Solution. Order the data: \[5,\ 8,\ 10,\ \mathbf {11},\ 11,\ 15,\ 15,\ 17,\ 20,\ 23,\ 29,\ 32,\ \mathbf {36},\ 42,\ 63,\ 145 .\] A distribution-free interval for the median is \(\left (X_{(k)},\ X_{(n+1-k)}\right )\), whose coverage is \[1 - 2\,P\!\left (\text {Bin}\!\left (n,\tfrac 12\right ) \leq k-1\right ).\] With \(n = 16\): \[\begin {array}{c|c|c} k & \text {interval} & \text {coverage}\\\hline 3 & (10,\ 42) & 0.9958\\ 4 & (11,\ 36) & 0.9787\\ 5 & (11,\ 32) & 0.9232 \end {array}\] Take the largest \(k\) whose coverage still reaches \(95\%\), namely \(k = 4\): \[\boxed {\left (X_{(4)},\ X_{(13)}\right ) = (11,\ 36) \text { minutes, with actual confidence } 97.87\% .}\]

Assumptions. Only that the sixteen observations are independent and come from a continuous distribution. Nothing is assumed about its shape — which is the point, since the value \(145\) is a severe outlier and would wreck any interval built on the mean and standard deviation.

Note on the confidence level. The attainable levels are discrete, so \(95\%\) cannot be hit exactly; one reports the interval and its true coverage rather than pretending it is \(95\%\). The next interval down, \((11,32)\), covers only \(92.3\%\) and is not a \(95\%\) interval at all.

Solution. Hypotheses. \(H_0 : \theta = 2\) against the one-sided \(H_1 : \theta > 2\), the direction fixed in advance by the pharmacology and not by the data.

Signs. Comparing each value with \(2\): \[+,\ -,\ +,\ +,\ +,\ -,\ +,\ +,\ +,\ +,\ 0 .\] The observation of exactly \(2\) gives no sign and is discarded, leaving \[n' = 10,\qquad N_{+} = 8,\qquad N_{-} = 2 .\]

Exact test. Under \(H_0\), \(N_{+} \sim B\!\left (10,\tfrac 12\right )\), so \[P\left (N_{+} \geq 8\right ) = \dfrac {\binom {10}{8}+\binom {10}{9}+\binom {10}{10}}{2^{10}} = \dfrac {45+10+1}{1024} = \dfrac {56}{1024} = 0.0547 .\]

Decision. Since \(0.0547 > 0.05\) we do not reject \(H_0\) at the \(5\%\) level. On this evidence the doctors’ doubts are not overturned.

But only just. Eight of ten moving the right way is suggestive, and the \(P\)-value misses the conventional threshold by \(0.0047\). The honest report is the \(P\)-value itself, not a verdict: a trial of even modest additional size would settle the question, and dismissing the drug on this result would be as wrong as adopting it.

Solution. The sample size is not \(85\). Only \(30 + 51 = 81\) houses are recorded above or below the claimed median, so \(4\) sold at exactly \(\$134{,}500\). Those give no sign and are discarded: \[n' = 81,\qquad N_{-} = 30,\qquad N_{+} = 51 .\]

Hypotheses. \(H_0 : \theta = 134{,}500\) against \(H_1 : \theta \neq 134{,}500\), two-sided, at \(\alpha = 0.10\).

Large-sample statistic. With \(R = \min (N_{+},N_{-}) = 30\), \[Z = \dfrac {n' - 2R - 1}{\sqrt {n'}} = \dfrac {81 - 60 - 1}{\sqrt {81}} = \dfrac {20}{9} = 2.2222 ,\] giving a two-sided \(P\)-value of \[2\left [1 - \Phi (2.2222)\right ] = 0.0263 .\] (The exact binomial two-sided value is \(0.0257\), so the approximation is good at this size.)

Decision. Since \(0.0263 < 0.10\) we reject \(H_0\): the data do not support the realtor’s claim, and the evidence points to a median above \(\$134{,}500\).

