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

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.

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.