistic has standard normal distribution. Write $\Phi$ for the standard normal distribution function. With $\bar{x}=1.2$, $\sigma =2$, $n=16$ and ${\mu }_0=0$, the observed test statistic is $z=1.2/0.5=2.4$. Its p-value isThus the p-value is a tail probability computed under the null hypothesis, including departures in both directions. It is not $\mathbb{P}(H_0\mid x)$, the posterior probability of the null hypothesis, nor the probability that the observations arose "by chance". The normal distribution test statistic and its calibration appear in NIST's teaching reference for testing a process mean; the tail interpretation is stated in Penn State's hypothesis-testing notes.

Choose the significance level $\alpha =0.05$ before observing the sample. The rejection region is $|Z|>z_{0.975}$, whose probability under $H_0$ is exactly 0.05. Here $2.4>1.9599639845\dots$, so reject $H_0$. This agrees with zero lying outside $[0.22,2.18]$. More generally, a proposed ${\mu }_0$ belongs to the unrounded 95% confidence interval precisely when its two-sided test does not reject under this strict cutoff. The equality boundary has probability zero in this continuous statistical model.

The significance level and statistical power describe different repeated-experiment probabilities. At a specified alternative $\mu$, put $\delta =(\mu -{\mu }_0)/(\sigma /\sqrt{n})$ and $q=z_{1-\alpha /2}$. ThenThe first is the Type I error probability; the second is the Type II error probability at that alternative. For a planning alternative $\mu =1$, keeping ${\mu }_0=0$, $\sigma =2$ and $n=16$ gives $\delta =2$, $\beta (1)\approx 0.483995$ and statistical power $0.516005$. At $n=64$, $\delta =4$ and statistical power rises to $0.979327$. These are model calculations for specified alternatives, not posterior probabilities inferred from a nonsignificant result. They explain why failure to reject cannot establish equality and why choosing sample size before an experiment is useful.

Historically, Ronald Fisher's 1925 book discusses the normal cutoff $1.96$ and tail probability 0.05 as a convenient convention on page 47, rather than a mathematically compulsory threshold: ^[ref].

There are 14 conservative-coded and 12 liberal-coded observations. The tweet summary has sample median 258.5, quartiles 204.5 and 323.2, and range 108 to 441, so the interquartile range is about 118.7 tweets. No measurement period is specified, so these should not be reinterpreted as tweets per day. The displayed overall sample mean is 261.5, subject to a further consistency issue below. These are small groups; the summary alone cannot establish normality, lack of outliers or an appropriate sampling design.

Let ${\mu }_C$ and ${\mu }_L$ be the two population means for the defined tweet measure. The null hypothesis is $H_0:{\mu }_C-{\mu }_L=0$, against the two-sided test alternative $H_1:{\mu }_C-{\mu }_L\ne 0$. Welch's t-test usesallowing unequal group variances. It requires independent observations and a suitable approximation to the sampling distribution of the mean difference; a large-sample justification cannot simply be assumed with group sizes 14 and 12. Its noninteger statistical degrees of freedom comes from the Welch approximation, not a fractional number of respondents.^[ref]

The two displayed sample means implyThe negative sign means the liberal-coded group has the larger observed mean. This is a difference in group averages, not a statement about every individual and not evidence that political orientation causes tweeting behaviour. Practical importance concerns this difference and its context; statistical significance does not itself measure its importance.

A p-value is calculated under the null hypothesis and the assumed sampling model. It is the probability of a result at least as incompatible with that hypothesis as the observed one, not the probability that the hypothesis is true or that this conclusion is mistaken. A prespecified significance level controls the long-run Type I error rate of rejecting a true null; a Type II error is nonrejection of a false null, and statistical power concerns rejection under a specified alternative. Neither a small p-value nor a nonrejection settles the effect size or a causal explanation.

Taken on its own, the printed p-value supports a nominal 5% rejection, but not a 1% rejection. Taken instead from the printed t statistic and statistical degrees of freedom, the two-sided tail is about 0.06380. These different 5% decisions cannot be reconciled by ordinary rounding. Conditional on trusting the group means, t and degrees of freedom, the implied standard error is $(-13.4881)/(-1.9436)\approx 6.93975$ and the corresponding 95% confidence interval isIt includes zero, consistently with that larger p-value. This calculation is a conditional consistency check, not recovery of the true raw-data interval.