The trap. Using \(n = 85\) instead of \(n' = 81\) gives \(Z = 2.6032\) — a different and wrong answer. Ties with the hypothesised median must be removed and the sample size reduced to match; keeping \(85\) in the denominator while having dropped the four observations from the counts is inconsistent.

Solution. (a). \(E(X) = n\pi = 215(0.9) = 193.5\) and \(\operatorname {var}(X) = n\pi (1-\pi ) = 215(0.9)(0.1) = 19.35\), so the standard deviation is \(4.399\). For \(X\) to be binomial, the guests must arrive independently of one another, each with the same probability \(0.9\). A party of several guests on one booking, who arrive or fail to arrive together, breaks the independence.

(b). All guests are accommodated when \(X\leq 200\). With the continuity correction, \[P(X\leq 200)\approx P\left (Z\leq \frac {200.5-193.5}{4.399}\right ) = P(Z\leq 1.59) = 0.9441.\] The exact binomial probability is \(0.9504\). Overbooking by \(15\) rooms therefore leaves the hotel about a \(5\%\) risk of turning someone away on a given night.

Solution. (a). \(X\sim B(400, 0.02)\), so \(E(X) = 400(0.02) = 8\) and \(\sigma = \sqrt {400(0.02)(0.98)} = \sqrt {7.84} = 2.8\).

(b). With the continuity correction, \begin {align*} P(X\geq 15)&\approx P\left (Z\geq \frac {14.5-8}{2.8}\right ) = P(Z\geq 2.32)\\ &= 1-0.9898 = 0.0102. \end {align*}

If these workers had the same risk as the general population, a count as high as \(15\) would occur only about once in a hundred samples. That is strong evidence that the illness is more common in this occupation, and it is a reason to investigate the working conditions. (The exact binomial tail is \(0.016\); the normal approximation is rough here because \(n\pi = 8\) is small, but the conclusion stands.)

Solution. (a). \(H_0: p = \frac 34\) is simple: it specifies the distribution of \(T\) completely, as \(B\left (925, \frac 34\right )\). \(H_1: p\neq \frac 34\) is composite: it contains infinitely many values of \(p\).

(b). This is the binomial test. With \(n = 925\), under \(H_0\), \[E(T) = 925(0.75) = 693.75,\qquad \sigma _T = \sqrt {925(0.75)(0.25)} = 13.17.\] The observed \(T = 682\), a proportion \(\frac {682}{925} = 0.737\) tall, is below expectation, so with the continuity correction \[z = \frac {682.5-693.75}{13.17} = -0.85,\] \[p\text {-value} = 2\,\Phi (-0.85) = 2(0.1977)\approx 0.39.\] Since \(0.39 > 0.05\), \(H_0\) is not rejected: a proportion of \(73.7\%\) tall is well within the chance variation expected from a \(75\%\) model, and the data are consistent with simple Mendelian inheritance. (The exact binomial \(p\)-value is also \(0.39\).)

Solution. The interval for the median generalises to any quantile, as the quartile test does. Let \(x_{0.75}\) be the upper quartile and let \(Y\) be the number of observations below it. Each observation falls below \(x_{0.75}\) with probability \(0.75\), so \(Y\sim B(16, 0.75)\), and the interval \(\left (X_{(r)}, X_{(s)}\right )\) covers \(x_{0.75}\) exactly when \(r\leq Y\leq s-1\). Hence \[P\left (X_{(r)} < x_{0.75} < X_{(s)}\right ) = P(Y\leq s-1)-P(Y\leq r-1).\] For about \(90\%\) we want each tail near \(0.05\). From the cumulative probabilities of \(B(16, 0.75)\), \[\begin {array}{c|cccccc} y & 8 & 9 & 10 & \cdots & 14 & 15\\\hline P(Y\leq y) & 0.0271 & 0.0796 & 0.1897 & \cdots & 0.9365 & 0.9900 \end {array}\] the values closest to \(0.05\) and \(0.95\) are \(P(Y\leq 8) = 0.0271\) and \(P(Y\leq 14) = 0.9365\). So \(r-1 = 8\) and \(s-1 = 14\), giving \(r = 9\) and \(s = 15\). The interval is \[\left (X_{(9)},\ X_{(15)}\right ) = (63.3,\ 73.3),\qquad \text {confidence } 0.9365-0.0271 = 0.9094.\] We are \(90.9\%\) confident that three quarters of the batch fail before somewhere between \(63.3\) and \(73.3\) hours. The interval is not centred on \(X_{(12)}\), the natural point estimate, because the binomial with \(p = 0.75\) is not symmetric.

