Solution. (a) The table gives areas to the left, which is what is wanted here. Reading \(0.0427\) back out of the body of the table gives \(k=-1.72\). The sign is worth a moment: the area is well under a half, so \(k\) must lie below the mean.

(b) \(P(Z>k)=0.2946\) means \(P(Z<k)=1-0.2946=0.7054\), and reading that back gives \[k=0.54.\]

(c) Split the interval at the mean’s side. From the table \(P(Z<-0.93)=0.1762\), so \[P(Z<k)=P(Z<-0.93)+P(-0.93<Z<k)=0.1762+0.7235=0.8997\] \[\implies \quad k=1.28.\]

Solution. Every part standardises with \(Z=\frac {X-18}{2.5}\).

(a) \(z=\frac {15-18}{2.5}=-1.2\), so \(P(X<15)=P(Z<-1.2)=0.1151\).

(b) \(P(Z<z)=0.2236\) gives \(z=-0.76\), so \[k=18+(-0.76)(2.5)=16.10.\]

(c) \(P(Z>z)=0.1814\) gives \(P(Z<z)=0.8186\) and \(z=0.91\), so \[k=18+0.91(2.5)=20.28.\]

(d) \(z_1=\frac {17-18}{2.5}=-0.4\) and \(z_2=\frac {21-18}{2.5}=1.2\), so \[P(17<X<21)=P(Z<1.2)-P(Z<-0.4)=0.8849-0.3446=0.5403.\]

Solution. Here \(X\sim N(200,\,15^2)\) and \(Z=\frac {X-200}{15}\).

(a) \(z=\frac {224-200}{15}=1.6\), so \[P(X>224)=1-P(Z<1.6)=1-0.9452=0.0548,\] about \(5.5\%\) of cups.

(b) \(z=\pm 0.6\), so \[P(191<X<209)=P(Z<0.6)-P(Z<-0.6)=0.7257-0.2743=0.4514.\]

(c) \(z=\frac {230-200}{15}=2.0\), so \(P(X>230)=1-0.9772=0.0228\). Out of \(1000\) cups \[1000\times 0.0228\approx 23\ \text {cups}\] would be expected to overflow.

(d) The lowest \(25\%\) is cut off at the first quartile, where \(z=-0.67\): \[k=200+(-0.67)(15)=189.9\ \text {ml}.\]

Solution. (a) A percentile rank of \(95\) means \(95\%\) of scores fall below it, so \(z=1.645\) and \[X=500+1.645(100)=664.5.\]

(b) The middle \(40\%\) leaves \(30\%\) in each tail, so the bounds are at \(P(Z<z)=0.30\) and \(0.70\), giving \(z=\pm 0.524\): \[500\pm 0.524(100)\quad \implies \quad 447.6\ \text {and}\ 552.4.\]

(c) \(z=\frac {650-500}{100}=1.5\), so \(P(X>650)=1-0.9332=0.0668\), and of \(1000\) students \[1000\times 0.0668\approx 67\ \text {students}.\]

Solution. These run the standardisation backwards: the \(z\) is known and a parameter is not.

(a) \(P(X<700)=0.10\) gives \(z=-1.28\), and \(z=\frac {700-\mu }{70}\), so \[-1.28=\frac {700-\mu }{70}\quad \implies \quad \mu =700+1.28(70)=789.6\ \text {hours}.\]

(b) \(P(X>850)=0.20\) gives \(P(Z<z)=0.80\) and \(z=0.84\), with \(z=\frac {850-800}{\sigma }\), so \[0.84=\frac {50}{\sigma }\quad \implies \quad \sigma =\frac {50}{0.84}=59.5\ \text {hours}.\]

Solution. This question is really about telling two distributions apart. Individual heights have standard deviation \(\sigma =2.5\); sample means of size \(100\) have standard error \[\sigma _{\overline {X}}=\frac {\sigma }{\sqrt {n}}=\frac {2.5}{\sqrt {100}}=0.25.\] Parts (a) and (c) are about \(\overline {X}\); part (b) is about a single player.

(a) \(z=\frac {71.5-71}{0.25}=2.0\), so \[P(\overline {X}>71.5)=1-0.9772=0.0228=2.28\%.\]

(b) A single height, so divide by \(\sigma \) and not by the standard error: \(z=\frac {70.5-71}{2.5}=-0.2\), giving \[P(X>70.5)=1-0.4207=0.5793=57.93\%.\]

(c) \(z=\frac {70.8-71}{0.25}=-0.8\) and \(z=\frac {71.4-71}{0.25}=1.6\), so \[P(70.8<\overline {X}<71.4)=0.9452-0.2119=0.7333=73.33\%.\]

Compare (a) with (b). Half an inch above the mean is unremarkable for one player, but for the mean of a hundred it is two standard errors out. That is the whole content of \(\sigma /\sqrt {n}\).

Solution. Nothing is said about the shape of the population, and nothing needs to be: \(n=64\) is large, so the Central Limit Theorem gives \(\overline {X}\) as approximately normal with \[\sigma _{\overline {X}}=\frac {1.6}{\sqrt {64}}=0.2.\]

(a) \(z=\frac {2.7-3.2}{0.2}=-2.5\), so \(P(\overline {X}\leq 2.7)=0.0062\).

(b) \(z=\frac {3.5-3.2}{0.2}=1.5\), so \(P(\overline {X}>3.5)=1-0.9332=0.0668\).

(c) \(z=0\) and \(z=\frac {3.4-3.2}{0.2}=1.0\), so \[P(3.2\leq \overline {X}<3.4)=0.8413-0.5000=0.3413.\]

Solution. Here \(n=9\) is small, but the population is stated to be normal, so \(\overline {X}\) is exactly normal — no appeal to the Central Limit Theorem is needed. The standard error is \[\sigma _{\overline {X}}=\frac {1}{\sqrt {9}}=\frac {1}{3}=0.3333.\]

(a) \(z=\frac {4.4-5}{1/3}=-1.8\) and \(z=\frac {5.2-5}{1/3}=0.6\), so \[P(4.4<\overline {X}<5.2)=0.7257-0.0359=0.6898.\]

(b) \(15\%\) to the right means \(P(Z<z)=0.85\) and \(z=1.04\): \[\overline {X}=5+1.04\left (\tfrac {1}{3}\right )=5.35\ \text {years}.\]

(c) A total is a mean in disguise. The nine lives total more than \(50\) exactly when their mean exceeds \(50/9=5.556\), so \[z=\frac {5.556-5}{1/3}=1.67\quad \implies \quad P=1-0.9525=0.0475.\]

Solution. \(\sigma \) is known, so this is a \(Z\) interval whatever the sample size.

(a) For \(96\%\) confidence, \(\alpha =0.04\) and \(\alpha /2=0.02\), so \(z_{0.02}=2.05\): \[\overline {X}\pm z_{\alpha /2}\frac {\sigma }{\sqrt {n}} =780\pm 2.05\times \frac {40}{\sqrt {30}}=780\pm 14.97\] \[\implies \quad (765.03,\ 794.97).\]

(b) ”Within \(10\) hours” fixes the margin of error at \(E=10\). Rearranging \(E=z_{\alpha /2}\frac {\sigma }{\sqrt {n}}\), \[n=\left (\frac {z_{\alpha /2}\,\sigma }{E}\right )^2=\left (\frac {2.05\times 40}{10}\right )^2=67.24 \quad \implies \quad n=68.\] Always round a sample size up. Rounding to \(67\) would leave the interval slightly wider than asked for, which is the one direction that fails the requirement.

