| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2017 |
| Session | June |
| Marks | 14 |
| 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. Part (a) requires basic expected frequency formula (row total × column total / grand total), part (b) is a routine hypothesis test with given significance level, and part (c) involves comparing a test statistic to critical values in tables. All steps are mechanical applications of the chi-squared procedure taught in S3, with no conceptual challenges or novel problem-solving required. Slightly easier than average due to the structured guidance and partial completion of expected frequencies. |
| Spec | 5.06a Chi-squared: contingency tables |
| \cline { 2 - 4 } \multicolumn{1}{c|}{} | Underweight | About right | Overweight |
| Male | 20 | 22 | 30 |
| Female | 16 | 28 | 34 |
| \cline { 2 - 4 } \multicolumn{1}{c|}{} | Underweight | About right | Overweight |
| Male | 17.28 | ||
| Female | 18.72 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{72 \times 50}{150} = 24,\ \frac{78 \times 50}{150} = 26\) | M1 | For one correct \(\frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}}\); can be implied by correct answers |
| \(\frac{72 \times 64}{150} = 30.72,\ \frac{78 \times 64}{150} = 33.28\) | A1 | 24, 26, 30.72, 33.28 only |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0\): Perceived (body) weight is independent of gender (no association) | B1 | Both hypotheses required; must mention "Perceived", "weight" and "gender" at least once; use of "relationship" or "correlation" or "connection" or "link" award B0 |
| \(H_1\): Perceived (body) weight is not independent of gender (association) | ||
| \(O\) | \(E\) | \(\frac{(O-E)^2}{E}\) |
| 20 | 17.28 | 0.428148 |
| 22 | 24 | 0.166667 |
| 30 | 30.72 | 0.016875 |
| 16 | 18.72 | 0.395214 |
| 28 | 26 | 0.153846 |
| 34 | 33.28 | 0.015577 |
| 150 | 150 | 1.176327 |
| M1 for at least 2 correct terms (as in 3rd or 4th column) or correct expressions | M1A1 | A1 for all correct; accept 2sf accuracy; allow truncation e.g. 1.17... |
| \(\sum \frac{(O-E)^2}{E}\) or \(\sum \frac{O^2}{E} - 150 = 1.18\) | A1 | awrt 1.18–1.19 |
| \(\nu = (3-1)(2-1) = 2,\ \chi^2_2(10\%) = 4.605\) | B1B1ft | 2 can be implied by 4.605 seen |
| Accept \(H_0\): Perceived (body) weight is independent of gender (no association) | A1ft | Correct comment in context – must mention "weight" and "gender"; condone "relationship" or "connection" here but not "correlation"; follow through from their test stat and cv, but hypotheses must be correct |
| Answer | Marks | Guidance |
|---|---|---|
| \(O\) | \(E\) | \(\frac{(O-E)^2}{E}\) |
| 36 | 50 | 3.92 |
| 50 | 50 | 0 |
| 64 | 50 | 3.92 |
| 150 | 150 | 7.84 |
| B1 for \(E_i = 50\), could be implied; M1 for combining values and for attempting \(\frac{(O-E)^2}{E}\) or \(\frac{O^2}{E}\) with at least 2 correct expressions or values | B1, M1A1 | A1 for all correct, can be implied by correct answer below |
| \(\sum \frac{(O-E)^2}{E}\) or \(\sum \frac{O^2}{E} - 150 = 7.84\) | A1 | awrt 7.84 |
| \(\nu = 2,\ \chi^2_2(2.5\%) = 7.378\) | A1 | 0.025 or 2.5% |
## Question 4:
### Part (a):
$\frac{72 \times 50}{150} = 24,\ \frac{78 \times 50}{150} = 26$ | M1 | For one correct $\frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}}$; can be implied by correct answers
$\frac{72 \times 64}{150} = 30.72,\ \frac{78 \times 64}{150} = 33.28$ | A1 | 24, 26, 30.72, 33.28 only
### Part (b):
$H_0$: Perceived (body) weight is independent of gender (no association) | B1 | Both hypotheses required; must mention "Perceived", "weight" and "gender" at least once; use of "relationship" or "correlation" or "connection" or "link" award B0
$H_1$: Perceived (body) weight is not independent of gender (association) | |
| $O$ | $E$ | $\frac{(O-E)^2}{E}$ | $\frac{O^2}{E}$ |
|-----|-----|-----|-----|
| 20 | 17.28 | 0.428148 | 23.14815 |
| 22 | 24 | 0.166667 | 20.16667 |
| 30 | 30.72 | 0.016875 | 29.29688 |
| 16 | 18.72 | 0.395214 | 13.67521 |
| 28 | 26 | 0.153846 | 30.15385 |
| 34 | 33.28 | 0.015577 | 34.73558 |
| 150 | 150 | 1.176327 | 151.1763 |
M1 for at least 2 correct terms (as in 3rd or 4th column) or correct expressions | M1A1 | A1 for all correct; accept 2sf accuracy; allow truncation e.g. 1.17...