Solution. Test \(H_0: M = 107\) against \(H_1: M > 107\). One observation equals \(107\) and is discarded, leaving \(n' = 14\). Of these, \(N_{+} = 8\) lie above \(107\) (\(135, 108, 111, 119, 127, 109, 117, 125\)) and \(N_{-} = 6\) below. Under \(H_0\), \(N_{+}\sim B\left (14, \frac 12\right )\), and large \(N_{+}\) supports \(H_1\), so \[p\text {-value} = P(N_{+}\geq 8) = \frac {1}{2^{14}}\sum _{j=8}^{14}\binom {14}{j} = \frac {6476}{16384} = 0.3953.\] Since \(0.3953 > 0.05\) we do not reject \(H_0\). Eight of fourteen above the hypothesised median is close to the seven expected by chance, and there is no evidence that the median IQ exceeds \(107\). The signed-rank test of the next section uses the same data more fully.

Solution. (a). Test \(H_0: M = 2\) against \(H_1: M > 2\). Four of the five prices exceed \(2\), so \(N_{+} = 4\), and under \(H_0\), \(N_{+}\sim B\left (5,\frac 12\right )\): \[p\text {-value} = P(N_{+}\geq 4) = \frac {\binom 54+\binom 55}{2^{5}} = \frac {6}{32} = 0.1875.\] Since \(0.1875 > 0.05\), \(H_0\) is not rejected: there is no evidence that the median has been exceeded. With only five houses this is almost inevitable. Even if all five had sold above K2 million, the \(p\)-value would be \(\frac {1}{32} = 0.031\), so only that one outcome could lead to rejection at the \(5\%\) level.

(b). The \(t\)-test assumes the prices come from a normal population. Five observations are far too few to check that, house prices are typically skewed to the right, and the sample contains an outlying \(1.13\) which would pull the mean down and inflate the standard deviation. The question is also about the median, the usual summary of house prices precisely because of that skewness, and the sign test addresses it directly with no assumption about shape.

Solution. The same people are asked twice, so the two proportions are paired and a two-sample comparison would be wrong. McNemar’s test (Section 2.7) handles pairing as the sign test does: the \(a+d = 75\) people who did not change carry no information about the direction of change and are set aside. Each of the \(b+c = 25\) who changed is a “plus” (Democrat to Republican) or a “minus” (Republican to Democrat), and under \(H_0\) (no change in overall alignment) each switch is equally likely to go either way: \[H_0:\ P(\text {D}\to \text {R}) = P(\text {R}\to \text {D}),\qquad b\sim B\left (b+c,\tfrac 12\right ).\] The usual form of the statistic is \[T = \frac {(b-c)^{2}}{b+c} = \frac {(21-4)^{2}}{25} = \frac {289}{25} = 11.56,\] referred to \(\chi ^{2}_{1}\). Since \(11.56 > \chi ^{2}_{0.05,1} = 3.841\), \(H_0\) is rejected. (With the continuity correction, \((|b-c|-1)^{2}/(b+c) = 10.24\); the exact binomial \(p\)-value is \(2P\left (B(25,\frac 12)\leq 4\right ) = 0.0009\). Every version gives the same decision.)

The debate changed the alignment significantly. Of the \(25\) people who switched, \(21\) moved away from the Democrat. The Democrat still leads after the debate, by \(67\) to \(33\) against \(84\) to \(16\) before, but the movement was overwhelmingly one way. The statistic \(T\) is the square of the large-sample sign-test statistic \(\frac {b-c}{\sqrt {b+c}}\), which is why McNemar’s test belongs with the binomial tests.