Solution. From the nine readings, \(\overline {X}=1.0056\) and \(\widehat {S}=0.0246\), so \(\widehat {S}^2=0.000603\).

(a) \(\sigma \) is unknown and \(n=9\) is small, so this is a \(t\) interval with \(v=n-1=8\) degrees of freedom and \(t_{8,\,0.005}=3.355\): \[1.0056\pm 3.355\times \frac {0.0246}{\sqrt {9}}=1.0056\pm 0.0275 \quad \implies \quad (0.978,\ 1.033).\]

(b) A variance is estimated with the chi-square distribution, and the interval is not symmetric: \[\left (\frac {(n-1)\widehat {S}^2}{\chi ^2_{v,\,\alpha /2}},\ \frac {(n-1)\widehat {S}^2}{\chi ^2_{v,\,1-\alpha /2}}\right )\] With \(v=8\) and \(\alpha =0.10\), the two critical values are \(\chi ^2_{8,\,0.05}=15.507\) and \(\chi ^2_{8,\,0.95}=2.733\), so \[\left (\frac {8(0.000603)}{15.507},\ \frac {8(0.000603)}{2.733}\right )=(0.00031,\ 0.00177).\] Note the larger critical value goes underneath the lower limit. Getting these the wrong way round is the commonest error here, and it produces an interval that does not contain the estimate.

Solution. (a) \(\sigma \) known, \(z_{0.025}=1.96\): \[2.25\pm 1.96\times \frac {0.15}{\sqrt {36}}=2.25\pm 0.049 \quad \implies \quad (2.201,\ 2.299).\]

(b) With \(E=0.09\), \[n=\left (\frac {1.96\times 0.15}{0.09}\right )^2=10.67\quad \implies \quad n=11.\] Fewer than the \(36\) already taken, because \(0.09\) is a looser requirement than the \(\pm 0.049\) the current sample achieves.

Solution. This runs a confidence interval backwards.

(a) The interval is symmetric about the sample mean, so \[\overline {X}=\frac {177.22+179.18}{2}=178.20\ \text {cm}.\] Its half-width is the margin of error, \[E=\frac {179.18-177.22}{2}=0.98=z_{0.025}\frac {\sigma }{\sqrt {100}}=1.96\frac {\sigma }{10}\] \[\implies \quad \sigma =\frac {0.98\times 10}{1.96}=5\ \text {cm}.\]

(b) For \(98\%\), \(z_{0.01}=2.33\): \[178.20\pm 2.33\times \frac {5}{10}=178.20\pm 1.165 \quad \implies \quad (177.04,\ 179.37).\] Wider than the \(95\%\) interval, as more confidence always is.

Solution. \(\sigma \) is unknown, so a \(t\) interval with \(v=32\) and \(t_{32,\,0.025}=2.037\).

(a) \[41.8\pm 2.037\times \frac {5.7}{\sqrt {33}}=41.8\pm 2.02 \quad \implies \quad (39.78,\ 43.82).\]

(b) The maximum error of estimate is the half-width just computed, \[E=2.037\times \frac {5.7}{\sqrt {33}}=2.02\ \text {calories}.\] It is the most \(\overline {X}\) is likely to be wrong by, at this level of confidence — not a guarantee, but a bound that holds \(95\) times in \(100\).

Solution. (a) \(t\) interval with \(v=27\) and \(t_{27,\,0.025}=2.052\): \[28.7\pm 2.052\times \frac {5.7}{\sqrt {28}}=28.7\pm 2.21 \quad \implies \quad (26.49,\ 30.91).\]

(b) Yes. The value \(36.2\) lies well outside \((26.49,\ 30.91)\) — not marginally, but about five calories beyond the upper limit. A confidence interval can be read as a test in this way: any value it excludes would be rejected as a null hypothesis at the corresponding significance level, here \(5\%\).

(c) For a standard deviation, build the interval for the variance first and take square roots at the end. With \(v=27\), \(\alpha =0.01\), \(\chi ^2_{27,\,0.005}=49.645\) and \(\chi ^2_{27,\,0.995}=11.808\): \[\left (\frac {27(5.7)^2}{49.645},\ \frac {27(5.7)^2}{11.808}\right )=(17.67,\ 74.29)\] and taking roots, \[(4.20,\ 8.62).\] Taking the root of each limit is legitimate because the square root is increasing, so it preserves the order of the endpoints.

Solution. (a) The sample proportion is \(\hat {P}=\frac {114}{200}=0.57\), and for a proportion the standard error is \(\sqrt {\frac {\hat {P}(1-\hat {P})}{n}}\). With \(z_{0.02}=2.05\): \[0.57\pm 2.05\sqrt {\frac {0.57(0.43)}{200}}=0.57\pm 0.0718 \quad \implies \quad (0.498,\ 0.642).\] Worth noticing: the interval straddles \(0.5\), so this sample does not establish that a majority favours the suit, even though \(57\%\) of those asked did.

(b) With \(E=0.02\), \[n=\left (\frac {z_{\alpha /2}}{E}\right )^2\hat {P}(1-\hat {P}) =\left (\frac {2.05}{0.02}\right )^2(0.57)(0.43)=2575.1\quad \implies \quad n=2576.\] Roughly thirteen times the current sample, to cut the margin from \(0.072\) to \(0.02\). Halving a margin costs four times the sample; this is the \(\sqrt {n}\) again.

Solution. (a) \(\hat {P}=\frac {24}{100}=0.24\) and \(z_{0.005}=2.575\): \[0.24\pm 2.575\sqrt {\frac {0.24(0.76)}{100}}=0.24\pm 0.110 \quad \implies \quad (0.130,\ 0.350).\]

(b) Note the change of confidence level: \(95\%\) here, not \(99\%\), so \(z_{0.025}=1.96\) and \[n=\left (\frac {1.96}{0.04}\right )^2(0.24)(0.76)=437.9\quad \implies \quad n=438.\] If no prior estimate of \(P\) were available, \(\hat {P}=0.5\) would be used instead, since \(P(1-P)\) is largest there and that gives the sample size that is safe whatever \(P\) turns out to be — here that would be \(601\).

Solution. Using the identity \(\sum (y-\overline {y})^2=\sum y^2-n\overline {y}^2\), \[E(s^2)=\frac {1}{n}\left [\sum E(y^2)-nE(\overline {y}^2)\right ].\] For each observation \(E(y^2)=\operatorname {var}(Y)+[E(Y)]^2=\sigma ^2+\mu ^2\), and for the sample mean \(E(\overline {y}^2)=\operatorname {var}(\overline {y})+\mu ^2=\frac {\sigma ^2}{n}+\mu ^2\). Substituting, \begin {align*} E(s^2)&=\frac {1}{n}\left [n(\sigma ^2+\mu ^2)-n\left (\frac {\sigma ^2}{n}+\mu ^2\right )\right ]\\ &=\frac {1}{n}\left [n\sigma ^2+n\mu ^2-\sigma ^2-n\mu ^2\right ]\\ &=\frac {n-1}{n}\,\sigma ^2. \end {align*}

So \(s^2\) is an estimator of \(\sigma ^2\), but a biased one: on average it falls short by a factor \(\frac {n-1}{n}\). The bias is \(-\frac {\sigma ^2}{n}\), which vanishes as \(n\) grows, so \(s^2\) is asymptotically unbiased.