The source lists its interval endpoints as 0.04582 and $-126.7904$, in that order. They are not ordered, and merely sorting them would still give midpoint about $-63.3723$, rather than the estimated difference $-13.4881$. The symmetric confidence-interval consistency identities therefore fail for the usual t interval. A frequentist 95% interval describes repeated-sample coverage of its procedure, not a 95% posterior probability for a fixed parameter; neither interpretation justifies silently repairing these endpoints.

The group summaries also imply pooled mean $(14\times 255.4286+12\times 268.9167)/26\approx 261.6539$, which cannot round to the printed 261.5. The pooled-mean compatibility under rounding identifies this additional mismatch for summaries of the same complete sample. Thus the observed direction is conditionally liberal-greater-than-conservative, but the inferential decision and interval require corrected coherent output. Nonrejection under the t/df reading would not establish equal population means. Random assignment, representativeness and possible confounding are not supplied by this test.

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.

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.

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.

Under the question's stipulated nonzero population difference, failing to reject is a type II error. A type I error would instead reject a true equal-mean null hypothesis. The premise concerns the underlying difference: another study's statistical significance, by itself, would not prove that its conclusion is correct.

For a specified alternative and sampling design, writeThe second probability is statistical power. Across repeated random samples under that same alternative and design, the long-run frequency of this kind of error is $\beta$.

No numerical value of $\beta$ follows from this output. It depends on the true effect size, group variances, sample sizes, decision rule and chosen significance level. In particular, it is neither $0.9561$, nor $1-0.9561$, nor automatically 5%. The reusable result that a p-value does not determine the type II error probability demonstrates this directly in a simple normal testing model.

A small effect size or an imprecise estimator can produce low statistical power. The present confidence interval describes the precision of the estimated difference; it does not supply the missing true alternative needed to calculate the power function. A specified 5% significance level concerns the type I error rate under the null, rather than a universal error rate whenever a result is nonsignificant.

Let ${\mu }_U$ and ${\mu }_V$ be the population means of the newspaper trust score for the unvaccinated and vaccinated groups. A two-sided hypothesis test comparesThe observed difference of sample means is $5.5-6.5=-1$: the vaccinated group has a one-unit higher average trust score. The negative test statistic agrees with subtraction in this order. A Student's t-test scales the estimated difference by its estimated standard error; $t=-3.6$ means that the observed difference is 3.6 estimated standard errors below the equal-mean null hypothesis.

Using the reported reference Student t-distribution with 747 statistical degrees of freedom,Thus the equal-mean null hypothesis is rejected at a prespecified 5%, 1% or 0.1% significance level, subject to the test's assumptions. A p-value is a probability of a result at least this extreme under the specified null hypothesis and sampling model. It is neither the probability that the null hypothesis is true nor the probability that this particular conclusion is mistaken. A chosen significance level controls the procedure's type I error probability under that model.

The estimated difference is more informative than statistical significance alone. If the reported statistic uses this one-unit contrast, its estimated standard error is $1/3.6\approx 0.2778$. The corresponding approximate 95% confidence interval isThis interval concerns a population mean difference under repeated sampling; it is not a range containing 95% of individual trust scores. Its practical interpretation also depends on the trust scale and the importance of a one-unit difference.

The groups are disjoint, so an independent-samples comparison is appropriate if the sampled observations are independent. The exact pooled Student's t-test additionally assumes common population variance and normal distributions; Welch's t-test permits unequal variances and uses an approximate reference distribution. A bounded or ordinal rating need not be exactly normal, so treating it numerically and using a large-sample approximation also requires judgment about group sizes and the score's meaning.

There is a bookkeeping qualification. The later outputs use 748 observations. If these are also the complete observations for this two-group comparison, the pooled test has $748-2=746$ statistical degrees of freedom, and the Welch degrees of freedom upper bound is also 746. The printed 747 therefore needs the original test call or a different sample count to reconcile it. The reported-calibration calculation above answers the stated test; it does not invent group sizes or silently change the reference distribution.

The evidence supports an association between higher newspaper trust and recorded vaccination, rather than a causal effect of trust. An observational study does not rule out confounding, selection bias or reverse causation. A statistical hypothesis test cannot supply those missing causal assumptions.

A null hypothesis specifies the population comparison to be tested, for example equality of the population means for women and their complement. The alternative hypothesis for a two-sided test is a nonzero difference. A Student's t-test measures an estimated difference in units of its estimated standard error:On the reported calibration, the sampling distribution under the null hypothesis is Student's t-distribution with $1225$ statistical degrees of freedom. Consequently the two-sided test has p-valueWith this many statistical degrees of freedom, the Student t-distribution is close to the standard normal distribution: $|t|=2.5$ exceeds the approximate $5\%$ two-sided critical value $1.96$, but falls short of the $1\%$ value $2.576$. Using the supplied calibration, reject at the $5\%$ significance level, but not at $1\%$. The large statistical degrees of freedom justify this comparison; a value $2.5$ need not be significant for an arbitrary very small number of statistical degrees of freedom.

