Sampling and Estimation

(i) — identify who is systematically excluded. Every person sampled comes from the tennis club's membership, so anyone in the town who isn't a member — including anyone who dislikes or doesn't play tennis — has no chance at all of being chosen. The sample is biased towards people who like tennis, not a fair cross-section of the town.

(ii) — describe a procedure that gives every adult in the town an equal chance, following the general method above: obtain a complete list of all adults living in the town (the sampling frame), assign each a unique number, and use random numbers to select the required sample size, rejecting any repeats.

Step 1 — find the label width. The population has $75$ members, so labels need $2$ digits: $01$ to $75$.

Step 2 — split the string into non-overlapping groups of 2, reading left to right. $08341629507712 \;\to\; 08,\ 34,\ 16,\ 29,\ 50,\ 77,\ 12$

Step 3 — work through the groups in order, checking range and repeats. $08$: valid, within $01$–$75$ — $1$st member. $34$: valid — $2$nd member. $16$: valid — $3$rd member.

Step 1 — find the label width. The population has $264$ members, so labels need $3$ digits: $001$ to $264$.

Step 2 — write the digit string with no gaps, and split into non-overlapping groups of 3. $10612118012147322759 \;\to\; 106,\ 121,\ 180,\ 121,\ 473,\ 227,\ 59\text{ (discarded, only 2 digits left)}$

Step 3 — work through the groups in order, checking range and repeats. $106$: valid, within $001$–$264$ — this is the $1$st member (matches the value given). $121$: valid — $2$nd member (also matches).

Step 4 — continue with the next groups. $180$: within range, not a repeat — the $3$rd member. $121$: within range, but a repeat of the $2$nd member already chosen — reject. $473$: outside the range $001$–$264$ — reject.

Step 5 — the next valid, non-repeated group. $227$: within range, not a repeat — the $4$th member.

Step 1 — multiply the random decimal by the population size. $0.512\times120 = 61.44$

Step 2 — round up to the next whole number. $61.44 \;\to\; 62$

Step 1 — multiply the random decimal by the population size. $0.401\times851 = 341.251$

Step 2 — round up to the next whole number, since the rule says "round up", not "round to the nearest".

Step 1 — write the range condition for member $m=7$, with $N=50$. $7-1 < 50d \leqslant 7 \quad\Longrightarrow\quad 6 < 50d \leqslant 7$

Step 2 — divide through by $50$ to get a range for $d$ itself. $\frac{6}{50} < d \leqslant \frac{7}{50} \quad\Longrightarrow\quad 0.12 < d \leqslant 0.14$

Step 3 — list the $2$-d.p. decimals the generator can actually produce inside that range. $0.13$ and $0.14$ qualify; $0.12$ does not, because the inequality on the left is strict. So two decimals give member $7$.

Step 4 — repeat for member $m=50$. $49 < 50d \leqslant 50 \quad\Longrightarrow\quad 0.98 < d \leqslant 1.00$

Step 5 — list the available decimals in this range. The generator stops at $0.99$, so $1.00$ is never produced, and $0.98$ is excluded by the strict inequality. Only $0.99$ survives — one decimal gives member $50$.

Step 1 — (b)(i): write the range condition for $m=680$, with $N=851$. $679 < 851d \leqslant 680 \quad\Longrightarrow\quad \frac{679}{851} < d \leqslant \frac{680}{851}$

Step 2 — evaluate both endpoints to enough decimal places to decide the $3$-d.p. values. $\frac{679}{851} = 0.797884\ldots, \qquad \frac{680}{851} = 0.799059\ldots$

Step 3 — list the $3$-d.p. decimals strictly inside that range. $0.798$ and $0.799$ both lie between $0.797884\ldots$ and $0.799059\ldots$, so both produce member $680$. Two random numbers give member $680$.

Step 4 — (b)(ii): repeat for $m=850$. $849 < 851d \leqslant 850 \quad\Longrightarrow\quad \frac{849}{851} < d \leqslant \frac{850}{851}$ $\frac{849}{851} = 0.997649\ldots, \qquad \frac{850}{851} = 0.998824\ldots$

Step 5 — list the $3$-d.p. decimals inside this range. Only $0.998$ lies between $0.997649\ldots$ and $0.998824\ldots$ — $0.997$ is below it and $0.999$ is above it. One random number gives member $850$.

Step 6 — (c): turn the two counts into the conclusion the mark is for. Member $680$ can be produced in two ways out of the $1000$ available random numbers, and member $850$ in only one, so $P(\text{member }680) = \frac{2}{1000}, \qquad P(\text{member }850) = \frac{1}{1000}$ The two members are not equally likely to be chosen, so the method does not give a random sample.

Step 1 — compare the number of outcomes with the number of people. The dice has $6$ equally likely outcomes and there are $4$ people. Since $6$ does not divide exactly by $4$, no assignment of all six outcomes can give the four people equal shares.

Step 2 — give each person one outcome, which is the largest equal share available. $1\to A, \quad 2\to B, \quad 3\to C, \quad 4\to D$

Step 3 — deal with the two leftover outcomes by rejecting them. If the dice shows $5$ or $6$, make no choice and throw again, repeating until a score of $1$ to $4$ appears.

Step 4 — check the probabilities really are equal. On each accepted throw the score is equally likely to be $1, 2, 3$ or $4$, so each person has probability $\frac14$. Assigning $5$ and $6$ to $A$ and $D$ instead would have given those two probability $\frac{2}{6}$ against $\frac16$ for $B$ and $C$ — not random.

Step 1 — list the equally likely outcomes of the device itself, which are the four ordered results of two coins, not the three values of $n$. $HH, \quad HT, \quad TH, \quad TT$

Step 2 — count how many of those four outcomes give each value of $n$. $n=0$ comes only from $TT$; $n=1$ comes from either $HT$ or $TH$; $n=2$ comes only from $HH$.

Step 3 — turn the counts into probabilities. $P(n=0) = \tfrac14, \qquad P(n=1) = \tfrac24 = \tfrac12, \qquad P(n=2) = \tfrac14$

Step 4 — state the comparison the mark is for. $P(Q)=\frac12$ but $P(P)=P(R)=\frac14$, so the three people are not equally likely to be chosen — indeed $Q$ is twice as likely as either of the others. The choice is therefore not random.