| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2013 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Standard 2×2 contingency table |
| Difficulty | Standard +0.3 This is a standard chi-squared test of independence followed by a routine two-proportion z-test. Both procedures are textbook applications with clearly structured data, requiring only mechanical calculation of expected frequencies, test statistics, and comparison with critical values. The question is slightly above average difficulty only due to the two-part structure and the need to recognize which test is appropriate in part (ii). |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.06a Chi-squared: contingency tables |
| Gender | |||
| \cline { 2 - 4 } | Female | Male | |
| \cline { 2 - 4 } Comply | Yes | 34 | 30 |
| \cline { 2 - 4 } | No | 11 | 25 |
| \cline { 2 - 4 } | |||
| \cline { 2 - 4 } | |||
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((H_0:\) attributes are independent, \(H_1:\) all alts) | ||
| Expected values: Y: \(28.8,\ 35.2\); N: \(16.2,\ 19.8\) | M1, A1 | Any one correct e.g. \(45\times64/100\). All correct. |
| \(\chi^2 = 4.7^2(28.8^{-1}+16.2^{-1}+35.2^{-1}+19.8^{-1}) = 3.87(381..)\) | M2, A1 | M2 with Yates and correct form. M1 without Yates and correct form or no modulus. A1 for 4.74 without Yates. A1 for 4.81, no modulus. |
| CV \(= 3.841\) | B1 | |
| \(3.87 > 3.841\) so reject \(H_0\) | M1 | Ft TS and CV |
| There is sufficient evidence (at the 5% SL) that gender and compliance with the advice are not independent. | A1 | cwo, other than rounding errors. |
| [8] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0: p_F = p_M,\ H_1: p_F > p_M\) | B1 | Or equivalent |
| \(p_s = (34+30)/(45+55)\) | B1 | |
| \(z = \dfrac{34/_{45} - 30/_{55}}{\sqrt{0.64\times0.36(45^{-1}+55^{-1})}}\) | M1 | Variance must contain 0.64 and 1-0.64 or denominators 45 and 55. e.g. \(\sqrt{\dfrac{0.64\times0.36}{100}}\) M1B0A0 |
| \(= 2.18\) | B1, A1 | For denominator |
| \(> 1.282\) | M1 | Or \(P(z>2.18)=0.015 < 0.10\) ft TS. |
| Reject \(H_0\), there is sufficient evidence at the \(2\tfrac{1}{2}\%\) significance level that more women than men comply with the advice. | A1 | SR if \(p_s\) not used. B1B0M1B0A1(for \(z=2.26(4)\)) M1A1. cwo, in context, not over-assertive. Max(5/7) |
| [7] |
# Question 8:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(H_0:$ attributes are independent, $H_1:$ all alts) | | |
| Expected values: Y: $28.8,\ 35.2$; N: $16.2,\ 19.8$ | M1, A1 | Any one correct e.g. $45\times64/100$. All correct. |
| $\chi^2 = 4.7^2(28.8^{-1}+16.2^{-1}+35.2^{-1}+19.8^{-1}) = 3.87(381..)$ | M2, A1 | M2 with Yates and correct form. M1 without Yates and correct form or no modulus. A1 for 4.74 without Yates. A1 for 4.81, no modulus. |
| CV $= 3.841$ | B1 | |
| $3.87 > 3.841$ so reject $H_0$ | M1 | Ft TS and CV |
| There is sufficient evidence (at the 5% SL) that gender and compliance with the advice are not independent. | A1 | cwo, other than rounding errors. |
| | **[8]** | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: p_F = p_M,\ H_1: p_F > p_M$ | B1 | Or equivalent |
| $p_s = (34+30)/(45+55)$ | B1 | |
| $z = \dfrac{34/_{45} - 30/_{55}}{\sqrt{0.64\times0.36(45^{-1}+55^{-1})}}$ | M1 | Variance must contain 0.64 and 1-0.64 or denominators 45 and 55. e.g. $\sqrt{\dfrac{0.64\times0.36}{100}}$ M1B0A0 |
| $= 2.18$ | B1, A1 | For denominator |
| $> 1.282$ | M1 | Or $P(z>2.18)=0.015 < 0.10$ ft TS. |
| Reject $H_0$, there is sufficient evidence at the $2\tfrac{1}{2}\%$ significance level that more women than men comply with the advice. | A1 | SR if $p_s$ not used. B1B0M1B0A1(for $z=2.26(4)$) M1A1. cwo, in context, not over-assertive. Max(5/7) |
| | **[7]** | |
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Please upload those pages and I'll be happy to format the content as requested.8 After contracting a particular disease, patients from a hospital are advised to have their blood tested monthly for a year. In order to test whether patients comply with this advice the hospital management commissioned a survey of 100 patients. A hospital statistician selected the patients randomly from records and asked the patients whether or not they had complied with the advice. The results classified by gender are as follows.
\begin{center}
\begin{tabular}{ l | c | c | c | }
& \multicolumn{2}{c}{Gender} & \\
\cline { 2 - 4 }
& & Female & Male \\
\cline { 2 - 4 }
Comply & Yes & 34 & 30 \\
\cline { 2 - 4 }
& No & 11 & 25 \\
\cline { 2 - 4 }
& & & \\
\cline { 2 - 4 }
\end{tabular}
\end{center}
(i) Test at the $5 \%$ significance level whether compliance with the advice is independent of gender.\\
(ii) A manager believed that a greater proportion of female patients than male patients comply with the advice. Carry out an appropriate test of proportions at the $10 \%$ significance level.
\hfill \mbox{\textit{OCR S3 2013 Q8 [15]}}