Solution. The pooled variance is \[ s_p^2 = \frac {11(3.1)^2 + 9(2.6)^2}{20} = \frac {105.71 + 60.84}{20} = 8.3275 , \qquad s_p = 2.886 , \] and the standard error is \(2.886\sqrt {1/12 + 1/10} = 1.2356\). Then \(t_0 = (48.2 - 45.1)/1.2356 = 2.509\). Since \(t_{0.025,20} = 2.086\), \(H_0\) is rejected (\(p = 0.021\)). The \(95\%\) interval is \(3.1 \pm 2.086(1.2356)\), that is \((0.52,\ 5.68)\).

Solution. \(E(S_i^2) = \sigma ^2\) for each sample, so \(E(S_p^2) = [(n_1 - 1)\sigma ^2 + (n_2 - 1)\sigma ^2]/(n_1 + n_2 - 2) = \sigma ^2\). If \(n_1 = n_2 = n\), then \(S_p^2 = (S_1^2 + S_2^2)/2\), and \[ S_p^2\Bigl (\frac 1n + \frac 1n\Bigr ) = \frac {S_1^2 + S_2^2}{n} = \frac {S_1^2}{n} + \frac {S_2^2}{n} , \] so the two denominators are equal and \(t_0^* = t_0\). Only the reference distribution differs: \(t_{2n-2}\) for the pooled test, \(t_\nu \) with \(\nu \le 2n - 2\) for Welch’s.

Solution. Write \(u_i = s_i^2/n_i > 0\) and \(f_i = n_i - 1\), so \(\nu = (u_1 + u_2)^2/(u_1^2/f_1 + u_2^2/f_2)\). For the lower bound, with \(f = \min (f_1, f_2)\), \[ \frac {u_1^2}{f_1} + \frac {u_2^2}{f_2} \le \frac {u_1^2 + u_2^2}{f} \le \frac {(u_1 + u_2)^2}{f} , \] so \(\nu \ge f\). For the upper bound, by the Cauchy–Schwarz inequality, \[ (u_1 + u_2)^2 = \Bigl (\frac {u_1}{\sqrt {f_1}}\sqrt {f_1} + \frac {u_2}{\sqrt {f_2}}\sqrt {f_2}\Bigr )^2 \le \Bigl (\frac {u_1^2}{f_1} + \frac {u_2^2}{f_2}\Bigr )(f_1 + f_2) , \] so \(\nu \le f_1 + f_2 = n_1 + n_2 - 2\), with equality when \(u_1/f_1 = u_2/f_2\).

Solution. With \(z_{0.025} = 1.960\) and \(z_{0.20} = 0.842\), \[ n = \frac {2(1.960 + 0.842)^2(25)}{16} = 24.53 , \] so \(25\) per treatment. The exact power from the non-central \(t\) distribution is \(0.791\) with \(n = 25\) and \(0.807\) with \(n = 26\), so \(26\) are required. As in the text, the approximation falls short by about one replicate, because it treats \(\sigma \) as known.

Solution. The reductions are \(d = (6, 6, -1, 11, 6, 6)\), with \(\bar d = 5.667\) and \(s_d = 3.830\). Then \[ t_0 = \frac {5.667}{3.830/\sqrt 6} = 3.624 , \] against \(t_{0.025,5} = 2.571\), so \(H_0\) is rejected (\(p = 0.015\)). The interval is \(5.667 \pm 2.571(1.5635)\), that is \((1.65,\ 9.69)\).

Each patient is a block, but the two “treatments” (before and after) cannot be randomised in order: after always follows before. Any change over the month, such as the season, the attention of being in a study, or regression towards the mean in patients selected for high pressure, is confounded with the drug. A control group taking a placebo, with patients allocated at random, would be needed to separate them.

Solution. The variances are \(2\sigma _y^2(1 - \rho )/n\) and \(2\sigma _y^2/n\), so pairing is better exactly when \(\rho > 0\); when units within a pair are negatively correlated it is worse. But the tests also differ in their error degrees of freedom, \(n - 1\) against \(2n - 2\), and so in the critical value \(t_{\alpha /2}\). For small \(n\) and small positive \(\rho \) the larger critical value of the paired test can outweigh its smaller variance. In practice pairs are formed because their members are alike, \(\rho \) is substantial, and the variance reduction dominates.