| Exam Board | WJEC |
|---|---|
| Module | Further Unit 2 (Further Unit 2) |
| Year | 2018 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared goodness of fit: Binomial |
| Difficulty | Standard +0.3 This is a standard chi-squared goodness of fit test with given parameters. Students must state hypotheses, calculate expected frequencies using binomial probabilities, compute the test statistic, and compare to critical values. All techniques are routine for Further Maths statistics; no novel insight required, making it slightly easier than average. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test5.06c Fit other distributions: discrete and continuous |
| Number of policies sold | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| Number of days | 2 | 2 | 9 | 12 | 15 | 9 | 1 |
| Number of policies sold | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| Observed | 2 | 2 | 9 | 12 | 15 | 9 | 1 |
| Expected | 0·205 | 1·843 | 6·912 | \(d\) | \(e\) | 9·331 | 2·333 |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0\): The data can be modelled by the Binomial distribution \(B(6,0.6)\) | B1 | Both |
| Answer | Marks |
|---|---|
| \((d = (P(X = 3) \times 50)\) \(d = 13.824\) | B1 |
| \((e = (P(X = 4) \times 50)\) \(e = 15.552\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Combine classes with expected frequencies less than 5 | M1 | SC for solution that does not combine classes or only combines some. (M0M1m0B1B1) |
| Number of policies sold | 0,1 or 2 | 3 |
| Observed | 13 | 12 |
| Expected | 8.96 | 13.824 |
| Use of \(\chi^2\) stat \(= \sum\frac{(O-E)^2}{E}\) or \(\sum\frac{O^2}{E} - N\) | M1 | \(= \frac{13^2}{8.96} + \frac{12^2}{13.824} + \frac{15^2}{15.552} + \frac{10^2}{11.664} - 50\) |
| \(= \frac{(13-8.96)^2}{8.96} + \frac{(12-13.824)^2}{13.824} + \frac{(15-15.552)^2}{15.552} + \frac{(10-11.664)^2}{11.664}\) | m1 | Dep on previous M1. FT their table. May see DF=6 FT their DF CV = 10.645 Since 17.394 > 10.645 Reject \(H_0\) There is sufficient evidence to reject the binomial model \(B(6, 0.6)\). Only award final B1 if previous 3 B1 awarded. |
| \(= 2.319(254605)\) | A1 | cao Example of SC FT their table. May see DF=6 FT their DF CV = 10.645 Since 17.394 > 10.645 Reject \(H_0\) |
| DF = 3 | B1 | There is sufficient evidence to reject the binomial model \(B(6,0.6)\) |
| 10% CV = 6.251 | B1 | |
| Since 2.319 < 6.251 do not reject \(H_0\). | B1 | |
| Insufficient evidence to reject the binomial model \(B(6, 0.6)\) | B1 | Only award final B1 if previous 3 B1 awarded. |
| Answer | Marks | Guidance |
|---|---|---|
| 6 is the number of clients she sees in one day AND 0.6 is the probability of selling a policy to each client. | E1 [12] | Must state one day. |
## 5(a)
$H_0$: The data can be modelled by the Binomial distribution $B(6,0.6)$ | B1 | Both
$H_1$: The data cannot be modelled by the Binomial distribution $B(6,0.6)$
## 5(b)(i)
Expected frequencies are
$(d = (P(X = 3) \times 50)$ $d = 13.824$ | B1 |
$(e = (P(X = 4) \times 50)$ $e = 15.552$ | B1 |
## 5(b)(ii)
Combine classes with expected frequencies less than 5 | M1 | SC for solution that does not combine classes or only combines some. (M0M1m0B1B1)
| Number of policies sold | 0,1 or 2 | 3 | 4 | 5 or 6 |
|---|---|---|---|---|
| Observed | 13 | 12 | 15 | 10 |
| Expected | 8.96 | 13.824 | 15.552 | 11.664 |
Use of $\chi^2$ stat $= \sum\frac{(O-E)^2}{E}$ or $\sum\frac{O^2}{E} - N$ | M1 | $= \frac{13^2}{8.96} + \frac{12^2}{13.824} + \frac{15^2}{15.552} + \frac{10^2}{11.664} - 50$
$= \frac{(13-8.96)^2}{8.96} + \frac{(12-13.824)^2}{13.824} + \frac{(15-15.552)^2}{15.552} + \frac{(10-11.664)^2}{11.664}$ | m1 | Dep on previous M1. FT their table. May see DF=6 FT their DF CV = 10.645 Since 17.394 > 10.645 Reject $H_0$ There is sufficient evidence to reject the binomial model $B(6, 0.6)$. Only award final B1 if previous 3 B1 awarded.
$= 2.319(254605)$ | A1 | cao Example of SC FT their table. May see DF=6 FT their DF CV = 10.645 Since 17.394 > 10.645 Reject $H_0$
DF = 3 | B1 | There is sufficient evidence to reject the binomial model $B(6,0.6)$
10% CV = 6.251 | B1 |
Since 2.319 < 6.251 do not reject $H_0$. | B1 |
Insufficient evidence to reject the binomial model $B(6, 0.6)$ | B1 | Only award final B1 if previous 3 B1 awarded.
## 5(c)
6 is the number of clients she sees in one day AND 0.6 is the probability of selling a policy to each client. | E1 [12] | Must state one day.
---A life insurance saleswoman investigates the number of policies she sells per day. The results for a random sample of 50 days are shown in the table below.
\begin{tabular}{|l|c|c|c|c|c|c|c|}
\hline
Number of policies sold & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
Number of days & 2 & 2 & 9 & 12 & 15 & 9 & 1 \\
\hline
\end{tabular}
She sees the same fixed number of clients each day. She would like to know whether the binomial distribution with parameters 6 and 0·6 is a suitable model for the number of policies she sells per day.
\begin{enumerate}[label=(\alph*)]
\item State suitable hypotheses for 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}{|l|c|c|c|c|c|c|c|}
\hline
Number of policies sold & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
Observed & 2 & 2 & 9 & 12 & 15 & 9 & 1 \\
\hline
Expected & 0·205 & 1·843 & 6·912 & $d$ & $e$ & 9·331 & 2·333 \\
\hline
\end{tabular}
\begin{enumerate}[label=(\roman*)]
\item Calculate the values of $d$ and $e$.
\item Carry out the test using a 10% level of significance and draw a conclusion in context. [10]
\end{enumerate}
\item What do the parameters 6 and 0·6 mean in this context? [1]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 2 2018 Q5 [12]}}