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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 $\delta =\sqrt{n}(\mu -{\mu }_0)/\sigma$.

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.

The fitted simple linear regression isIts regression slope is a predicted decrease of approximately 0.006314 trust-score units per extra year, or 0.06314 units per decade. This is an estimated statistical association between age and the score. A cross-sectional study does not establish that the same person's trust decreases at this rate as they age.

The estimated standard error of the slope is 0.005427. Testing a zero regression slope gives $t\approx -1.163$ on 746 residual degrees of freedom and $p\approx 0.245$. An approximate 95% confidence interval isThere is insufficient evidence of a nonzero linear age association at the 5% level. The interval includes both small negative and small positive slopes; failure to reject does not establish an exactly zero relationship or rule out a nonlinear one.

The regression intercept 6.388920 predicts the score at age zero. That age is outside the observed 19–89 range, so interpreting it substantively would be extrapolation. Its very small reported p-value tests a zero intercept, rather than the age hypothesis. The 746 residual degrees of freedom correspond to 748 observations minus the intercept and slope.

The coefficient of determination is $R^2=0.001811$, so the fitted line explains approximately 0.1811% of the observed centered variation, rather than 18.11%. The adjusted coefficient of determination is approximately 0.0004732. The residual standard error is 2.602 score units, close to the reported response standard deviation; most variation remains unexplained by the fitted line. With one predictor and an intercept, the slope's F-test is the square of its Student's t-test statistic, up to printed rounding, so $F\approx 1.354$ and $p\approx 0.245$ agree with the slope test.

Output 6 is a residual-versus-fitted plot: its vertical coordinates should be regression residuals $e_i=y_i-{\hat{y}}_i$. The fitted values range over only approximately 5.83–6.27, as the weak slope predicts over ages 19–89. The printed scatter shows no obvious strong mean curvature or clear widening funnel, but it has marked horizontal concentrations near its extreme residual levels. An absence of obvious curvature does not prove a correct linear conditional mean, and the plot alone cannot establish homoscedasticity, independent errors or a normal distribution. A normal probability plot and residuals against observation order address different aspects. Large residuals also do not by themselves establish influential observations; regression leverage and Cook's distance matter.

There is an additional inconsistency in the original figure. The descriptive output reports observed responses between zero and ten, so residual constraints from a bounded response requireYet the printed graphic contains points around $(\hat{y},e)=(6.25,4.2)$, which would imply $y\approx 10.45$, above the stated observed maximum. Some lower-edge points likewise imply negative responses. This is present in the original PDF, rather than introduced by the transcription. The graphic therefore cannot be treated as a fully consistent diagnostic of the reported dataset. The observations about its visible shape above describe the supplied picture; reliable model checking would require the actual fitted values and residuals.

The printed code has separate typographical problems. The displayed model call supports the intended formula with ~, and the plot needs one closing parenthesis:

model1 <- lm(t_guard ~ age, data = exam)
plot(fitted(model1), residuals(model1),
     xlab = "Fitted values", ylab = "Residuals")
abline(h = 0)

These calls correct the syntax; they cannot be rerun on the original observations because the individual-level dataset is not supplied. The regression interpretation above uses the available numerical summaries and states the diagnostic limitation explicitly.

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.

The ordinary least squares fit estimates the conditional mean of the continuous safety response as a linear function of the income coding:The regression coefficient for income is positive. Since larger response values mean greater safety, one additional coded income unit corresponds to an estimated increase of $0.2369$ safety-scale units in the fitted mean. Following the question's stated units, a unit is a £10,000 income step. The observed coding also uses income bands with open end categories, so this is a linear trend per coded band step; it is not evidence of a known constant effect for every extra pound, especially within the open top band. The output's abbreviated variable label does not override the stated units.

The intercept is the fitted mean at income code zero. It is estimated as $6.5933$, but zero is outside the observed income codes, so that value is an extrapolated model baseline, not an observed estimate for a particular income category. Its small p-value tests a zero intercept; it is not the main test of the income association.

For the income coefficient, the standard error is $0.0604$. The reported test statistic and two-sided test giveproviding strong evidence against the null hypothesis of zero linear slope under the stated model assumptions. A $95\%$ Student t confidence interval, using $1727$ residual statistical degrees of freedom, isThis is an interval for the population linear coefficient under the model, not an interval containing $95\%$ of individual safety scores. The large-sample estimate is precise enough to detect a positive association, but that does not establish substantial individual predictive power.