The p-value is the probability, under the specified null hypothesis and sampling assumptions, of a test statistic at least as discrepant as the observed one. It is not the probability that the null hypothesis is true, nor the probability that the observed difference happened “by chance”. A significance level of $5\%$ controls the long-run type I error rate of the testing procedure when its assumptions hold. Failing to reject at $1\%$ does not prove equality. Statistical significance also does not measure the substantive size or importance of a difference.

The reported sample means place the women's mean $1.4$ scale units above the overall mean. The negative test statistic can reflect the order of subtraction; it does not reverse that observed ordering. An appropriate pre-specified directional one-sided test would have a tail p-value approximately $0.00627$ on the supplied calibration, but one cannot choose that direction after inspecting the results simply to halve a p-value.

There is a design issue behind the numerical calculation: the women are included in the overall sample. Comparing a subgroup mean with its containing sample mean requires accounting for shared observations. If $w$ is the women's sample fraction and ${\overline{x}}_O$ the mean for the remaining respondents, thenThese two estimates are not independent. An ordinary independent-two-sample Student's t-test of women against the entire respondent sample would therefore use an inappropriate standard error. The interpretation above accepts the supplied test statistic and its stated null calibration; a valid analysis of the underlying data should compare disjoint groups, or calculate the shared-sample covariance correctly. The summary alone cannot establish which calculation was used. Simple random sampling supports sampling inference under the stated design, but an observed gender association does not establish a causal effect.

Each row of the contingency table is a conditional distribution of the safety category among respondents with that experience status. Thus the relevant comparison uses row percentages, rather than interpreting the larger number of respondents with some bad experiences as itself an effect on safety.

The sample proportions reporting feeling very safe are $39.59\%$ without bad experiences and $32.78\%$ with some bad experiences: the latter is lower by about $6.81$ percentage points. Feeling fairly safe rises from $28.49\%$ to $34.63\%$, a difference of about $6.14$ percentage points. Feeling unsafe changes only from $31.92\%$ to $32.59\%$, about $0.67$ percentage points. The main descriptive shift is therefore from very safe toward fairly safe, rather than a large increase in the unsafe category.

Pearson's chi-squared test tests the null hypothesis that experience status and safety category are independent. If that were true, the expected cell count would be its row total times column total divided by $n$. The Pearson chi-squared statistic sums squared observed-minus-expected count differences divided by their expected counts. HereThe expected counts are comfortably large, with minimum approximately $234.7$, so the ordinary large-sample chi-squared distribution approximation is appropriate under the sampling assumptions. There is evidence against independence even at the $1\%$ significance level. The reported likelihood-ratio result gives the same conclusion.

Magnitude is a separate question. Cramér's V giveswhich describes a weak statistical association. It does not mean that $7.7\%$ of safety is explained. Nor does the p-value measure the size of that association.

For ordinal direction, order the experience categories as none, some, and the safety categories from very safe toward unsafe. Goodman and Kruskal's gamma compares concordant pairs with discordant pairs after excluding tied pairs:Its small positive sign then corresponds to a weak tendency toward less safety among respondents with bad experiences. The sign would reverse if exactly one category ordering were reversed. Because this contingency table has many ties, the response-direction Somers' D, which retains response ties among pairs with different predictor values, is smaller:These coefficients describe ordered association; the omnibus Pearson's chi-squared test tests any departure from independence and does not supply a separate p-value for gamma.

The column-dependent Goodman and Kruskal's lambda makes the weak predictive improvement tangible. Predicting the overall most frequent safety category makes $1165$ errors in this table. Using a separate most frequent category within each experience row makes $1145$, giving $\lambda =20/1165\simeq 0.0172$, only about a $1.7\%$ reduction in those classification errors.

The sample supports a statistically detectable but weak association, chiefly a reduced sample proportion feeling very safe. This observational study does not establish that bad experiences cause the change: reverse direction, reporting differences or confounding could also contribute. The numerical p-value also presumes the sampling and complete-case assumptions appropriate to the observations used.

Conditioning on marital status produces five contingency tables. In each, compare the conditional row percentages. The following differences are the sample proportion with some bad experiences minus the sample proportion without them, in percentage points; negative very-safe differences mean fewer respondents feeling very safe in the former group.

