| Exam Board | WJEC |
|---|---|
| Module | Further Unit 2 (Further Unit 2) |
| Year | 2023 |
| Session | June |
| Marks | 20 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | State test assumptions or distributions |
| Difficulty | Standard +0.8 This is a substantial Further Maths statistics question requiring chi-squared goodness of fit test with binomial probabilities, followed by an expectation calculation involving overselling. While the individual techniques are standard (calculating binomial probabilities, chi-squared test, expected value), the multi-part nature, the need to handle the overselling scenario with compensation, and the context of Further Maths places it moderately above average difficulty. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test |
| Number of boats not taken | 0 | 1 | 2 | 3 | 4 | 5 or more |
| Frequency | 10 | 35 | 29 | 25 | 8 | 3 |
| Number of boats not taken | 0 | 1 | 2 | 3 | 4 | 5 or more |
| Observed | 10 | 35 | 29 | 25 | 8 | 3 |
| Expected | \(f\) | 29·72 | \(g\) | 20·91 | 9·88 | 4·75 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (a) \(H_0\): The data can be modelled by the binomial distribution \(B(20, 0.1)\). \(H_1\): The data cannot be modelled by the binomial distribution \(B(20, 0.1)\). | B1 | or equivalent, must state \(B(20, 0.1)\) |
| (b)(i) Expected frequencies are \((f = (P(X = 0) \times 110)\) \(f = 13.37\) | B1 | Accept 13.4 |
| \((g = (P(X = 2) \times 110)\) \(g = 31.37\) | B1 | Accept 31.4 |
| (ii) Combine classes with expected frequencies less than 5 | M1 | SC for solution that does not combine classes (M0M1m1A0B1) |
| Table: | ||
| Number of boats not taken out: 0, 1, 2, 3, 4 or more | ||
| Obs: 10, 35, 29, 25, 11 | ||
| Exp: 13.37, 29.72, 31.37, 20.91, 14.63 | ||
| Use of \(\chi^2_{\text{stat}} = \sum \frac{(O-E)^2}{E}\) OR \(\sum \frac{O^2}{E} - N\) | M1 | Must see at least 2 terms added |
| \(= \frac{10^2}{13.37} + \frac{35^2}{29.72} + \frac{29^2}{31.37} + \frac{25^2}{20.91} + \frac{11^2}{14.63} - 110\) | m1 | |
| \(= 3.667...\) | A1 | cao. 3.66760 if unrounded values used. |
| \(DF = 4\) | B1 | |
| 5% CV = 9.488 | B1 | Accept other test levels: 1% CV = 13.277; 10% CV = 7.779 |
| Since 3.667 < 9.488 there is insufficient evidence to reject \(H_0\). | m1 | Dep on 2nd M1 |
| Insufficient evidence to reject the suggestion that the number of boats not taken out each day can be modelled by the binomial model \(B(20, 0.1)\) | A1 | cso provided B1 awarded in (a); A0 for categorical statements |
| Total [20] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (c)(i) Let the random variable \(Y\) be the number groups that turn up expecting to take a boat out. \(Y \sim B(22, 0.9)\) | B1 | May be seen or implied in (ii) |
| (ii) Let the random variable \(S\) be the income of the company in pounds. Values for \(s\) are 330, 310 and 290 | B1 | All 3 values and no others |
| \(P(S = 330) = P(Y \leq 20) = 0.6608\) | B1 | Recognising link between income and the probability of the number of groups who turn up to take a boat, si. If using \(W \sim B(22, 0.1)\), probabilities are \(P(W \geq 2) = 0.6608\); \(P(W = 1) = 0.2407\); \(P(W = 0) = 0.0985\) |
| \(P(S = 310) = P(Y = 21) = 0.2407\) | ||
| \(P(S = 290) = P(Y = 22) = 0.0985\) | B2 | B2 for three correct probabilities. B1 for one correct probability. |
| \(E(S) = 330 \times 0.6608 + 310 \times 0.2407 + 290 \times 0.0985\) | M1A1 | M1 for one correct term and addition. |
