| Exam Board | OCR MEI |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | June |
| Marks | 20 |
| 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 straightforward two-part question testing standard chi-squared independence test and hypothesis testing procedures. Part (a) requires calculating expected frequencies, computing chi-squared statistic, and comparing to critical value—all routine steps. Part (b) is a standard one-tailed z-test with given summary statistics. Both parts follow textbook procedures with no conceptual challenges or novel insights required, making this slightly easier than average. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.06a Chi-squared: contingency tables |
|
| |||||
| Student smokes | 21 | 27 | ||||
| Student does not smoke | 17 | 55 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0\): no association between student smoking and parent smoking; \(H_1\): some association between student smoking and parent smoking | B1 | Correct hypotheses in context; NB if \(H_0\), \(H_1\) reversed do not award first B1 or final B1dep*; Allow hypotheses in terms of independence; Do not allow "relationship" or "correlation" for "association" |
| Expected frequencies: Student smokes/Parent smokes = 15.2; Student smokes/Parent does not smoke = 32.8; Student does not smoke/Parent smokes = 22.8; Student does not smoke/Parent does not smoke = 49.2 | B1, B1 | For at least one row/column of expected values correct; All correct |
| Contributions: Student smokes/Parent smokes = 2.213; Student smokes/Parent does not smoke = 1.026; Student does not smoke/Parent smokes = 1.475; Student does not smoke/Parent does not smoke = 0.684 | M1, A1 | For valid attempt at \(\frac{(O-E)^2}{E}\); All correct to 3 d.p.; NB these M1A1 marks cannot be implied by a correct final value of \(X^2\) |
| \(X^2 = 5.398\) | B1 | Allow awrt 5.40; Do not penalise use of Yates correction, giving \(X^2 = 4.51\) |
| Refer to \(\chi^2_1\); Critical value at 5% level \(= 3.841\) | B1, B1 | For 1 degree of freedom; CAO for cv; \(p\) value \(= 0.02016\); No further marks from here if wrong or omitted, unless \(p\)-value used instead |
| Result is significant | B1* | For significant oe; FT their test statistic |
| There is sufficient evidence to suggest that there is association between student smoking and parent smoking | B1dep* | NB if \(H_0\), \(H_1\) reversed do not award first B1 or final B1dep*; For non-assertive conclusion in context; Do not allow "relationship" or "correlation" for "association" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\bar{x} = 88.2/100 = 0.882\) | B1 | For 0.882 seen |
| \(s = \sqrt{\frac{78.68 - (88.2)^2/100}{99}} = \sqrt{\frac{0.8876}{99}} = \sqrt{0.0089657}\) | M1 | For correctly structured calculation for the sample standard deviation or variance |
| \(= 0.0947\) (allow 0.095 www) | A1 | Allow A1 for \(s^2 = 0.0089657\); or 0.00897 (allow 0.0090) |
| \(H_0: \mu = 0.87\); \(H_1: \mu > 0.87\); where \(\mu\) denotes the mean nicotine content (of cigarettes of this brand in the population) | B1, B1 | For both correct; For definition of \(\mu\) in context; Hypotheses in words only must refer to population; Do not allow other symbols unless clearly defined as population mean |
| Test statistic \(= \frac{0.882 - 0.87}{0.0947/\sqrt{100}} = \frac{0.012}{0.00947} = 1.267\) | M1*, A1 | Including correct use of \(\sqrt{100}\); CAO; FT their \(s\) |
| Upper 1% level 1-tailed critical value of \(z = 2.326\) | B1 | For 2.326 No further marks from here if wrong |
| \(1.267 < 2.326\) (Not significant) | M1dep* | For sensible comparison leading to a conclusion |
| There is insufficient evidence to suggest that the mean nicotine content of this brand is greater than 0.87mg | A1 | For non-assertive conclusion in words and in context; FT only candidate's test statistic |
