| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Session | Specimen |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Standard 2×2 contingency table |
| Difficulty | Moderate -0.3 This is a standard chi-squared test of independence with a 2×2 contingency table, requiring routine calculation of expected frequencies, test statistic, and comparison to critical values. Part (ii) adds a straightforward connection to proportions testing. While it involves multiple steps, all procedures are textbook applications with no novel insight required, making it slightly easier than average. |
| Spec | 5.06a Chi-squared: contingency tables |
| \cline { 2 - 4 } \multicolumn{1}{c|}{} | Age of shopper | ||
| \cline { 2 - 4 } \multicolumn{1}{c|}{} | Under 35 | 35 and over | Total |
| In favour of change | 187 | 161 | 348 |
| Not in favour of change | 283 | 369 | 652 |
| Total | 470 | 530 | 1000 |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(H_0\): shoppers' views and age are independent, \(H_1\): shoppers' views and age are not independent | B1 | For stating both hypotheses |
| Expected frequencies under \(H_0\) are \(\begin{matrix} 163.56 & 184.44 \\ 306.44 & 345.56 \end{matrix}\) | M1 | For correct method for expected frequencies |
| A1 | For all four correct | |
| Test statistic is \(\frac{22.94^2}{163.56} + \frac{22.94^2}{184.44} + \frac{22.94^2}{306.44} + \frac{22.94^2}{345.56}\) | M1 | For correct calculation process, inc Yates |
| \(= 9.31\ldots\) | A1 | For correct value of the test statistic |
| This is greater than the critical 0.5% value of 7.879 | M1 | For a relevant (1 df) comparison |
| Hence there is very strong evidence to reject \(H_0\) and conclude that views about changing to metric units are not independent of age | A1 | 7 |
| (ii) \(H_0: p_1 = p_2, H_1: p_1 \neq p_2\) | B1 | For both hypotheses stated |
| Under \(H_0\) the sample value of the common proportion is \(\frac{187 + 161}{1000} = 0.348\) | B1 | For correct value of estimated \(p\) |
| Test statistic is \(\frac{\frac{187}{470} - \frac{161}{530}}{\sqrt{0.348 \times 0.652 \times \left(\frac{1}{470} + \frac{1}{530}\right)}}\) | M1 | For num \(p_1 - p_2\) and denom using attempted s.d. based on a common estimate of \(p\) |
| \(= 3.118\) | A1 | For completely correct expression |
| A1 | For correct value of the test statistic | |
| This is greater than the 0.2% (two-tail) critical value of 3.090 | M1 | For a relevant comparison using the normal distribution |
| Hence this test supports the conclusion of part (i) | A1 | 7 |
**(i)** $H_0$: shoppers' views and age are independent, $H_1$: shoppers' views and age are not independent | B1 | For stating both hypotheses
Expected frequencies under $H_0$ are $\begin{matrix} 163.56 & 184.44 \\ 306.44 & 345.56 \end{matrix}$ | M1 | For correct method for expected frequencies
| A1 | For all four correct
Test statistic is $\frac{22.94^2}{163.56} + \frac{22.94^2}{184.44} + \frac{22.94^2}{306.44} + \frac{22.94^2}{345.56}$ | M1 | For correct calculation process, inc Yates
$= 9.31\ldots$ | A1 | For correct value of the test statistic
This is greater than the critical 0.5% value of 7.879 | M1 | For a relevant (1 df) comparison
Hence there is very strong evidence to reject $H_0$ and conclude that views about changing to metric units are not independent of age | A1 | 7 | For correctly justifying the given answer (the final two marks remain available if Yates' correction is omitted)
**(ii)** $H_0: p_1 = p_2, H_1: p_1 \neq p_2$ | B1 | For both hypotheses stated
Under $H_0$ the sample value of the common proportion is $\frac{187 + 161}{1000} = 0.348$ | B1 | For correct value of estimated $p$
Test statistic is $\frac{\frac{187}{470} - \frac{161}{530}}{\sqrt{0.348 \times 0.652 \times \left(\frac{1}{470} + \frac{1}{530}\right)}}$ | M1 | For num $p_1 - p_2$ and denom using attempted s.d. based on a common estimate of $p$
$= 3.118$ | A1 | For completely correct expression
| A1 | For correct value of the test statistic
This is greater than the 0.2% (two-tail) critical value of 3.090 | M1 | For a relevant comparison using the normal distribution
Hence this test supports the conclusion of part (i) | A1 | 7 | For any relevant comparison or comment6 Certain types of food are now sold in metric units. A random sample of 1000 shoppers was asked whether they were in favour of the change to metric units or not. The results, classified according to age, were as shown in the table.
\begin{center}
\begin{tabular}{ | l | c c | r | }
\cline { 2 - 4 }
\multicolumn{1}{c|}{} & \multicolumn{2}{c|}{Age of shopper} & \\
\cline { 2 - 4 }
\multicolumn{1}{c|}{} & Under 35 & 35 and over & Total \\
\hline
In favour of change & 187 & 161 & 348 \\
Not in favour of change & 283 & 369 & 652 \\
\hline
Total & 470 & 530 & 1000 \\
\hline
\end{tabular}
\end{center}
(i) Use a $\chi ^ { 2 }$ test to show that there is very strong evidence that shoppers' views about changing to metric units are not independent of their ages.\\
(ii) The data may also be regarded as consisting of two random samples of shoppers; one sample consists of 470 shoppers aged under 35 , of whom 187 were in favour of change, and the second sample consists of 530 shoppers aged 35 or over, of whom 161 were in favour of change. Determine whether a test for equality of population proportions supports the conclusion in part (i).
\hfill \mbox{\textit{OCR S3 Q6 [14]}}