The coefficient of determination is $R^2=0.008828$: the fitted line accounts for about $0.883\%$ of the in-sample variation about the response sample mean. The adjusted coefficient of determination is about $0.825\%$. Thus income alone explains very little of the observed variation, despite the small slope p-value. The residual standard error of $3.997$ response units describes substantial scatter about the fitted line. This is an estimated residual standard deviation, rather than the standard error of the slope or a complete individual prediction interval.

The overall F-test has $F=15.38$ on $1,1727$ statistical degrees of freedom and $p=9.129\times 10^{-5}$. With one predictor and an intercept, it tests the same zero-slope hypothesis as the coefficient Student's t-test; $F=t^2$, up to rounding. It is not an additional independent confirmation. In this simple linear regression, the Pearson correlation coefficient has the slope's sign and magnitude $\sqrt{R^2}\simeq 0.0940$, another description of the weak linear association.

The regression uses $1729$ complete observations and estimates two coefficients, giving $1729-2=1727$ residual statistical degrees of freedom. The deletion of $88$ observations is complete-case analysis; extending its conclusions to the full target population requires suitable assumptions about the missing-data mechanism. Classical reported standard errors also rely on an adequate linear conditional mean, independent errors and homoscedasticity; a normal linear model gives the finite-sample Student t-distribution and F-test calibration, while large-sample approximations can justify similar tests under weaker distributional assumptions. The printed summary alone does not establish these assumptions. Regression diagnostics and, where appropriate, heteroskedasticity-consistent standard errors address different parts of that assessment.

Finally, this unadjusted observational study estimates an association. Other covariates, such as age or experience status, may be related to both income and reported safety. Without additional design or causal identification assumptions, the positive fitted slope does not establish that increasing income would cause an increase in safety.

Use the account of Student's t-test in part c of the Experimental Psychology paper, choosing the one-sample, independent-samples or paired form to match the study.

Assume independent samples $X_1,\dots ,X_n$ and $Y_1,\dots ,Y_m$, each consisting of independent normal random variables, with respective means ${\mu }_1,{\mu }_2$ and a common unknown variance ${\sigma }^2>0$. Take $n,m\ge 2$. Normality, independence and equal variances are the assumptions giving the exact pooled Student's t-test; unequal variances would require a different test. We test $H_0:{\mu }_1={\mu }_2$ against an unrestricted difference.

Put $N=n+m$ andThe normal likelihood is proportional to $({\sigma }^2{)}^{-N/2}\exp [-S/(2{\sigma }^2)]$, where $S$ is the residual sum of squares. Under the unrestricted model, the maximizing means are $\bar{X},\bar{Y}$ and $S=W$. Under $H_0$, the maximizing common mean is $(n\bar{X}+m\bar{Y})/N$ and decomposition about the sample means gives $S=W+B$. For either model, maximizing over variance gives ${\hat{\sigma }}^2=S/N$. Thus the generalized likelihood-ratio test statistic isSmall $\Lambda$ is therefore equivalent to large $|T|$.

Under the null, $Z=(\bar{X}-\bar{Y})/[\sigma \sqrt{1/n+1/m}]$ is standard normal. In each sample an orthogonal change of Gaussian coordinates separates the sample-mean coordinate from the centered residual coordinates; those coordinates are independent. Consequently $V=W/{\sigma }^2$ has a chi-squared distribution with $N-2$ degrees of freedom and is independent of $Z$. Therefore $T=Z/\sqrt{V/(N-2)}$ has the Student's t-distribution with $N-2$ degrees of freedom. At significance level $\alpha$, reject exactly when $\boxed{|T|>t_{N-2,1-\alpha /2}}$, the two-sample pooled t-test.

The first output uses the equal-variance, pooled Student's t-test, as shown by its seven statistical degrees of freedom. Its exact model assumes two independent samples from normal distributions, with means ${\mu }_X,{\mu }_Y$ and a common unknown variance ${\sigma }^2$. The hypotheses are $H_0:{\mu }_X={\mu }_Y$ against $H_1:{\mu }_X<{\mu }_Y$. The sample means are $\bar{x}=4.35$ and $\bar{y}=9.88$, and the unbiased within-sample sample variances areThe pooled sample variance and its resulting standard error areHenceTo justify its reference distribution, under the null $(\bar{X}-\bar{Y})/(\sigma \sqrt{1/4+1/5})$ is standard normal, while $(3s_X^2+4s_Y^2)/{\sigma }^2$ is ${\chi }_7^2$, independently of the two sample means. Dividing the first variable by the square root of the second divided by seven gives the exact pooled two-sample t statistic, with Student t-distribution $t_7$. Its one-sided p-value is the lower tailThe associated one-sided 95% confidence interval for ${\mu }_X-{\mu }_Y$ isThe printed missing lower bound means that this is a one-sided upper confidence limit, not that a finite lower endpoint should be estimated. With such small samples the normality and common-variance assumptions matter; a Welch analysis would use a different variance estimate and generally different degrees of freedom.