| \(= £321.25\) | A1 | |
| Alternative solution: Let the random variable \(T\) be the loss of the company in pounds. Values for \(t\) are \((0,) 20\) and 40 | (B1) | All 3 values and no others. |
| \(P(T = 0) = P(Y \leq 20) = 0.6608\) | (B1) | Recognising link between loss and the probability of the number of groups who turn up to take a boat. If using \(W \sim B(22, 0.1)\), probabilities are \(P(W \geq 2) = 0.6608\); \(P(W = 1) = 0.2407\); \(P(W = 0) = 0.0985\) |
| \(P(T = 20) = P(Y = 21) = 0.2407\) | ||
| \(P(T = 40) = P(Y = 22) = 0.0985\) | (B2) | B2 for three correct probabilities. B1 for one correct probability. |
| \(E(T) = (0 \times 0.6608+) 20 \times 0.2407 + 40 \times 0.0985 = £8.75\) | (M1A1) | M1 for one correct term and addition. |
| Net income (\(= £330 - £8.75) = £321.25\) | (A1) | |
| (d) Valid reason. e.g. The manager is justified because the expected income is greater than £15 × 20 = £300 | E1 | Accept "Not justified" with appropriate reason such as "not a long term strategy because it will harm the brand." FT their \(E(S)\) |
| Total [20] |
| Answer/Working | Mark | Guidance |
|---|---|---|
| **(a)** $H_0$: The data can be modelled by the binomial distribution $B(20, 0.1)$. $H_1$: The data cannot be modelled by the binomial distribution $B(20, 0.1)$. | B1 | or equivalent, must state $B(20, 0.1)$ |
| **(b)(i)** Expected frequencies are $(f = (P(X = 0) \times 110)$ $f = 13.37$ | B1 | Accept 13.4 |
| $(g = (P(X = 2) \times 110)$ $g = 31.37$ | B1 | Accept 31.4 |
| **(ii)** Combine classes with expected frequencies less than 5 | M1 | SC for solution that does not combine classes (M0M1m1A0B1) |
| **Table:** | | |
| Number of boats not taken out: 0, 1, 2, 3, 4 or more | | |
| Obs: 10, 35, 29, 25, 11 | | |
| Exp: 13.37, 29.72, 31.37, 20.91, 14.63 | | |
| Use of $\chi^2_{\text{stat}} = \sum \frac{(O-E)^2}{E}$ OR $\sum \frac{O^2}{E} - N$ | M1 | Must see at least 2 terms added |
| $= \frac{10^2}{13.37} + \frac{35^2}{29.72} + \frac{29^2}{31.37} + \frac{25^2}{20.91} + \frac{11^2}{14.63} - 110$ | m1 | |
| $= 3.667...$ | A1 | cao. 3.66760 if unrounded values used. |
| $DF = 4$ | B1 | |
| 5% CV = 9.488 | B1 | Accept other test levels: 1% CV = 13.277; 10% CV = 7.779 |
| Since 3.667 < 9.488 there is insufficient evidence to reject $H_0$. | m1 | Dep on 2nd M1 |
| Insufficient evidence to reject the suggestion that the number of boats not taken out each day can be modelled by the binomial model $B(20, 0.1)$ | A1 | cso provided B1 awarded in (a); A0 for categorical statements |
| **Total [20]** | | |
# Question 6 (continued)
| Answer/Working | Mark | Guidance |
|---|---|---|
| **(c)(i)** Let the random variable $Y$ be the number groups that turn up expecting to take a boat out. $Y \sim B(22, 0.9)$ | B1 | May be seen or implied in (ii) |
| **(ii)** Let the random variable $S$ be the income of the company in pounds. Values for $s$ are 330, 310 and 290 | B1 | All 3 values and no others |
| $P(S = 330) = P(Y \leq 20) = 0.6608$ | B1 | Recognising link between income and the probability of the number of groups who turn up to take a boat, si. If using $W \sim B(22, 0.1)$, probabilities are $P(W \geq 2) = 0.6608$; $P(W = 1) = 0.2407$; $P(W = 0) = 0.0985$ |
| $P(S = 310) = P(Y = 21) = 0.2407$ | | |
| $P(S = 290) = P(Y = 22) = 0.0985$ | B2 | B2 for three correct probabilities. B1 for one correct probability. |