Multiplying through by \(\frac {n}{n-1}\) removes the bias entirely and gives \[\widehat {S}^2=\frac {\sum (y-\overline {y})^2}{n-1},\] which is why the divisor \(n-1\) appears wherever a variance is estimated from a sample. The reason it must be smaller than \(n\): the deviations are measured from \(\overline {y}\), which was itself computed from the same data and therefore sits closer to the observations than the true \(\mu \) does.

Solution. The sample variance is \(256\), so \(s=16\), and \(n=64\) is large, so a \(Z\) interval serves. \[\text {S.E}=\frac {s}{\sqrt {n}}=\frac {16}{8}=2,\qquad z_{0.05}=1.645\] \[33\pm 1.645(2)=33\pm 3.29\quad \implies \quad (29.71,\ 36.29)\ \text {minutes}.\]

Solution. \[\hat {P}_1=\frac {12}{50}=0.24,\qquad \hat {P}_2=\frac {12}{60}=0.20,\qquad \hat {P}_1-\hat {P}_2=0.04.\] For a difference of proportions the variances add: \[\text {S.E}=\sqrt {\frac {\hat {P}_1(1-\hat {P}_1)}{n_1}+\frac {\hat {P}_2(1-\hat {P}_2)}{n_2}} =\sqrt {\frac {0.24(0.76)}{50}+\frac {0.20(0.80)}{60}}=0.0795.\] With \(z_{0.01}=2.33\), \[0.04\pm 2.33(0.0795)=0.04\pm 0.185\quad \implies \quad (-0.145,\ 0.225).\]

The interval contains zero, so at this level of confidence the data give no evidence that the two brands differ in reliability. The observed gap of four percentage points is comfortably within what sampling variation alone would produce from samples this size.

Solution. ”Error less than \(0.04\) with probability \(0.90\)” sets \(E=0.04\) at \(90\%\) confidence, so \(z_{0.05}=1.645\): \[n=\left (\frac {z_{\alpha /2}}{E}\right )^2 P(1-P) =\left (\frac {1.645}{0.04}\right )^2(0.6)(0.4)=405.8\quad \implies \quad n=406.\]

Using the prior guess \(P\approx 0.6\) is what makes this smaller than it would otherwise be. With no information at all, \(P=0.5\) would be assumed, since \(P(1-P)\) is largest there, giving \(n=423\) — the sample size that is safe whatever \(P\) turns out to be.

Solution. From the eight readings, \(\overline {X}=2958.75\) and \(\widehat {S}=39.26\).

The sample is small and \(\sigma \) unknown, so this is a \(t\) interval with \(v=7\) and \(t_{7,\,0.025}=2.365\): \[2958.75\pm 2.365\times \frac {39.26}{\sqrt {8}}=2958.75\pm 32.82\] \[\implies \quad (2925.93,\ 2991.57)\ \text {m/s}.\]

Solution. Here \(n=6\), \(\overline {X}=3.683\) and \(\widehat {S}=1.905\). With \(v=5\) and \(t_{5,\,0.05}=2.015\): \[3.683\pm 2.015\times \frac {1.905}{\sqrt {6}}=3.683\pm 1.567\] \[\implies \quad (2.12,\ 5.25)\ \text {hundred Kwacha}.\]

The interval is very wide — it spans more than three hundred Kwacha — because six observations with this much scatter simply cannot pin the mean down. That width is information, not a failure of the method.

Solution. A common variance is assumed, so the two sample variances are pooled into one estimate: \[S_p^2=\frac {(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2} =\frac {10(31.4)+10(44.8)}{20}=\frac {762}{20}=38.1\] \[\implies \quad S_p=6.17.\] The pooled estimate carries \(n_1+n_2-2=20\) degrees of freedom, and \(t_{20,\,0.005}=2.845\): \[\text {S.E}=S_p\sqrt {\frac {1}{n_1}+\frac {1}{n_2}}=6.17\sqrt {\frac {2}{11}}=2.632\] \[(60.4-65.3)\pm 2.845(2.632)=-4.9\pm 7.49 \quad \implies \quad (-12.39,\ 2.59).\]

Again the interval contains zero, so there is no evidence at the \(1\%\) level that the two population means differ — despite the sample means being nearly five apart. Eleven observations each, with variances of this size, is not enough to resolve a gap that small.

Solution. (a) This is a whole population of five, so divide by \(N\) and not by \(N-1\): \[\mu =\frac {10+6+0+4+0}{5}=\frac {20}{5}=4\ \text {days}\] \[\sigma ^2=\frac {(10-4)^2+(6-4)^2+(0-4)^2+(4-4)^2+(0-4)^2}{5}=\frac {36+4+16+0+16}{5}=\frac {72}{5}=14.4.\]

(b)(i) There are \(\binom {5}{3}=10\) equally likely samples. Listing their totals and means:

The probabilities total \(1\), which is the check.

(b)(ii) \[E(\overline {X})=\sum \overline {X}\,P(\overline {X}) =\frac {1}{10}\left (\frac {4}{3}\right )+\frac {1}{10}(2)+\frac {3}{10}\left (\frac {10}{3}\right ) +\frac {2}{10}\left (\frac {14}{3}\right )+\frac {2}{10}\left (\frac {16}{3}\right )+\frac {1}{10}\left (\frac {20}{3}\right )=4.\] Since \(E(\overline {X})=\mu =4\), the sample mean is an unbiased estimator — and note that this holds even though the sampling is without replacement.

(b)(iii) \[\operatorname {var}(\overline {X})=\sum \left (\overline {X}-4\right )^2 P(\overline {X})=\frac {12}{5}=2.4.\] With replacement the variance would be \[\frac {\sigma ^2}{n}=\frac {14.4}{3}=4.8,\] and indeed \(2.4=\tfrac {1}{2}(4.8)\), as the question claims.

The factor of one half is not a coincidence. Sampling without replacement from a finite population multiplies the variance by the finite population correction \[\frac {N-n}{N-1}=\frac {5-3}{5-1}=\frac {2}{4}=\frac {1}{2},\] and \(4.8\times \frac {1}{2}=2.4\). Drawing three of only five employees uses up a large part of the population, so the sample mean varies less than it would if the same employee could be picked twice.

Solution. (a)

(i).

A statistic is any quantity computed from the sample alone, such as \(\overline {X}\) or \(\widehat {S}^2\). It contains no unknowns, so it can always be evaluated.

(ii).

A parameter is a fixed numerical property of the population, such as \(\mu \) or \(\sigma ^2\). It is generally unknown, which is why it must be estimated.

(iii).

An estimator \(\widehat {\theta }\) is unbiased for \(\theta \) if \(E(\widehat {\theta })=\theta \): averaged over all possible samples it hits the target. It says nothing about any single sample, only about the long run.