$\sum \frac{(O-E)^2}{E}$ or $\sum \frac{O^2}{E} - 150 = 1.18$ | A1 | awrt 1.18–1.19
$\nu = (3-1)(2-1) = 2,\ \chi^2_2(10\%) = 4.605$ | B1B1ft | 2 can be implied by 4.605 seen
Accept $H_0$: Perceived (body) **weight** is independent of **gender** (no association) | A1ft | Correct comment in context – must mention "weight" and "gender"; condone "relationship" or "connection" here but **not** "correlation"; follow through from their test stat and cv, but **hypotheses must be correct**
### Part (c):
| $O$ | $E$ | $\frac{(O-E)^2}{E}$ | $\frac{O^2}{E}$ |
|-----|-----|-----|-----|
| 36 | 50 | 3.92 | 25.92 |
| 50 | 50 | 0 | 50 |
| 64 | 50 | 3.92 | 81.92 |
| 150 | 150 | 7.84 | 157.84 |
B1 for $E_i = 50$, could be implied; M1 for combining values and for attempting $\frac{(O-E)^2}{E}$ or $\frac{O^2}{E}$ with at least 2 correct expressions or values | B1, M1A1 | A1 for all correct, can be implied by correct answer below
$\sum \frac{(O-E)^2}{E}$ or $\sum \frac{O^2}{E} - 150 = 7.84$ | A1 | awrt 7.84
$\nu = 2,\ \chi^2_2(2.5\%) = 7.378$ | A1 | 0.025 or 2.5%
---4. A psychologist carries out a survey of the perceived body weight of 150 randomly chosen people. He asks them if they think they are underweight, about right or overweight. His results are summarised in the table below.
\begin{center}
\begin{tabular}{ | l | c | c | c | }
\cline { 2 - 4 }
\multicolumn{1}{c|}{} & Underweight & About right & Overweight \\
\hline
Male & 20 & 22 & 30 \\
\hline
Female & 16 & 28 & 34 \\
\hline
\end{tabular}
\end{center}
The psychologist calculates two of the expected frequencies, to 2 decimal places, for a test of independence between perceived body weight and gender. These results are shown in the table below.
\begin{center}
\begin{tabular}{ | l | c | c | c | }
\cline { 2 - 4 }
\multicolumn{1}{c|}{} & Underweight & About right & Overweight \\
\hline
Male & 17.28 & & \\
\hline
Female & 18.72 & & \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Complete the table of expected frequencies shown above.
\item Test, at the $10 \%$ level of significance, whether or not perceived body weight is independent of gender. State your hypotheses clearly.
The psychologist now combines the male and female data to test whether or not body weight types are chosen equally.
\item Find the smallest significance level, from the tables in the formula booklet, for which there is evidence of a preference.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2017 Q4 [14]}}