The second output is the Wilcoxon rank sum test for two independent continuous samples. Under a common-distribution null hypothesis, labels are exchangeable conditional on the pooled values, so the four ranks assigned to the first sample form a uniformly chosen four-element subset of $\{1,\dots ,9\}$. The first sample receives ranks $2,1,3,5$, giving raw rank sumEquivalently the Mann–Whitney U test statistic is $U_X=W_X-4\cdot 5/2=1$, the number of cross-sample pairs with an $X$ observation larger than a $Y$ observation. This output reports the raw rank sum, rather than the shifted statistic used by some software.

A first sample tending to be smaller has a small rank sum. Under the null, all $\left(\genfrac{}{}{0ex}{}{9}{4}\right)=126$ rank allocations have equal probability. Only $\{1,2,3,4\}$ and $\{1,2,3,5\}$ have rank sum at most 11. By the exact rank-sum tail one above the minimum, the exact lower-tail p-value isNo large-sample normal approximation or continuity correction is involved; the absence of ties makes this simple enumeration valid. For a common-shape location-shift family, the test's alternative is a negative location shift of $X$ relative to $Y$. Without that model, the null is equality of distributions and the rank comparison is not simply a test of equality of means or medians.

Both tests reject at the 5% level in the direction of smaller $X$ observations; only the pooled t-test rejects at 1%. The t-test uses the numerical spacing and targets a mean difference under its normal model. The rank-sum test uses ordering and an exact exchangeability calculation, avoiding the normality and common-normal-variance assumptions. Their different p-values therefore reflect both different statistics and different modelling assumptions, not a numerical contradiction.

Write $Q={\sum }_i(X_i-\bar{X}{)}^2$ and $Q_0={\sum }_i(X_i-{\mu }_0{)}^2=Q+n(\bar{X}-{\mu }_0{)}^2$. The normal-sample likelihood isWithout the mean restriction, its maximum has $\hat{\mu }=\bar{X}$ and ${\hat{\sigma }}^2=Q/n$. Under the null, the maximizing variance is $Q_0/n$. Substituting them gives the generalized likelihood-ratio test statisticwhere $S^2=Q/(n-1)$ and $T=\sqrt{n}(\bar{X}-{\mu }_0)/S$. Assume $n\ge 2$; $Q>0$ almost surely for positive population variance.

The ratio decreases strictly as $|T|$ increases, so rejecting for small $\Lambda$ is equivalent to rejecting for large $|T|$. Under $H_0$, $T$ has Student's t-distribution with $n-1$ degrees of freedom. Therefore the exact size-$\alpha$ rule isThis is the two-sided Student's t-test. Equivalently reject when $\Lambda <(1+t_{n-1,1-\alpha /2}^2/(n-1){)}^{-n/2}$. No large-sample chi-squared approximation is needed for this normal-mean likelihood ratio with unknown variance.

Write $R_j=I-P_{-j}$. Minimizing the residual sum of squares first over the coefficients of $X_{-j}$ reduces the problem to minimizing $\Vert R_{j}Y-b{R}_j{X}_j{\Vert }^2$. Full column rank ensures $R_j{X}_j\ne 0$. Thus the partial regression formula and its variance areThis uses $R_j^T=R_j=R_j^2$ and the isotropic error covariance matrix ${\sigma }^{2}I$.

Since the span of $X_k$ is contained in the column space of $X_{-j}$, orthogonal projection onto the latter removes at least as much squared norm. Consequentlyand inversion proves the requested lower bound. Full rank makes the denominator positive.

Adding $c{1}_n$ to column $j\ge 2$ leaves the column space unchanged. Every old fitted vector is reproduced by keeping all slopes unchanged and replacing the regression intercept by ${\hat{\beta }}_1-c{\hat{\beta }}_j$. Uniqueness of the ordinary least squares estimators then proves that slope $j$ is unchanged. In particular all columns can be centered by such changes. Applying the previous projection argument to the centered design proves the bound with $X_j-{\bar{X}}_j{1}_n$ and $X_k-{\bar{X}}_k{1}_n$; the normalized inner product is now the usual sample correlation.