(iv).

The standard error is the standard deviation of the sampling distribution of a statistic — for the sample mean, \(\sigma /\sqrt {n}\). It measures how much the statistic would vary from sample to sample, and is not the same as \(\sigma \), which measures variation between individual observations.

(b) Note carefully that the deviations here are taken about \(\mu \), the true mean, and not about \(\overline {X}\). For each observation \[E\left [(x_i-\mu )^2\right ]=\operatorname {var}(X)=\sigma ^2\] directly from the definition of variance. Hence \[E\left [\frac {1}{n}\sum _{i=1}^{n}(x_i-\mu )^2\right ] =\frac {1}{n}\sum _{i=1}^{n}E\left [(x_i-\mu )^2\right ]=\frac {1}{n}\left (n\sigma ^2\right )=\sigma ^2.\]

So with \(\mu \) known, dividing by \(n\) is correct. This is worth setting beside the earlier result that \(\frac {1}{n}\sum (x_i-\overline {X})^2\) is biased, with expectation \(\frac {n-1}{n}\sigma ^2\). The difference is entirely in the centre used. \(\overline {X}\) is computed from the same data and therefore sits closer to the observations than \(\mu \) does, so deviations about it are systematically too small — and one degree of freedom must be given up to compensate. When \(\mu \) is known, nothing has been estimated and nothing is lost.

Solution. The current sample gives \(\hat {P}=0.40\) and a \(95\%\) margin of error \[E=z_{0.025}\sqrt {\frac {\hat {P}(1-\hat {P})}{n}}=1.96\sqrt {\frac {0.4(0.6)}{100}}=1.96(0.049)=0.0960.\] To hold that same margin at \(99\%\) confidence, where \(z_{0.005}=2.576\), \[n=\left (\frac {z_{0.005}}{E}\right )^2\hat {P}(1-\hat {P}) =\left (\frac {2.576}{0.0960}\right )^2(0.24)=172.7\quad \implies \quad n=173.\]

There is a quicker route worth knowing. Holding \(E\) and \(\hat {P}\) fixed, \(n\) is proportional to \(z^2\), so \[n_{99}=n_{95}\left (\frac {z_{0.005}}{z_{0.025}}\right )^2=100\left (\frac {2.576}{1.96}\right )^2=100(1.727)=172.7,\] the same answer without recomputing the margin. Buying the extra confidence costs about \(73\%\) more data — a reminder that confidence is not free, and that the jump from \(95\%\) to \(99\%\) is a more expensive one than it sounds.

Solution. (a)(i) Insert \(\overline {x}\) and expand: \begin {align*} \sum (x_i-c)^2 &=\sum \left [(x_i-\overline {x})+(\overline {x}-c)\right ]^2\\ &=\sum (x_i-\overline {x})^2+2(\overline {x}-c)\sum (x_i-\overline {x})+n(\overline {x}-c)^2. \end {align*}

The middle term vanishes because \(\sum (x_i-\overline {x})=0\), and \((\overline {x}-c)^2=(c-\overline {x})^2\), giving \[\sum (x_i-c)^2=\sum (x_i-\overline {x})^2+n(c-\overline {x})^2.\]

(a)(ii) The first term does not involve \(c\) and the second is \(n(c-\overline {x})^2\geq 0\), vanishing only when \(c=\overline {x}\). So the sum of squared deviations is least when taken about \(\boxed {c=\overline {x}}\) — which is precisely the property that makes the mean the natural centre for a least-squares method.

(b)(i) \[\overline {x}=\frac {5472}{12}=456.\] The constant in the second sum is \(456\), which is exactly \(\overline {x}\), so by part (a) the correction term \(n(c-\overline {x})^2\) is zero and \[\sum (x_i-\overline {x})^2=1620.\] The unbiased estimate of the variance is therefore \[\widehat {\sigma }^2=\frac {1620}{n-1}=\frac {1620}{11}=147.27,\qquad \widehat {\sigma }=12.14.\]

(b)(ii) With \(\sigma \) unknown and \(n=12\), use \(t\) with \(11\) degrees of freedom, \(t_{11,\,0.025}=2.201\): \[456\pm 2.201\times \frac {12.14}{\sqrt {12}}=456\pm 7.71 \quad \implies \quad (448.29,\ 463.71).\]

(b)(iii) Run the same construction backwards. The interval \((451,463)\) is symmetric about its centre, so \[\overline {x}=\frac {451+463}{2}=457,\] and its half-width is the margin of error: \[6=t_{11,\,0.025}\frac {\widehat {\sigma }}{\sqrt {12}}=2.201\frac {\widehat {\sigma }}{\sqrt {12}} \quad \implies \quad \widehat {\sigma }=\frac {6\sqrt {12}}{2.201}=9.44\] \[\implies \quad \widehat {\sigma }^2=89.2.\] The second sample is centred slightly higher and is appreciably less variable than the first.

Solution. (a) \[\overline {x}=\frac {2.57}{10}=0.257\ \text {kg}\] \[\sum (x-\overline {x})^2=\sum x^2-\frac {(\sum x)^2}{n}=0.6610-\frac {2.57^2}{10}=0.6610-0.66049=0.00051\] \[\widehat {S}^2=\frac {0.00051}{9}=0.0000567\quad \implies \quad \widehat {S}=0.00753\ \text {kg}.\] With \(\sigma \) unknown and \(n=10\), use \(t\) with \(9\) degrees of freedom, \(t_{9,\,0.025}=2.262\): \[0.257\pm 2.262\times \frac {0.00753}{\sqrt {10}}=0.257\pm 0.00538 \quad \implies \quad (0.2516,\ 0.2624)\ \text {kg}.\] Note how much cancellation there is in \(\sum x^2-\frac {(\sum x)^2}{n}\): two numbers near \(0.66\) leave a difference of \(0.00051\). Rounding either total early would destroy the answer, which is a good reason to carry full precision to the last step.

(b) The requirement is on the width, not the margin. The interval runs \(\overline {x}\pm E\), so its width is \(2E\): \[2E<0.002\quad \implies \quad E<0.001.\] With \(\sigma \) now known, use \(z_{0.025}=1.96\): \[n=\left (\frac {z_{\alpha /2}\,\sigma }{E}\right )^2=\left (\frac {1.96\times 0.008}{0.001}\right )^2=245.9 \quad \implies \quad n=246.\] Reading ”width” as ”margin” would give \(n=62\), a quarter of the correct figure — the single commonest error in sample-size questions.

Solution.

1.

\(H_0:\ \mu =15\) (the manager’s claim); \(H_1:\ \mu >15\). He would like the figure to be higher and is asserting it is not, so the alternative of interest is that it exceeds \(15\).

2.

\(\alpha =5\%\), one-tailed.

3.

The variance is \(9\), so \(s=3\), and with \(n=36\) \[\text {S.E}=\frac {3}{\sqrt {36}}=0.5,\qquad Z=\frac {17-15}{0.5}=4.0.\]

4.

Critical region: \(Z>1.645\).

5.

\(4.0>1.645\), so we reject \(H_0\). The evidence contradicts the manager’s claim: the salesmen are averaging significantly more than \(15\) contacts a week.