The p-values test independence within the respective strata; the magnitudes and directions come from the conditional percentages and association measures. Differences in the table are calculated from the counts before rounding.

Among single respondents, the very-safe sample proportion changes from $33.95\%$ to $29.52\%$, while the unsafe sample proportion changes from $36.28\%$ to $38.10\%$. The direction resembles the weak pooled pattern, but $p=0.6114$ supplies little evidence against independence in this stratum. This is not evidence that the population association is exactly zero.

Among married respondents, the very-safe sample proportion changes from $41.18\%$ to $32.75\%$ and the fairly-safe sample proportion rises from $27.68\%$ to $35.39\%$. Unsafe remains almost unchanged, $31.14\%$ versus $31.87\%$. Thus the within-stratum association again chiefly shifts very safe toward fairly safe. Its $p=0.0257$ is below the $5\%$ significance level, while $V=0.092$ still indicates a small association.

For respondents living with a partner, very safe changes from $41.46\%$ to $28.32\%$, fairly safe from $30.08\%$ to $40.71\%$, and unsafe from $28.46\%$ to $30.97\%$. The very-safe difference is larger than in the married stratum. The printed p-value $0.04885$ would be just below the conventional $5\%$ significance level. However, this particular output is internally inconsistent: recalculating Pearson's chi-squared test from its six displayed counts gives $X^2=4.910389$, $p=0.08585$, and $V=0.14425$. The reported $V=0.144$ agrees with the counts, whereas the printed $X^2=6.900389$ would itself imply $p=0.03174$ for two statistical degrees of freedom, rather than the printed $0.04885$. Using the displayed counts, this stratum does not reject independence at $5\%$. The source PDF itself contains these incompatible figures; this is not resolved by rounding.

Among divorced respondents, very safe changes from $41.18\%$ to $37.17\%$, but unsafe also decreases, from $36.76\%$ to $30.97\%$, as fairly safe increases from $22.06\%$ to $31.86\%$. This is a redistribution toward the middle category, not a simple shift toward greater unsafety. The near-zero gamma, $-0.013$, reflects the lack of a consistent ordinal direction, while $V=0.106$ can still register distributional differences without a direction. The p-value $0.3583$ does not reject independence.

Among widowed respondents, very safe changes from $51.43\%$ to $42.47\%$, and unsafe from $14.29\%$ to $27.40\%$. The ordinal association is positive on the chosen ordering, with gamma $0.220$. Its lack of conventional statistical significance, $p=0.3158$, does not imply a smaller estimated association than in the married group: the widowed stratum has only $108$ complete observations, versus $857$ in the married stratum, and therefore less precision. All five tables have sufficiently large expected counts for the reported approximations; even the smallest expected count is about $8.10$.

The plot uses large-sample Wald confidence intervals for the difference of independent sample proportions within each stratum. They concern the individual category contrasts, whereas the tabulated Pearson's chi-squared tests compare the full three-category distributions. The intervals are unadjusted for multiple hypothesis testing; the pooled row reuses the stratum observations, so it is not a sixth independent sample.

In the elaboration model, the persistence of a generally weak positive ordinal pattern within four strata is closer to partial replication than to a complete explanation of the pooled association by marital status. The differences in estimated patterns suggest possible specification: the divorced group's shift toward the middle category differs from the widowed group's shift toward unsafe. They do not by themselves establish population effect modification. In particular, significant and nonsignificant results need not differ significantly; compare the effects using an interaction contrast or a fitted model with the appropriate interaction terms, rather than comparing their separate p-values.^[ref]

One additional omnibus check can be derived from the supplied tables without assuming a common effect. Recalculating the five statistics from the counts, including the corrected living-with-partner statistic, the joint conditional independence test for stratified contingency tables givesUnder independent-stratum sampling this tests independence within every stratum jointly; it does not test whether the association is homogeneous across strata. It does not reject at $5\%$, but this alone does not establish that stratification removed the population association: its ten-degree-of-freedom alternative is less focused than the pooled two-degree-of-freedom test, and nonrejection is not proof of equality.

The elaboration model also distinguishes an antecedent confounder from an intervening causal mediator. These cross-sectional tables do not establish the relevant time ordering or identify either role for marital status. Five separate stratum tests further require caution about multiple hypothesis testing: for example a Bonferroni correction at familywise $5\%$ would use $0.01$ per test, below all five individual p-values. That caution changes the interpretation of selected significance labels, not the observed conditional percentages.