Each reported coefficient test is the two-sided Student's t-test of $H_0:{\beta }_j=0$ against $H_1:{\beta }_j\ne 0$, conditional on the other predictors. Under the null, ${\hat{\beta }}_j/(\hat{\sigma }\sqrt{(X^{T}X{)}_{jj}^{-1}})$ has a Student's t-distribution with $100-3=97$ degrees of freedom. At five percent the intercept is significant, but neither individual test slope is. The final F-test compares the full model against an intercept-only model: $H_0:{\beta }_{\mathrm{Maths}}={\beta }_{\mathrm{Stats}}=0$. It has $(2,97)$ degrees of freedom and rejects at five percent, since its $p$-value is $0.003166$. Thus the predictors jointly explain variation, without strong evidence separating their individual conditional effects.

The two test predictors have sample correlation $0.9371630$, although each has only moderate correlation with the final exam response. The variance inflation factor is $1/(1-0.9371630^2)\simeq 8.22$. Their nearly shared direction produces substantial multicollinearity, inflating the individual slope standard errors. The joint F-test can detect this common predictive direction even though either coefficient, given the other, has a large $p$-value.

The quasibinomial regression retains the logit link and mean function $p_i=(1+e^{-x_i^T\beta }{)}^{-1}$ but uses the working variance function $\mathrm{Var}(Y_i)=\varphi p_i(1-p_i)$. Its quasi-score equation isso multiplying by the common dispersion parameter does not change the coefficient estimates. Iteratively reweighted least squares therefore gives the same fitted means and coefficients as the binomial model. The usual Pearson dispersion estimator isThe estimated coefficient covariance matrix is $\hat{\varphi }(X^{T}WX{)}^{-1}$, with $W_{ii}={\hat{p}}_i(1-{\hat{p}}_i)$. From the intercept standard errors, $(1.08921/1.13031{)}^2\simeq 0.92860$; the rounded slope errors give approximately the same factor. The display uses approximate Student's t-tests with 118 residual degrees of freedom instead of fixed-dispersion normal tests.

For a genuinely individual binary response, the Bernoulli distribution identity $Y_i^2=Y_i$ forces $\mathrm{Var}(Y_i)=p_i(1-p_i)$. Thus $\varphi \ne 1$ is a working quasi-likelihood specification, not a different independent binary distribution with that mean. The mild estimated underdispersion does not establish an improved model, and the mean predictions are unchanged. In particular, a scalar rescaling does not model correlation among users, and usual likelihood-based Akaike information criterion comparisons are unavailable for a family without a specified probability likelihood. The output gives no convincing reason to prefer model3 to model2.

Use the usual normal linear model $Y=X\beta +\epsilon$, with fixed full-column-rank design $X\in {\mathbb{R}}^{n\times p}$, $n>p$, and $\epsilon \sim N(0,{\sigma }^2{I}_n)$. These Gaussian distribution and constant-variance hypotheses are necessary for an exact finite-sample Student's t-test. Put $V=(X^{T}X{)}^{-1}$ and $\nu =n-p$. The vector of ordinary least squares estimators satisfiesWith $s^2=\Vert Y-X\hat{\beta }{\Vert }^2/\nu$, the test statistic isUnder the null hypothesis, $T$ has Student's t-distribution with $\nu$ degrees of freedom. A two-sided level-$5\%$ Student's t-test rejects when $|T|>t_{\nu ,0.975}$. If the intended test is one-sided against ${\beta }_0>0$, it rejects when $T>t_{\nu ,0.95}$; the statistical power conclusion below holds for either convention.

The command performs a nested-model F-test for $H_0:{\beta }_1=0$ against $H_1:{\beta }_1\ne 0$, retaining the intercept and $x_2$ coefficient in both models. Write $RS{S}_2$ and $RS{S}_3$ for the residual sum of squares in the reduced and full fits. Under $H_0$ and the independent normal-error assumptions,By Cochran's theorem, the numerator and denominator are independent scaled chi-squares with one and $n-3$ degrees of freedom. Equivalently, for this single-coefficient restriction, $F=t_{{\beta }_1}^2$, so its p-value equals that of the two-sided coefficient Student's t-test in model3.

Using the printed $t$ value gives $F=0.800^2=0.640$ and $p\approx 0.43$. The printed estimate and standard error instead giveThis discrepancy is present in the PDF and is too large to be explained by rounding at the shown precision; the underlying ANOVA output is omitted, so an exact unique statistic cannot be recovered. The univariate correlation and $t$ value imply $n\approx 2+2.575^2(1-0.2517297^2)/0.2517297^2\approx 100.006$, consistent with $n=100$ after rounding. With denominator degrees of freedom 97, the two versions give p-values approximately $0.4257$ and $0.4298$, respectively. Either way the partial effect is not significant at 5%.