\(Z=4.0\) gives a \(P\)-value of about \(0.00003\), so this is not a marginal call. Note that \(n=36\) allows \(Z\) to be used even though \(\sigma \) is estimated by \(s\).

Solution. (a) The errors are assumed to

(i).

have mean zero, \(E(\varepsilon _i)=0\), so the line is right on average;

(ii).

have constant variance, \(\operatorname {var}(\varepsilon _i)=\sigma ^2\) for every \(i\) — homoscedasticity, meaning the scatter about the line does not widen along it;

(iii).

be independent of one another, so \(\operatorname {cov}(\varepsilon _i,\varepsilon _j)=0\) for \(i\neq j\);

(iv).

be normally distributed — needed for the \(t\) tests and confidence intervals, though not for the estimates themselves.

(b) Least squares chooses \(\alpha \) and \(\beta \) to minimise \[S=\sum _{i=1}^{n}\left (y_i-\alpha -\beta x_i\right )^2.\] Differentiating and setting each derivative to zero: \[\frac {\partial S}{\partial \alpha }=-2\sum \left (y_i-\alpha -\beta x_i\right )=0 \quad \implies \quad \sum y_i=n\alpha +\beta \sum x_i\] \[\frac {\partial S}{\partial \beta }=-2\sum x_i\left (y_i-\alpha -\beta x_i\right )=0 \quad \implies \quad \sum x_iy_i=\alpha \sum x_i+\beta \sum x_i^2.\] These are the normal equations. The first gives \(\widehat {\alpha }=\overline {y}-\widehat {\beta }\,\overline {x}\) immediately — so the fitted line passes through \((\overline {x},\overline {y})\) — and substituting into the second and rearranging yields \[\widehat {\beta }=\frac {\sum x_iy_i-\frac {\sum x_i\sum y_i}{n}}{\sum x_i^2-\frac {(\sum x_i)^2}{n}} =\frac {S_{XY}}{S_{XX}}.\]

(c) For the five points, \(n=5\) with \[\sum x=4,\quad \sum y=-1,\quad \sum x^2=6,\quad \sum xy=5\] \[S_{XX}=6-\frac {16}{5}=2.8,\qquad S_{XY}=5-\frac {(4)(-1)}{5}=5.8\] \[\widehat {\beta }=\frac {5.8}{2.8}=2.071,\qquad \widehat {\alpha }=-0.2-2.071(0.8)=-1.857\] \[\implies \quad \widehat {Y}=-1.857+2.071X.\]

Solution. Heights are given to the nearest cm, so the true boundaries are \(119.5, 129.5, \ldots , 179.5\) and the mid-values are \(124.5, 134.5, \ldots , 174.5\). Here \(n=57\).

(a) \[\overline {X}=\frac {\sum fX}{\sum f}=\frac {8426.5}{57}=147.83\ \text {cm}\] \[\sigma ^2=\frac {\sum f(X-\overline {X})^2}{\sum f}=\frac {6866.7}{57}=120.47 \quad \implies \quad \sigma =10.98\ \text {cm}.\]

(b), (c)

The cumulative frequency curve is plotted against the upper boundaries \(129.5, 139.5, \ldots \) Interpolating within the class containing each position: \begin {align*} Q_1\ (0.25\times 57=14.25):&\quad 139.5+\frac {14.25-14}{17}(10)=139.6\ \text {cm}\\ Q_3\ (0.75\times 57=42.75):&\quad 149.5+\frac {42.75-31}{18}(10)=156.0\ \text {cm}\\ D_7\ (0.70\times 57=39.9):&\quad 149.5+\frac {39.9-31}{18}(10)=154.4\ \text {cm}\\ P_{90}\ (0.90\times 57=51.3):&\quad 159.5+\frac {51.3-49}{7}(10)=162.8\ \text {cm} \end {align*}

The interquartile range is \(156.0-139.6=16.4\) cm, so the middle half of the class lies within about \(16\) cm of height.

Solution. With \(n=50\) the sample is large, so a \(Z\) interval serves: \[\text {S.E}=\frac {0.75}{\sqrt {50}}=0.1061,\qquad z_{0.025}=1.96\] \[9.8\pm 1.96(0.1061)=9.8\pm 0.208 \quad \implies \quad (9.592,\ 10.008)\ \text {m/s}^2.\]

The accepted value of \(g\), about \(9.81\) m/s\(^2\), lies comfortably inside, so these measurements are consistent with it. The interval is wide, though — a spread of \(0.4\) across an estimate of \(9.8\) — because a standard deviation of \(0.75\) is large for this measurement. The apparatus, not the sample size, is the limiting factor.

Solution. (a)(i) The target population is the whole group about which conclusions are wanted; the study population is the part of it actually available to be sampled. For a survey of Zambian university students, the target population is every such student, while the study population might be students at the institutions that agreed to take part. Any gap between the two limits how far the conclusions carry, and it is a gap no amount of extra sampling can close.

(a)(ii) A statistic is computed from a sample and varies from sample to sample — it is a random variable with its own distribution. A constant has one fixed value that does not depend on the sample, such as \(\pi \), or the divisor \(n\). Note that a parameter is also fixed, but is a property of the population and generally unknown; a constant is known.

(b)(i) Taking the tens digit as the stem:

(b)(ii) With \(n=40\), using \(\tfrac {k}{4}(n+1)\): \[Q_1=10.25\text {th}=68,\qquad Q_2=20.5\text {th}=79,\qquad Q_3=30.75\text {th}=88.75.\]

(b)(iii) \[\overline {X}=\frac {3102}{40}=77.55.\] The mean is below the median of \(79\), so the distribution is slightly skewed to the left. The stem plot shows why: the leaves pile up at stems \(7\) and \(8\) and thin out downwards through stem \(5\).

(b)(iv) With \(X\sim N(80,12^2)\), \[z=\frac {77.55-80}{12}=-0.20\quad \implies \quad P(X>77.55)=1-P(Z<-0.20)=1-0.4207=0.579.\] About \(58\%\) of applicants would be expected to score above this group’s mean — unsurprising, since \(77.55\) sits a little below the population mean of \(80\).

Solution. (a) A frequency distribution should have classes that are exhaustive — every observation falls in one — and mutually exclusive, so none falls in two. In practice that means stating the boundaries unambiguously and leaving no gaps. Classes of equal width are usually preferred as well, since unequal widths force the use of frequency density.

The advantage of a summary is that it makes the shape of a large set of data visible at once, which a list of \(134\) numbers does not. The disadvantage is that the individual values are lost: everything in a class is thereafter treated as sitting at the mid-point, so any figure computed from the table is an estimate.

(b)(i) Mid-points \(35, 45, \ldots , 95\): \[\overline {X}=\frac {\sum fX}{\sum f}=\frac {9000}{134}=67.16\ \text {seconds}.\]

(b)(ii) \[\sigma ^2=\frac {\sum f(X-\overline {X})^2}{\sum f}=\frac {40\,272.4}{134}=300.5 \quad \implies \quad \sigma =17.34\ \text {seconds}.\]

(b)(iii) The plain standard error is \[\frac {\sigma }{\sqrt {n}}=\frac {17.34}{\sqrt {134}}=1.50\ \text {seconds}.\] But the population is only \(500\) workers and the sample of \(134\) is more than a quarter of it, so the finite population correction should be applied: \[\text {S.E}=\frac {\sigma }{\sqrt {n}}\sqrt {\frac {N-n}{N-1}} =1.50\sqrt {\frac {500-134}{499}}=1.50(0.856)=1.28\ \text {seconds}.\] Mentioning the \(500\) is the whole point of part (iii). Without it the sample would be treated as drawn from an unlimited population and the standard error overstated by about \(17\%\).