| $E(S) = 330 \times 0.6608 + 310 \times 0.2407 + 290 \times 0.0985$ | M1A1 | M1 for one correct term and addition. |
| $= £321.25$ | A1 | |
| **Alternative solution:** Let the random variable $T$ be the loss of the company in pounds. Values for $t$ are $(0,) 20$ and 40 | (B1) | All 3 values and no others. |
| $P(T = 0) = P(Y \leq 20) = 0.6608$ | (B1) | Recognising link between loss and the probability of the number of groups who turn up to take a boat. If using $W \sim B(22, 0.1)$, probabilities are $P(W \geq 2) = 0.6608$; $P(W = 1) = 0.2407$; $P(W = 0) = 0.0985$ |
| $P(T = 20) = P(Y = 21) = 0.2407$ | | |
| $P(T = 40) = P(Y = 22) = 0.0985$ | (B2) | B2 for three correct probabilities. B1 for one correct probability. |
| $E(T) = (0 \times 0.6608+) 20 \times 0.2407 + 40 \times 0.0985 = £8.75$ | (M1A1) | M1 for one correct term and addition. |
| Net income ($= £330 - £8.75) = £321.25$ | (A1) | |
| **(d)** Valid reason. e.g. The manager is justified because the expected income is greater than £15 × 20 = £300 | E1 | Accept "Not justified" with appropriate reason such as "not a long term strategy because it will harm the brand." FT their $E(S)$ |
| **Total [20]** | | |A company has 20 boats to hire out. Payment is always taken in advance and all 20 boats are hired out each day.
A manager at the company notices that 10% of groups do not turn up to take the boats, despite having already paid to hire them. The manager wishes to investigate whether the numbers of boats that do not get taken each day can be modelled by the binomial distribution B$(20, 0 \cdot 1)$. The numbers of boats that were not taken for 110 randomly selected days are given below.
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
Number of boats not taken & 0 & 1 & 2 & 3 & 4 & 5 or more \\
\hline
Frequency & 10 & 35 & 29 & 25 & 8 & 3 \\
\hline
\end{tabular}
\begin{enumerate}[label=(\alph*)]
\item State suitable hypotheses to carry out a goodness of fit test. [1]
\item Here is part of the table for a $\chi^2$ goodness of fit test on the data.
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
Number of boats not taken & 0 & 1 & 2 & 3 & 4 & 5 or more \\
\hline
Observed & 10 & 35 & 29 & 25 & 8 & 3 \\
\hline
Expected & $f$ & 29·72 & $g$ & 20·91 & 9·88 & 4·75 \\
\hline
\end{tabular}
\begin{enumerate}[label=(\roman*)]
\item Calculate the values of $f$ and $g$.
\item By completing the test, give the conclusion the manager should reach. [10]
\end{enumerate}
\end{enumerate}
The cost of hiring a boat is £15. Since demand is high and the proportion of groups that do not turn up is also relatively high, the manager decides to take payment for 22 boats each day. She would give £20 (a full refund and some compensation) to any group that has paid and turned up, but cannot take a boat out due to the overselling. Assume that the proportion of groups not turning up stays the same.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumii}{2}
\item \begin{enumerate}[label=(\roman*)]
\item Suggest a binomial model that the manager could use for the number of groups arriving expecting to hire a boat.
\item Hence calculate the expected daily net income for the company following the manager's decision. [8]
\end{enumerate}
\item Is the manager justified in her decision? Give a reason for your answer. [1]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 2 2023 Q6 [20]}}