| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Statistics (Further Paper 3 Statistics) |
| Year | 2020 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Cell combining required |
| Difficulty | Standard +0.3 This is a straightforward chi-squared test question requiring recognition of a common error (expected frequency < 5 requiring pooling of categories), followed by a standard hypothesis test completion using given test statistic and critical value lookup. The conceptual demand is low for Further Maths Statistics students, though it requires careful attention to detail. |
| Spec | 5.06a Chi-squared: contingency tables |
| Number of snacks eaten | ||||
| \cline { 2 - 5 }\cline { 2 - 4 } | 0 | 1 | 2 or more | |
| \cline { 2 - 4 } Time of Day | 23.68 | 21.05 | 5.26 | |
| \cline { 2 - 5 } | Night | 21.32 | 18.95 | 4.74 |
| \cline { 2 - 5 } | ||||
| \cline { 2 - 5 } | ||||
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| There is an expected frequency less than 5 so two columns will need to be merged. | E1 | Deduces that two columns will need to be merged as there is an expected frequency less than 5 |
| This means that the test will only have 1 degree of freedom. This means that Yates' correction will need to be used and so the test statistic will be \(\sum\frac{(\ | O-E\ | -0.5)^2}{E}\) rather than \(\sum\frac{(O-E)^2}{E}\) |
| Yates' correction will need to be used | E1 | Explains that Yates' correction will need to be used |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\chi^2\) cv for 1 df \(= 6.635\); \(8.74 > 6.635\) | R1 | Evaluates \(\chi^2\)-test statistic by comparing the correct critical value with the test statistic |
| Reject \(H_0\) | E1F | Infers \(H_0\) rejected. FT 'their' critical value. |
| Significant evidence to suggest that there is an association between time of day and number of snacks eaten. | E1F | Concludes in context. Conclusion must not be definite. FI their incorrect acceptance of \(H_0\) if stated or 'their' comparison if not. |
## Question 8(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| There is an expected frequency less than 5 so two columns will need to be merged. | E1 | Deduces that two columns will need to be merged as there is an expected frequency less than 5 |
| This means that the test will only have 1 degree of freedom. This means that Yates' correction will need to be used and so the test statistic will be $\sum\frac{(\|O-E\|-0.5)^2}{E}$ rather than $\sum\frac{(O-E)^2}{E}$ | E1 | Deduces that there is only 1 degree of freedom or that this will form a 2×2 table. Allow if only seen in 8(b). |
| Yates' correction will need to be used | E1 | Explains that Yates' correction will need to be used |
---
## Question 8(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\chi^2$ cv for 1 df $= 6.635$; $8.74 > 6.635$ | R1 | Evaluates $\chi^2$-test statistic by comparing the correct critical value with the test statistic |
| Reject $H_0$ | E1F | Infers $H_0$ rejected. FT 'their' critical value. |
| Significant evidence to suggest that there is an association between time of day and number of snacks eaten. | E1F | Concludes in context. Conclusion must not be definite. FI their incorrect acceptance of $H_0$ if stated or 'their' comparison if not. |
**Total: 6 marks**
---8 Ray is conducting a hypothesis test with the hypotheses\\
$\mathrm { H } _ { 0 }$ : There is no association between time of day and number of snacks eaten\\
$\mathrm { H } _ { 1 }$ : There is an association between time of day and number of snacks eaten\\
He calculates expected frequencies correct to two decimal places, which are given in the following table.
\begin{center}
\begin{tabular}{ l | l | c | c | c | }
\multicolumn{2}{c}{} & \multicolumn{2}{c}{Number of snacks eaten} & \\
\cline { 2 - 5 }\cline { 2 - 4 }
& & 0 & 1 & 2 or more \\
\cline { 2 - 4 }
Time of Day & 23.68 & 21.05 & 5.26 & \\
\cline { 2 - 5 }
& Night & 21.32 & 18.95 & 4.74 \\
\cline { 2 - 5 }
& & & & \\
\cline { 2 - 5 }
\end{tabular}
\end{center}
Ray calculates his test statistic using $\sum \frac { ( O - E ) ^ { 2 } } { E }$\\
8
\begin{enumerate}[label=(\alph*)]
\item State, with a reason, the error Ray has made and describe any changes Ray will need to make to his test.\\
8
\item Having made the necessary corrections as described in part (a), the correct value of the test statistic is 8.74
Complete Ray's hypothesis test using a $1 \%$ level of significance.
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Statistics 2020 Q8 [6]}}