Solution. (a) Randomisation protects against bias from sources nobody has thought of. Known nuisance factors can be controlled by blocking, but there are always others — the order units were processed, small differences in equipment, the time of day. Allocating treatments at random means such factors are as likely to favour one treatment as another, so they enter the error term rather than masquerading as a treatment effect. It is also what justifies the probability statements the tests rest on.

Three methods of sampling: simple random sampling, where every subset of the required size is equally likely; systematic sampling, taking every \(k\)th item from an ordered list; and stratified sampling, dividing the population into strata and sampling within each. Cluster sampling is a fourth.

(b)(i) The same ten cars appear in both rows. The two columns are not independent: a car that is thirsty before a tune-up is thirsty after it. The data are ten matched pairs, and must be analysed as the differences.

(b)(ii) Taking \(d=\text {after}-\text {before}\):

1.

\(H_0:\ \mu _d=0\); \(H_1:\ \mu _d\neq 0\).

2.

\(\alpha =5\%\), two-tailed.

3.

\(\overline {d}=1.9\) and \(\widehat {S}_d=2.079\), so \[t=\frac {1.9}{2.079/\sqrt {10}}=\frac {1.9}{0.657}=2.89\] on \(9\) degrees of freedom.

4.

Critical region: \(|t|>t_{9,\,0.025}=2.262\).

5.

\(2.89>2.262\), so we reject \(H_0\). There is evidence at the \(5\%\) level that the tune-up changes fuel consumption, improving it by about \(1.9\) km per litre on average.

Here the pairing does not merely sharpen the test — it decides the outcome. Treated wrongly as two independent samples, the statistic would be \(t=0.61\) with a \(P\)-value of \(0.55\), and the tune-up would be pronounced useless. The car-to-car variation is enormous, from \(7\) to \(31\) km per litre, and it swamps an effect of \(1.9\). Pairing removes that variation entirely, because each car is compared only with itself.

Solution. This is a two-way layout with one observation per cell: both fertilizer and variety are factors of interest, so neither is a nuisance to be blocked away. With no replication there is no separate estimate of interaction, and the residual serves as the error term.

With \(N=12\) and \(T=1009\), \[CF=\frac {T^2}{N}=\frac {1009^2}{12}=84\,840.08.\] The column of \(\sum y_{ij}^2\) totals \(28\,125+31\,955+25\,155=85\,235\), so \begin {align*} SS_{\text {Total}} &=85\,235-84\,840.08=394.92\\ SS_{\text {Fertilizer}} &=\frac {335^2+357^2+317^2}{4}-CF=200.67\\ SS_{\text {Variety}} &=\frac {253^2+259^2+241^2+256^2}{3}-CF=62.25\\ SS_{\text {Error}} &=394.92-200.67-62.25=132.00 \end {align*}

Each total is divided by the number of observations behind it — fertilizer totals by the \(4\) varieties, variety totals by the \(3\) fertilizers.

(b)(i) Fertilizers. \(F_{2,6,\,0.05}=5.14\). Since \(4.56<5.14\) we fail to reject \(H_0\): there is no evidence at the \(5\%\) level of a difference between the three fertilizers.

(b)(ii) Varieties. \(F_{3,6,\,0.05}=4.76\). Since \(0.94<4.76\) we fail to reject \(H_0\): no evidence of a difference between the four varieties.

The fertilizer result deserves a sentence rather than a bare verdict. At \(F=4.56\) against a critical \(5.14\) the \(P\)-value is about \(0.062\) — outside \(5\%\), but only just, and the fertilizer totals of \(335\), \(357\) and \(317\) do differ noticeably. Reporting this as ”no effect” would overstate what the test shows. With only one observation per cell there are just \(6\) error degrees of freedom, and a design with replication would very likely settle the question. The variety result, at \(F=0.94\), needs no such caution: it is close to the value of \(1\) expected when a factor does nothing at all.

Solution. (a). The students were paired on IQ and scholastic performance, so the two rows are not independent samples. The interval must be built from the seven differences, not from the two means separately.

With \(n=7\), \(\sum d=15\) and \(\sum d^2=79\): \[\overline {d}=\frac {15}{7}=2.143,\qquad \hat {S}_d^2=\frac {\sum d^2-(\sum d)^2/n}{n-1}=\frac {79-225/7}{6}=\frac {46.857}{6}=7.810,\] so \(\hat {S}_d=2.795\). This is one sample of seven differences with \(\sigma \) unknown, so the multiplier is \(t\) on \(n-1=6\) degrees of freedom, \(t_{6,\,0.025}=2.447\): \[\overline {d}\pm t_{6,\,0.025}\frac {\hat {S}_d}{\sqrt {n}} =2.143\pm 2.447\times \frac {2.795}{\sqrt {7}} =2.143\pm 2.447(1.056) =2.143\pm 2.585.\] \[-0.44<\mu _d<4.73\]

(b). The interval contains zero. A mean difference of \(0\) — the two programmes performing identically — is among the values the data leave plausible, so at the \(5\%\) level there is no evidence of a difference between the programmes.

That verdict deserves qualifying rather than simply recording. The point estimate is \(+2.14\) marks in favour of the special programme, and six of the seven differences are positive; the interval fails to exclude zero mainly because seven pairs is very little data. The honest reading is not “the programmes are the same” but “seven pairs cannot tell them apart”. Note also what the pairing bought: the raw scores range from \(60\) to \(96\), while the differences range only from \(-2\) to \(6\). Almost all the variation is between students, and pairing removes it from the comparison entirely.

Solution. (a). Both samples are large (\(45\) and \(50\)), so the sample variances may be used in place of the unknown population variances and the multiplier is \(Z\): \[(\overline {X}_1-\overline {X}_2)\pm Z_{\alpha /2} \sqrt {\frac {\hat {S}_1^2}{n_1}+\frac {\hat {S}_2^2}{n_2}}.\] With \(\overline {X}_1-\overline {X}_2=2.10-2.45=-0.35\) and \(Z_{0.025}=1.96\): \[\sqrt {\frac {0.64}{45}+\frac {0.70}{50}}=\sqrt {0.014222+0.014000}=\sqrt {0.028222}=0.1680,\] \[-0.35\pm 1.96(0.1680)=-0.35\pm 0.3293.\] \[-0.679<\mu _1-\mu _2<-0.021\]

(b). The interval lies entirely below zero, so zero is not a plausible value for \(\mu _1-\mu _2\). The data do support the administrator: at the \(5\%\) level there is a difference, and its direction is settled too — the whole interval is negative, so the female mean is the larger, by somewhere between \(0.02\) and \(0.68\) of a grade point.

