| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Marks | 16 |
| 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 a binomial distribution, following a routine template. Part (a) requires calculating expected frequencies from B(16, 0.1), combining classes, and performing the test—all mechanical steps. Part (b) is even simpler as the test statistic is given. The only mild challenge is remembering to adjust degrees of freedom when a parameter is estimated, but this is a standard textbook procedure for S3 level. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test5.06c Fit other distributions: discrete and continuous |
| No. of Left-handed Players | 0 | 1 | 2 | 3 | 4 | \(\geq 5\) |
| No. of Tournaments | 4 | 12 | 18 | 11 | 5 | 0 |
| Answer | Marks |
|---|---|
| (a) \(H_0: B(16, 0.1)\) is a suitable model; \(H_1: B(16, 0.1)\) is not a suitable model | B1 |
| Answer | Marks |
|---|---|
| \(P(3) = \frac{16 \times 15 \times 14}{3 \times 2}(0.1)^3(0.9)^{13} = 0.1423\) | M1 A2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\times 50\) to give exp. freqs then freq of \(\geq 5\) is \((50 - \text{sum of others})\) \(\therefore\) exp. freqs are 9.27, 16.47, 13.73, 7.12, 2.57, 0.84 | M1 A1, M1 | |
| O | E | (O-E) |
| 4 | 9.27 | -5.27 |
| 12 | 16.47 | -4.47 |
| 18 | 13.73 | 4.27 |
| 16 | 10.53 | 5.47 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\therefore \sum \frac{(O-E)^2}{E} = 8.379\) | M1 A2 | |
| \(\nu = 4 - 1 = 3, \chi^2_{\text{crit}}(5\%) = 7.815\) | M1 A1 | |
| \(8.379 > 7.815 \therefore\) reject \(H_0\). \(B(16, 0.1)\) is not a suitable model | A1 | |
| (b) \(\nu = 5 - 2 = 3, \chi^2_{\text{crit}}(5\%) = 7.815\). \(2.127 < 7.815 \therefore\) do not reject \(H_0\), the binomial dist. is a suitable model | M1 A1, A1 | Total: 16 marks |
(a) $H_0: B(16, 0.1)$ is a suitable model; $H_1: B(16, 0.1)$ is not a suitable model | B1 |
$P(0) = (0.9)^{16} = 0.1853$
$P(1) = 16(0.1)(0.9)^{15} = 0.3294$
$P(2) = \frac{16 \times 15}{2}(0.1)^2(0.9)^{14} = 0.2745$
$P(3) = \frac{16 \times 15 \times 14}{3 \times 2}(0.1)^3(0.9)^{13} = 0.1423$ | M1 A2 |
$P(4) = \frac{16 \times 15 \times 14 \times 13}{4 \times 3 \times 2}(0.1)^4(0.9)^{12} = 0.0514$
$\times 50$ to give exp. freqs then freq of $\geq 5$ is $(50 - \text{sum of others})$ $\therefore$ exp. freqs are 9.27, 16.47, 13.73, 7.12, 2.57, 0.84 | M1 A1, M1 |
| O | E | (O-E) | $\frac{(O-E)^2}{E}$ |
|----|----|-------|-------------------|
| 4 | 9.27 | -5.27 | 2.9960 |
| 12 | 16.47 | -4.47 | 1.2132 |
| 18 | 13.73 | 4.27 | 1.3280 |
| 16 | 10.53 | 5.47 | 2.8415 |
Combining groups $\geq 3$
$\therefore \sum \frac{(O-E)^2}{E} = 8.379$ | M1 A2 |
$\nu = 4 - 1 = 3, \chi^2_{\text{crit}}(5\%) = 7.815$ | M1 A1 |
$8.379 > 7.815 \therefore$ reject $H_0$. $B(16, 0.1)$ is not a suitable model | A1 |
(b) $\nu = 5 - 2 = 3, \chi^2_{\text{crit}}(5\%) = 7.815$. $2.127 < 7.815 \therefore$ do not reject $H_0$, the binomial dist. is a suitable model | M1 A1, A1 | Total: 16 marks7. A student collects data on whether competitors in local tennis tournaments are right, or left-handed. The table below shows the number of left-handed players who reached the last 16 for fifty tournaments.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
No. of Left-handed Players & 0 & 1 & 2 & 3 & 4 & $\geq 5$ \\
\hline
No. of Tournaments & 4 & 12 & 18 & 11 & 5 & 0 \\
\hline
\end{tabular}
\end{center}
The student believes that a binomial distribution with $n = 16$ and $p = 0.1$ could be a suitable model for these data.
\begin{enumerate}[label=(\alph*)]
\item Stating your hypotheses clearly test the student's model at the $5 \%$ level of significance.\\
(13 marks)\\
To improve the model the student decides to estimate $p$ using the data in the table. Using this value of $p$ to calculate expected frequencies the student had 5 classes after combining and calculated that $\sum \frac { ( O - E ) ^ { 2 } } { E } = 2.127$
\item Test at the $5 \%$ level of significance whether or not the binomial distribution is a suitable model for the number of left-handed players who reach the last 16 in local tennis tournaments.
\section*{END}
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 Q7 [16]}}