| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Standard 2×3 contingency table |
| Difficulty | Standard +0.3 This is a standard chi-squared test of independence with a 2×3 contingency table. Students must state hypotheses, calculate expected frequencies, compute the test statistic using the formula, find degrees of freedom (2), and compare to critical value. While it requires multiple steps and careful arithmetic, it follows a completely routine procedure taught explicitly in S3 with no novel problem-solving or conceptual insight required. Slightly above average difficulty only due to the computational burden and multiple steps involved. |
| Spec | 5.06a Chi-squared: contingency tables |
| Happiness | \multirow{2}{*}{Total} | ||||
| \cline { 3 - 5 } \multicolumn{2}{|c|}{} | Not happy | Fairly happy | Very happy | ||
| \multirow{2}{*}{Gender} | Female | 9 | 43 | 34 | 86 |
| \cline { 2 - 6 } | Male | 13 | 25 | 16 | 54 |
| Total | 22 | 68 | 50 | 140 | |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Expected values calculated: Female: 13.51, 41.77, 30.71; Male: 8.49, 26.23, 19.29 | M1, A1 | M1 for use of \(\frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}}\); A1 awrt 13.5, 41.8, 30.7, 8.5, 26.2, 19.3 |
| \(H_0\): Happiness and gender are independent/not associated; \(H_1\): Happiness and gender are not independent/associated | B1 | Must mention "happiness" and "gender" at least once; "relationship"/"correlation"/"connection" is B0 |
| \(\frac{(O-E)^2}{E}\) values: 1.508, 0.0361, 0.351, 2.402, 0.0575, 0.560 | dM1, A1, A1 | dM1 for at least 2 correct terms; dependent on 1st M1; 2nd A1 all correct (2sf); 3rd A1 awrt 4.91 |
| \(\sum \frac{(O-E)^2}{E} = 4.91\) or \(\sum \frac{O^2}{E} - N = 144.91 - 140 = 4.91\) | Condone 4.915 | |
| \(\nu = (3-2)(2-1) = 2\) | B1 | 2nd B1 for correct degrees of freedom |
| \(\sum \frac{(O-E)^2}{E} < 5.991\) | B1ft | 3rd B1ft for cv following from their degrees of freedom |
| \(4.91 < 5.991\) so insufficient evidence to reject \(H_0\) / Accept \(H_0\) | M1 | 3rd M1 for correct statement linking test statistic and cv; contradictory statements score M0 |
| No association between gender and happiness | A1 | Must mention "gender" and "happiness"; condone "relationship" or "connection" but not "correlation" |
# Question 3:
| Answer | Mark | Guidance |
|--------|------|----------|
| Expected values calculated: Female: 13.51, 41.77, 30.71; Male: 8.49, 26.23, 19.29 | M1, A1 | M1 for use of $\frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}}$; A1 awrt 13.5, 41.8, 30.7, 8.5, 26.2, 19.3 |
| $H_0$: Happiness and gender are independent/not associated; $H_1$: Happiness and gender are not independent/associated | B1 | Must mention "happiness" and "gender" at least once; "relationship"/"correlation"/"connection" is B0 |
| $\frac{(O-E)^2}{E}$ values: 1.508, 0.0361, 0.351, 2.402, 0.0575, 0.560 | dM1, A1, A1 | dM1 for at least 2 correct terms; dependent on 1st M1; 2nd A1 all correct (2sf); 3rd A1 awrt 4.91 |
| $\sum \frac{(O-E)^2}{E} = 4.91$ or $\sum \frac{O^2}{E} - N = 144.91 - 140 = 4.91$ | | Condone 4.915 |
| $\nu = (3-2)(2-1) = 2$ | B1 | 2nd B1 for correct degrees of freedom |
| $\sum \frac{(O-E)^2}{E} < 5.991$ | B1ft | 3rd B1ft for cv following from their degrees of freedom |
| $4.91 < 5.991$ so insufficient evidence to reject $H_0$ / Accept $H_0$ | M1 | 3rd M1 for correct statement linking test statistic and cv; contradictory statements score M0 |
| No association between gender and happiness | A1 | Must mention "gender" and "happiness"; condone "relationship" or "connection" but **not** "correlation" |
---3. A number of males and females were asked to rate their happiness under the headings "not happy", "fairly happy" and "very happy".
The results are shown in the table below
\begin{center}
\begin{tabular}{ | c | l | c | c | c | c | }
\hline
\multicolumn{2}{|c|}{} & \multicolumn{3}{|c|}{Happiness} & \multirow{2}{*}{Total} \\
\cline { 3 - 5 }
\multicolumn{2}{|c|}{} & Not happy & Fairly happy & Very happy & \\
\hline
\multirow{2}{*}{Gender} & Female & 9 & 43 & 34 & 86 \\
\cline { 2 - 6 }
& Male & 13 & 25 & 16 & 54 \\
\hline
\multicolumn{2}{|c|}{Total} & 22 & 68 & 50 & 140 \\
\hline
\end{tabular}
\end{center}
Stating your hypotheses, test at the $5 \%$ level of significance, whether or not there is evidence of an association between happiness and gender. Show your working clearly.
\hfill \mbox{\textit{Edexcel S3 2014 Q3 [10]}}