The lower end of that range is worth a second look. An interval that only just excludes zero is statistically significant and practically almost empty: a difference of \(0.02\) of a grade point would matter to nobody. Significance answers “is there a difference?”, not “is the difference large enough to care about?”, and the interval — unlike a bare test — lets you answer both.

Solution. (a). The two samples are small and the population variances unknown, so this is the pooled \(t\) interval. From the data, \[\overline {X}_A=\frac {705}{10}=70.5,\qquad \overline {X}_B=\frac {713}{10}=71.3,\] \[\hat {S}_A^2=60.944,\qquad \hat {S}_B^2=48.456.\] Pooling, \[S_p^2=\frac {(n_A-1)\hat {S}_A^2+(n_B-1)\hat {S}_B^2}{n_A+n_B-2} =\frac {9(60.944)+9(48.456)}{18}=\frac {109.400}{2}=54.700,\] so \(S_p=7.396\). With \(t_{18,\,0.05}=1.734\) and \(\overline {X}_B-\overline {X}_A=0.8\): \[0.8\pm 1.734\times 7.396\sqrt {\tfrac {1}{10}+\tfrac {1}{10}} =0.8\pm 1.734(3.308)=0.8\pm 5.736.\] \[-4.94<\mu _B-\mu _A<6.54\]

(b). The interval contains zero, so there is no evidence at the \(10\%\) level that the two methods differ. The estimated advantage of method B is \(0.8\) words per minute, against an interval more than eleven words wide — the estimate is swamped by its own uncertainty.

(c). Three assumptions:

(i).

Both populations of typing speeds are approximately normal. With \(n=10\) in each sample the Central Limit Theorem is no help, so this is a real requirement.

(ii).

The population variances are equal, \(\sigma _A^2=\sigma _B^2\) — this is what licenses pooling. The rough check holds: \(\hat {S}_A^2/\hat {S}_B^2=60.944/48.456=1.26\), comfortably inside \((\frac {1}{2},2)\).

(iii).

The two samples are independent.

Assumption (iii) is the one this question is really about, and it is false. The same ten students were tested under both methods, so the data are ten pairs, not two independent samples, and the paired interval of Problem 1 is the correct tool. Treating them as independent leaves the enormous student-to-student variation — speeds run from \(58\) to \(86\) — inside the standard error, which is exactly why the interval came out so wide. The question asks you to state the assumptions precisely so that you notice one of them does not hold.

Solution. Both population standard deviations are known, so the multiplier is \(Z\) regardless of the sample sizes: \[(\overline {X}_1-\overline {X}_2)\pm Z_{\alpha /2} \sqrt {\frac {\sigma _1^2}{n_1}+\frac {\sigma _2^2}{n_2}}.\] For \(94\%\) confidence, \(\alpha =0.06\) and \(\alpha /2=0.03\), so we need the value cutting off \(3\%\) in the upper tail: \(Z_{0.03}=1.88\). Then \[\sqrt {\frac {25}{25}+\frac {9}{36}}=\sqrt {1+0.25}=\sqrt {1.25}=1.118,\] \[(80-75)\pm 1.88(1.118)=5\pm 2.10.\] \[2.90<\mu _1-\mu _2<7.10\] The interval is wholly positive, so the first population mean is the larger.

Two points of technique. First, the confidence level is \(94\%\), not one of the three familiar ones, and \(Z_{0.03}=1.88\) must be read from the body of the normal table rather than recalled. Second, \(n_1=25\) is small, but that does not call for \(t\): \(t\) exists to pay for estimating \(\sigma \), and here nothing is estimated. What the small sample does require is the stated normality of the first population, and the question supplies it.

Solution. Equal variances are assumed and both are unknown, so the pooled \(t\) interval applies: \[S_p^2=\frac {(n_1-1)\hat {S}_1^2+(n_2-1)\hat {S}_2^2}{n_1+n_2-2} =\frac {13(1.5)+15(1.8)}{28}=\frac {19.5+27.0}{28}=\frac {46.5}{28}=1.6607,\] so \(S_p=1.2887\) on \(n_1+n_2-2=28\) degrees of freedom, and \(t_{28,\,0.005}=2.763\). \[S_p\sqrt {\frac {1}{n_1}+\frac {1}{n_2}}=1.2887\sqrt {\frac {1}{14}+\frac {1}{16}} =1.2887\sqrt {0.13393}=1.2887(0.36596)=0.4716.\] With \(\overline {X}_2-\overline {X}_1=19-17=2\), \[2\pm 2.763(0.4716)=2\pm 1.303.\] \[0.70<\mu _2-\mu _1<3.30\]

The whole interval is positive, so even at \(99\%\) confidence medication 2 takes longer: patients on it recover between about \(0.7\) and \(3.3\) days later than those on medication 1. Note the direction the question asked for — \(\mu _2-\mu _1\), not \(\mu _1-\mu _2\). Reversing the subtraction reverses the interval to \((-3.30,-0.70)\), which says exactly the same thing; what would be wrong is to quote \((0.70,3.30)\) while describing medication 1 as the slower. Note also that the pooled estimate \(S_p^2=1.66\) lies between the two sample variances \(1.5\) and \(1.8\), as it must: it is a weighted average of them, leaning towards \(1.8\) because the second sample is the larger.

Solution. The sample proportions are \[\hat {P}_1=\frac {250}{1000}=0.25\quad \text {(males)},\qquad \hat {P}_2=\frac {275}{1000}=0.275\quad \text {(females)}.\] All four counts are far above \(5\), so the normal approximation is sound and \[(\hat {P}_1-\hat {P}_2)\pm Z_{\alpha /2} \sqrt {\frac {\hat {P}_1(1-\hat {P}_1)}{n_1}+\frac {\hat {P}_2(1-\hat {P}_2)}{n_2}}.\] \[\sqrt {\frac {0.25(0.75)}{1000}+\frac {0.275(0.725)}{1000}} =\sqrt {0.0001875+0.00019938}=\sqrt {0.00038688}=0.01967,\] \[(0.25-0.275)\pm 1.96(0.01967)=-0.025\pm 0.0386.\] \[-0.0636<P_1-P_2<0.0136\]

The interval contains zero, so there is no evidence at the \(5\%\) level that the disorder strikes the sexes at different rates. Observe that here — unlike a test of \(P_1=P_2\) — each sample supplies its own \(\hat {P}\) to the standard error. There is no pooling, because a confidence interval makes no null hypothesis of equality to compute under; it simply estimates a difference. Pooling belongs to the test of Sheet 4, not to the interval.

