| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Standard 3×3 contingency table |
| Difficulty | Standard +0.3 This is a standard chi-squared test of independence with clearly structured data requiring routine application of the test procedure: state hypotheses, calculate expected frequencies, compute test statistic, compare to critical value, and conclude. The calculation is straightforward with 3×3 contingency table (4 degrees of freedom). Part (b) requires minimal interpretation. Slightly above average difficulty due to being a complete hypothesis test rather than just calculation, but still a textbook exercise with no novel problem-solving required. |
| Spec | 5.06a Chi-squared: contingency tables |
| ITV | Channel 4 | Channel 5 | |
| Family Saloon | 69 | 35 | 28 |
| Sports Car | 20 | 28 | 18 |
| Off-road Vehicle | 12 | 22 | 8 |
| Answer | Marks |
|---|---|
| giving expected freqs \(\begin{matrix} 55.55 & 46.75 & 29.70 \\ 27.78 & 23.38 & 14.84 \\ 17.67 & 14.87 & 9.46 \end{matrix}\) | M1, A2, M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_1\): difference in proportion of adverts on different channels | B1 | |
| O | E | (O - E) |
| 69 | 55.55 | 13.45 |
| 35 | 46.75 | −11.75 |
| 28 | 29.70 | −1.7 |
| 20 | 27.78 | −7.78 |
| 28 | 23.38 | 4.62 |
| 18 | 14.84 | 3.16 |
| 12 | 17.67 | −5.67 |
| 22 | 14.87 | 7.13 |
| 8 | 9.46 | −1.46 |
| \(\sum \frac{(O-E)^2}{E} = 15.535\) | M1, A3 | |
| \(\nu = 4\), \(\chi^2_{crit}(5\%) = 9.488\) | M1, A1 | |
| 15.535 > 9.488 ∴ significant there is evidence of different proportion of adverts on different channels | A1 | |
| (b) e.g. advertisers perception of the type of people who watch each channel | B1 | (14) |
**(a)** expected freq. family/ITV = $\frac{101×32}{240} = 55.55$
family/Ch4 = $\frac{85×32}{240} = 46.75$
sports/ITV = $\frac{101×60}{240} = 27.78$
sports/Ch4 = $\frac{85×60}{240} = 23.38$
giving expected freqs $\begin{matrix} 55.55 & 46.75 & 29.70 \\ 27.78 & 23.38 & 14.84 \\ 17.67 & 14.87 & 9.46 \end{matrix}$ | M1, A2, M1, A1 |
$H_0$: no difference in proportion of adverts on different channels
$H_1$: difference in proportion of adverts on different channels | B1 |
| O | E | (O - E) | $\frac{(O-E)^2}{E}$ |
|---|---|---------|------------|
| 69 | 55.55 | 13.45 | 3.2566 |
| 35 | 46.75 | −11.75 | 2.9532 |
| 28 | 29.70 | −1.7 | 0.0973 |
| 20 | 27.78 | −7.78 | 2.1788 |
| 28 | 23.38 | 4.62 | 0.9129 |
| 18 | 14.84 | 3.16 | 0.6729 |
| 12 | 17.67 | −5.67 | 1.8194 |
| 22 | 14.87 | 7.13 | 3.4188 |
| 8 | 9.46 | −1.46 | 0.2253 |
$\sum \frac{(O-E)^2}{E} = 15.535$ | M1, A3 |
$\nu = 4$, $\chi^2_{crit}(5\%) = 9.488$ | M1, A1 |
15.535 > 9.488 ∴ significant there is evidence of different proportion of adverts on different channels | A1 |
**(b)** e.g. advertisers perception of the type of people who watch each channel | B1 | (14) |
---6. A market researcher recorded the number of adverts for vehicles in each of three categories on ITV, Channel 4 and Channel 5 over a period of time. The results are shown in the table below.
\begin{center}
\begin{tabular}{|l|l|l|l|}
\hline
& ITV & Channel 4 & Channel 5 \\
\hline
Family Saloon & 69 & 35 & 28 \\
\hline
Sports Car & 20 & 28 & 18 \\
\hline
Off-road Vehicle & 12 & 22 & 8 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Stating your hypotheses clearly, test at the $5 \%$ level of significance whether or not there is evidence of the proportion of adverts for each type of vehicle being dependent on the channel.
\item Suggest a reason for your result in part (a).
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 Q6 [14]}}