| Exam Board | WJEC |
|---|---|
| Module | Further Unit 2 (Further Unit 2) |
| Year | 2022 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Expected frequencies partially provided |
| Difficulty | Standard +0.3 This is a standard chi-squared test of independence with straightforward calculations. All steps are routine: stating hypotheses, calculating expected frequencies using the standard formula, computing chi-squared contributions, and interpreting results. The question guides students through each step with partial answers provided, making it easier than a typical A-level statistics question that requires independent setup and execution. |
| Spec | 5.06a Chi-squared: contingency tables |
| Age in years | |||||
| Use social media | 18-29 | 30-49 | 50-64 | 65 or older | Total |
| Yes | 310 | 412 | 348 | 196 | 1266 |
| No | 42 | 116 | 196 | 333 | 687 |
| Total | 352 | 528 | 544 | 529 | 1953 |
| Expected values | Age in years | |||
| Use social media | 18-29 | 30-49 | 50-64 | 65 or older |
| Yes | \(228 \cdot 18\) | \(342 \cdot 27\) | 352.64 | 342.92 |
| No | 123.82 | 185.73 | 191.36 | 186.08 |
| Chi-squared contributions | Age in years | |||
| Use social media | 18-29 | 30-49 | 50-64 | 65 or older |
| Yes | 29.34 | \(s\) | 0.06 | 62.94 |
| No | 54.07 | 26-18 | 0.11 | 115.99 |
| Answer | Marks | Guidance |
|---|---|---|
| 6(a) | \(H_0\): Social media usage is independent of age. \(H_1\): Social media usage is not independent of age | B1 |
| 6(b) | \(\frac{1266 \times 352}{1953} = 228.18\) *ag | B1 |
| 6(c) | \(s = \frac{(412 - 342.27)^2}{342.27}\) | M1 |
| \(s = 14.2(0595699...)\) | A1 | |
| 6(d) | \((4 - 1) \times (2 - 1) = 3\) degrees of freedom. | B1 |
| 5% CV = 7.815 | B1 | |
| Add \(\chi^2\) contributions \(29.34 + 14.21 + 0.06 + 62.94 + 54.07 + 26.18 + 0.11 + 115.99 = 302.90\) | M1 | M1A1 if statement along the lines of "one contribution is > 7.815" |
| A1 | ||
| Since 302.91 > 7.815 we can reject \(H_0\). There is (strong) evidence to suggest that social media usage is not independent of age. | B1 E1 | FT provided \(\chi^2 > 7.815\) Only award E1 if previous three B1 awarded and part (a) correct |
| 6(e) | Valid explanation. e.g. The \(p\) value would not lead to rejecting \(H_0\), which is the incorrect conclusion. | E1 |
| Total [11] |
6(a) | $H_0$: Social media usage is independent of age. $H_1$: Social media usage is not independent of age | B1 |
6(b) | $\frac{1266 \times 352}{1953} = 228.18$ *ag | B1 | oe
6(c) | $s = \frac{(412 - 342.27)^2}{342.27}$ | M1 |
| $s = 14.2(0595699...)$ | A1 |
6(d) | $(4 - 1) \times (2 - 1) = 3$ degrees of freedom. | B1 |
| 5% CV = 7.815 | B1 |
| Add $\chi^2$ contributions $29.34 + 14.21 + 0.06 + 62.94 + 54.07 + 26.18 + 0.11 + 115.99 = 302.90$ | M1 | M1A1 if statement along the lines of "one contribution is > 7.815"
| | A1 |
| Since 302.91 > 7.815 we can reject $H_0$. There is (strong) evidence to suggest that social media usage is not independent of age. | B1 E1 | FT provided $\chi^2 > 7.815$ Only award E1 if previous three B1 awarded and part (a) correct
6(e) | Valid explanation. e.g. The $p$ value would not lead to rejecting $H_0$, which is the incorrect conclusion. | E1 |
| **Total [11]** |
---6. An online survey on the use of social media asked the following question:
\begin{displayquote}
"Do you use any form of social media?"
\end{displayquote}
The results for a total of 1953 respondents are shown in the table below.
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|}
\hline
& \multicolumn{4}{|c|}{Age in years} & \\
\hline
Use social media & 18-29 & 30-49 & 50-64 & 65 or older & Total \\
\hline
Yes & 310 & 412 & 348 & 196 & 1266 \\
\hline
No & 42 & 116 & 196 & 333 & 687 \\
\hline
Total & 352 & 528 & 544 & 529 & 1953 \\
\hline
\end{tabular}
\end{center}
To test whether there is a relationship between social media use and age, a significance test is carried out at the $5 \%$ level.
\begin{enumerate}[label=(\alph*)]
\item State the null and alternative hypotheses.
\item Show how the expected frequency $228 \cdot 18$ is calculated in the table below.
\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
Expected values & \multicolumn{4}{|c|}{Age in years} \\
\hline
Use social media & 18-29 & 30-49 & 50-64 & 65 or older \\
\hline
Yes & $228 \cdot 18$ & $342 \cdot 27$ & 352.64 & 342.92 \\
\hline
No & 123.82 & 185.73 & 191.36 & 186.08 \\
\hline
\end{tabular}
\end{center}
\item Determine the value of $s$ in the table below.
\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
Chi-squared contributions & \multicolumn{4}{|c|}{Age in years} \\
\hline
Use social media & 18-29 & 30-49 & 50-64 & 65 or older \\
\hline
Yes & 29.34 & $s$ & 0.06 & 62.94 \\
\hline
No & 54.07 & 26-18 & 0.11 & 115.99 \\
\hline
\end{tabular}
\end{center}
\item Complete the significance test, showing all your working.
\item A student, analysing these data on a spreadsheet, obtains the following output.\\
\includegraphics[max width=\textwidth, alt={}, center]{77fd7ad7-f5a3-4947-afc6-e5ef45bef7a8-5_202_1271_445_415}
Explain why the student must have made an error in calculating the $p$-value.
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 2 2022 Q6 [11]}}