Solution. (a). Pooling with \(n_1=12\), \(\hat {S}_1=1.0\), \(n_2=10\), \(\hat {S}_2=0.8\): \[S_p^2=\frac {11(1.00)+9(0.64)}{20}=\frac {11.00+5.76}{20}=\frac {16.76}{20}=0.838, \qquad S_p=0.9154,\] on \(20\) degrees of freedom, with \(t_{20,\,0.05}=1.725\). \[S_p\sqrt {\frac {1}{12}+\frac {1}{10}}=0.9154\sqrt {0.18333}=0.9154(0.42817)=0.3920,\] \[(16-11)\pm 1.725(0.3920)=5\pm 0.676.\] \[4.32<\mu _1-\mu _2<5.68\]

(b). A ratio of variances calls for the \(F\)-distribution. The \(100(1-\alpha )\%\) interval for \(\sigma _1^2/\sigma _2^2\) is \[\frac {\hat {S}_1^2}{\hat {S}_2^2}\cdot \frac {1}{F_{\alpha /2,\,v_1,v_2}} \;<\;\frac {\sigma _1^2}{\sigma _2^2}\;<\; \frac {\hat {S}_1^2}{\hat {S}_2^2}\cdot F_{\alpha /2,\,v_2,v_1},\] with \(v_1=n_1-1=11\) and \(v_2=n_2-1=9\) — and note that the degrees of freedom swap in the upper limit. Here \(\hat {S}_1^2/\hat {S}_2^2=1.00/0.64=1.5625\), and for \(98\%\) confidence \(\alpha /2=0.01\), so \(F_{0.01,\,11,9}=5.178\) and \(F_{0.01,\,9,11}=4.632\): \[\frac {1.5625}{5.178}<\frac {\sigma _1^2}{\sigma _2^2}<1.5625(4.632) \quad \implies \quad 0.302<\frac {\sigma _1^2}{\sigma _2^2}<7.237.\] Taking square roots throughout — legitimate because every quantity is positive — \[0.549<\frac {\sigma _1}{\sigma _2}<2.690\]

(c). Yes. The interval in (b) contains 1, and \(\sigma _1/\sigma _2=1\) is precisely the statement \(\sigma _1=\sigma _2\). Equal variances are therefore consistent with the data, and the pooling in (a) is justified.

The width of that interval is itself the lesson: from \(0.55\) to \(2.69\), one standard deviation could plausibly be half the other or nearly three times it. Variances are estimated far less precisely than means from the same data, which is why the equal-variance assumption is so rarely rejected on small samples — and why part (c) is a check that the assumption is not contradicted, rather than a demonstration that it is true.

Solution. Pooling the two variances, with \(n_1=n_2=12\): \[S_p^2=\frac {11(5000)^2+11(6100)^2}{22}=\frac {(5000)^2+(6100)^2}{2} =\frac {25{,}000{,}000+37{,}210{,}000}{2}=31{,}105{,}000,\] so \(S_p=5577.2\) km on \(22\) degrees of freedom, and \(t_{22,\,0.025}=2.074\). \[S_p\sqrt {\frac {1}{12}+\frac {1}{12}}=5577.2\sqrt {0.16667}=5577.2(0.40825)=2276.9,\] \[(36{,}300-38{,}100)\pm 2.074(2276.9)=-1800\pm 4722.\] \[-6522<\mu _1-\mu _2<2922\ \text {km}\]

The interval contains zero, so there is no evidence that the two brands differ in average life. Brand B outlasted brand A by \(1800\) km in the samples, but with tyre lives varying by five or six thousand kilometres within each brand, twelve tyres apiece cannot establish a difference of that size.

When the sample sizes are equal the pooled variance simplifies to the plain average of the two sample variances, as above — the weights \(n_1-1\) and \(n_2-1\) are the same and cancel. That shortcut only works for \(n_1=n_2\); with unequal samples the larger one must carry the greater weight.

Solution. (a). The sample proportions needing service are \[\hat {P}_A=\frac {20}{150}=0.1333,\qquad \hat {P}_B=\frac {18}{180}=0.1000.\] Each sample has at least five successes and five failures, so \[\sqrt {\frac {0.1333(0.8667)}{150}+\frac {0.1000(0.9000)}{180}} =\sqrt {0.00077037+0.00050000}=\sqrt {0.00127037}=0.03564,\] \[(0.1333-0.1000)\pm 1.96(0.03564)=0.0333\pm 0.0699.\] \[-0.0365<P_A-P_B<0.1032\]

(b). The interval contains zero, so there is no evidence of a difference between the two types of mount at the \(5\%\) level. Type A needed service a third more often in relative terms (\(13.3\%\) against \(10.0\%\)), but with only \(20\) and \(18\) failures observed, that gap is well within sampling variation.

This is worth contrasting with Problem 6. There the difference in proportions was smaller (\(0.025\) against \(0.033\)) yet the conclusion was the same only by a narrow margin — because the samples were \(1000\) each rather than \(150\) and \(180\). Precision in estimating a proportion is governed by the number of events, not by how striking the percentages look; \(20\) failures is simply not many.

Solution. (a). The summary statistics are \[n_1=5,\quad \overline {X}_1=98.4,\quad \hat {S}_1^2=76.30;\qquad n_2=7,\quad \overline {X}_2=110.71,\quad \hat {S}_2^2=1035.90.\] Taking the equal-variance assumption as instructed, \[S_p^2=\frac {4(76.30)+6(1035.90)}{10}=\frac {305.2+6215.4}{10}=652.06, \qquad S_p=25.54,\] on \(10\) degrees of freedom with \(t_{10,\,0.05}=1.812\). \[S_p\sqrt {\frac {1}{5}+\frac {1}{7}}=25.54\sqrt {0.34286}=25.54(0.58554)=14.95,\] \[(98.4-110.71)\pm 1.812(14.95)=-12.31\pm 27.10.\] \[-39.41<\mu _1-\mu _2<14.79\ \text {minutes}\]

(b). With \(\hat {S}_1^2/\hat {S}_2^2=76.30/1035.90=0.07366\), \(v_1=4\), \(v_2=6\), and \(\alpha /2=0.05\) for a \(90\%\) interval: \(F_{0.05,\,4,6}=4.534\) and \(F_{0.05,\,6,4}=6.163\). \[\frac {0.07366}{4.534}<\frac {\sigma _1^2}{\sigma _2^2}<0.07366(6.163)\] \[0.016<\frac {\sigma _1^2}{\sigma _2^2}<0.454\]

(c). No. The interval in (b) lies entirely below 1, so \(\sigma _1^2=\sigma _2^2\) is not a plausible value: company II’s running times are between about twice and sixty times as variable as company I’s. Pooling was therefore not justified, and the interval in (a) rests on a false assumption.

The data show why without any arithmetic. Company I’s five films run from \(87\) to \(110\) minutes; company II’s seven run from \(82\) to \(175\), with one film nearly an hour longer than any other. Pooling forces a single \(S_p^2=652\) on both companies — eight times company I’s own variance — which inflates the standard error for company I and deflates it for company II.

The proper remedy is the Smith–Satterthwaite (Welch) interval, which does not pool: \[(\overline {X}_1-\overline {X}_2)\pm t_{v,\,\alpha /2} \sqrt {\frac {\hat {S}_1^2}{n_1}+\frac {\hat {S}_2^2}{n_2}},\qquad v=\frac {\left (\hat {S}_1^2/n_1+\hat {S}_2^2/n_2\right )^2} {\dfrac {(\hat {S}_1^2/n_1)^2}{n_1-1}+\dfrac {(\hat {S}_2^2/n_2)^2}{n_2-1}}.\] Here \(v=7.19\), rounded down to \(7\), giving \(-12.31\pm 1.895(12.78)\), or \((-36.5,\ 11.9)\) — a narrower interval than the pooled one, and the defensible answer. This estimate of \(v\) is generally not a whole number and is rounded down, which is the conservative choice.