The individual t-tests only test each regression coefficient after adjusting for the other covariate. Here neither covariate is individually significant at conventional levels, but from the abbreviated coefficient table alone one cannot conclude that neither explains the response: strong multicollinearity can make both standard errors large even when the covariates are jointly useful.

The intercept-only model has $49$ residual degrees of freedom, so the sample size is $50$. The one-covariate model and full model therefore have respectively $48$ and $47$ residual degrees of freedom. The completed sequential analysis of variance table hasThus the missing sums of squares are $133.617$ and $0.014$, and the corresponding F-statistics are approximately $203.8$ and $0.0214$. The first p-value is below $2\times 10^{-16}$; the second is approximately $0.8845$. The tiny second increment is consistent with the reported t-test because, for one added coefficient, the partial F-statistic equals the squared t-statistic up to displayed rounding.

For fixed $n$, the likelihood ratio is strictly decreasing as $t^2$ increases. A likelihood-ratio test rejects for small $\lambda$, so its rejection rule is equivalently large $|t|$. Under the null hypothesis, the standard normal distribution and chi-squared distribution results give independent quantitiesTheir ratio therefore has Student's t-distribution with $n-1$ statistical degrees of freedom. The exact level-$\alpha$ rule isThis is precisely the two-sided Student's t-test; choosing the corresponding cutoff for $\lambda$ yields the same rejection region and the same p-value.

The equal-variance two-sample Student's t-test pools the within-group sums of squares. If $s_A^2$ and $s_B^2$ are the two unbiased sample variances, the required positive square root isFor ten observations in each group, $S^2=(s_A^2+s_B^2)/2$ and the usual unpaired reference Student's t-distribution has $18$ statistical degrees of freedom, provided the assumed independence between groups is justified.

TODO: Attachment C is absent from the held release. Recover it to check any supplied critical-value table, significance convention or additional instructions. This leaf remains incomplete against that missing input. The printed summary statistics nevertheless allow the following conditional numerical answer, using a two-sided $5\%$ Student's t-test and independent observations with common variance.

Test $H_0:{\mu }_W={\mu }_S$ against $H_1:{\mu }_W\ne {\mu }_S$. The pooled sample variance isThe test statistic and statistical degrees of freedom areUsing the Student t-distribution directly gives a two-sided p-value $p\simeq 0.00250$ and a $5\%$ critical value $1.96076$. Thus reject equal seasonal means at $5\%$ and $1\%$; the winter mean is higher by $0.36$ concentration units. This does not establish a large physical effect or its cause.

The two variances are similar, and Welch's t-test gives essentially the same result: $t=3.02344$, $\nu \simeq 2958.36$, $p\simeq 0.00252$. Both calculations assume the counts represent independent observations. Serial dependence, repeated measurements and strong non-normality can invalidate nominal p-values; the summary table alone cannot diagnose them. The unequal means are a conclusion under these statistical assumptions, pending recovery of Attachment C.

In a fixed full-rank normal linear model, the studentized estimator of one coefficient has a noncentral t-distribution with noncentrality equal to that coefficient divided by its standard deviation. The fitted and residual orthogonal projections of the Gaussian noise are independent. Therefore the statistical power of its exact Student's t-test does not depend on the other regression coefficients, with design and variance fixed.

Student's t-test calibrates a standardized mean contrast using a Student's t-distribution under a specified sampling model. For independent $X_1,\dots ,X_n\sim N(\mu ,{\sigma }^2)$ with $n\ge 2$ and unknown positive variance, the one-sample statistic isUnder the null hypothesis $\mu ={\mu }_0$, $T\sim t_{n-1}$ exactly.

The exact law uses two facts: $Z=\sqrt{n}(\bar{X}-{\mu }_0)/\sigma$ is standard normal, and $V=(n-1)S^2/{\sigma }^2\sim {\chi }_{n-1}^2$ independently of $Z$. Hence $T=Z/\sqrt{V/(n-1)}$, the defining construction of the Student's t-distribution. A normally distributed numerator divided by an arbitrary estimated standard error need not have this law.

For a two-sided level-$\alpha$ test, reject when $|T|>t_{n-1,1-\alpha /2}$; a directional alternative uses the corresponding one-sided quantile. The paired t-test applies the one-sample construction to independent pairs' differences. The exact pooled two-sample t statistic uses independent normal samples with a common variance; Welch's t-test instead estimates unequal variances and uses an approximate degrees-of-freedom calibration. The selected model and statistical degrees of freedom determine which reference distribution is justified.