Represent a possibly randomized statistical test by a function $\phi (Y)$ taking values in $[0,1]$, its conditional probability of rejection. Its power function is $\pi (\rho )={\mathbb{E}}_{\rho }\phi (Y)$, and its size of a statistical test is the supremum of this probability over the null hypothesis. At a prescribed significance level $\alpha$, a uniformly most powerful test has at least as much power as every other level-$\alpha$ test at every parameter in the alternative.

Put $S={\sum }_{j=1}^n{Y}_j$. For any fixed alternative $0<{\rho }_1<{\rho }_0$, the likelihood ratio iswhich is strictly increasing in $S$. The Neyman-Pearson lemma therefore selects the upper tail $S>c_{\alpha }$. Its cutoff is determined solely by the null distribution, not by ${\rho }_1$, so the same statistical test is most powerful against every allowed alternative and hence uniformly most powerful.

Under rate $\rho$, $S$ has a gamma distribution with shape $n$ and rate $\rho$. For completeness, convolution starts with $\rho e^{-\rho s}$ and, if the $n$-term density is ${\rho }^n{s}^{n-1}{e}^{-\rho s}/(n-1)!$, convolving once more gives ${\rho }^{n+1}{s}^n{e}^{-\rho s}/n!$. Integrating this density by parts repeatedly gives its upper-tail probability. Consequently reject precisely when $S>c_{\alpha }$, whereand the power function isFor $0<\alpha <1$ the continuous strictly decreasing null tail gives a unique positive cutoff, with exact size $\alpha$ and no boundary randomization.

Use a conventional two-sided test at significance level $0.05$. Let $n=4000$ per arm, $p_C=0.005$, $p_T=0.0025$ and $\Delta =p_C-p_T=0.0025$. Under the specified alternative, the normal approximation to the difference of independent binomial proportions has standard deviationFor planning the pooled null standard error, use $\bar{p}=(p_C+p_T)/2=0.00375$, giving ${\sigma }_0=\sqrt{2\bar{p}(1-\bar{p})/n}=0.00136674$. The rejection threshold is approximately $1.96{\sigma }_0=0.00267880$, exceeding the mean alternative difference. With $\Phi$ the standard normal distribution function, the approximate statistical power isNo: the usual two-sided 5% calculation gives about 45% power, below 50%. The expected death counts are only 20 and 10, so an exact discrete design calculation could refine this approximation.

The question does not specify the tail convention. If a beneficial direction is prespecified and a one-sided test at 5% is justified, replace $1.96$ by approximately $1.645$; the resulting statistical power is about $0.573$, and the answer is then yes. This change of testing convention must be declared.

A simple hypothesis specifies the entire probability distribution. A critical region $C$ is the set of observations for which the null hypothesis is rejected. The size of a statistical test is ${\sup }_{\theta \in H_0}{\mathrm{Pr}}_{\theta }(X\in C)$, which reduces to ${\mathrm{Pr}}_0(C)$ for a simple hypothesis. The power function is ${\mathrm{Pr}}_{\theta }(C)$ as a function of the true parameter or distribution. At a specified alternative, the Type II error probability is ${\mathrm{Pr}}_{\theta }(C^c)=1-{\mathrm{Pr}}_{\theta }(C)$, the chance of failing to reject a false null hypothesis. The Type I error probability is the corresponding rejection probability under the null hypothesis.

The Neyman-Pearson lemma says that, for two simple hypotheses, a test rejecting where $f_1>k{f}_0$, accepting where $f_1<k{f}_0$, and if necessary randomizing on equality to achieve prescribed size $\alpha$, has maximal statistical power among tests of size at most $\alpha$. Conversely a most powerful test can be taken in this likelihood ratio form, up to equality sets and null sets.

Here the likelihood ratio for $x\ne 0$ isIts value is infinite at $0$, which has probability zero under both continuous distributions. Large likelihood ratios correspond to small $|x|$. Thus the best critical region is $|X|<c_{\alpha }$, whereThere is no boundary randomization to worry about. The complete answer is

For the minimum sum of errors in a simple hypothesis test, write the error sum as ${\int }_C{f}_0+{\int }_{C^c}{f}_1$. At each observation its smaller contribution is obtained by rejecting exactly where $f_1>f_0$. This proves optimality over all tests, including randomized ones, rather than just optimizing within an assumed family. Here it gives $c_{\ast }=\sqrt{2/\pi }$, and thereforeThe minimum error sum is $1-e^{-1/\pi }+2[1-\Phi (\sqrt{2/\pi })]$. This criterion weights the two error probabilities equally; it does not impose a conventional $5\%$ significance level.