## Question 4:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0$: no association between student smoking and parent smoking; $H_1$: some association between student smoking and parent smoking | B1 | Correct hypotheses in context; NB if $H_0$, $H_1$ reversed do not award first B1 or final B1dep*; Allow hypotheses in terms of independence; Do not allow "relationship" or "correlation" for "association" |
| Expected frequencies: Student smokes/Parent smokes = 15.2; Student smokes/Parent does not smoke = 32.8; Student does not smoke/Parent smokes = 22.8; Student does not smoke/Parent does not smoke = 49.2 | B1, B1 | For at least one row/column of expected values correct; All correct |
| Contributions: Student smokes/Parent smokes = 2.213; Student smokes/Parent does not smoke = 1.026; Student does not smoke/Parent smokes = 1.475; Student does not smoke/Parent does not smoke = 0.684 | M1, A1 | For valid attempt at $\frac{(O-E)^2}{E}$; All correct to 3 d.p.; NB these M1A1 marks cannot be implied by a correct final value of $X^2$ |
| $X^2 = 5.398$ | B1 | Allow awrt 5.40; Do not penalise use of Yates correction, giving $X^2 = 4.51$ |
| Refer to $\chi^2_1$; Critical value at 5% level $= 3.841$ | B1, B1 | For 1 degree of freedom; CAO for cv; $p$ value $= 0.02016$; No further marks from here if wrong or omitted, unless $p$-value used instead |
| Result is significant | B1* | For significant oe; FT their test statistic |
| There is sufficient evidence to **suggest** that there is **association between student smoking and parent smoking** | B1dep* | NB if $H_0$, $H_1$ reversed do not award first B1 or final B1dep*; For **non-assertive** conclusion in context; Do not allow "relationship" or "correlation" for "association" |
**[10 marks]**
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\bar{x} = 88.2/100 = 0.882$ | B1 | For 0.882 seen |
| $s = \sqrt{\frac{78.68 - (88.2)^2/100}{99}} = \sqrt{\frac{0.8876}{99}} = \sqrt{0.0089657}$ | M1 | For correctly structured calculation for the sample standard deviation or variance |
| $= 0.0947$ (allow 0.095 www) | A1 | Allow A1 for $s^2 = 0.0089657$; or 0.00897 (allow 0.0090) |
| $H_0: \mu = 0.87$; $H_1: \mu > 0.87$; where $\mu$ denotes the **mean nicotine** content (of cigarettes of this brand in the population) | B1, B1 | For both correct; For definition of $\mu$ in context; Hypotheses in words only must refer to population; Do not allow other symbols unless clearly defined as population mean |
| Test statistic $= \frac{0.882 - 0.87}{0.0947/\sqrt{100}} = \frac{0.012}{0.00947} = 1.267$ | M1*, A1 | Including correct use of $\sqrt{100}$; CAO; FT their $s$ |
| Upper 1% level 1-tailed critical value of $z = 2.326$ | B1 | For 2.326 **No further marks from here if wrong** |
| $1.267 < 2.326$ (Not significant) | M1dep* | For sensible comparison leading to a conclusion |
| There is **insufficient evidence to suggest** that the **mean nicotine content** of this brand is **greater** than 0.87mg | A1 | For non-assertive conclusion in words and in context; FT only candidate's test statistic |
**[10 marks]**
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\begin{enumerate}[label=(\alph*)]
\item As part of an investigation into smoking, a random sample of 120 students was selected. The students were asked whether they were smokers, and also whether either of their parents were smokers. The results are summarised in the table below. Test, at the $5 \%$ significance level, whether there is any association between the smoking habits of the students and their parents.
\begin{center}
\begin{tabular}{ | l | c | c | }
\hline
& \begin{tabular}{ c }
At least one \\
parent smokes \\
\end{tabular} & \begin{tabular}{ c }
Neither parent \\
smokes \\
\end{tabular} \\
\hline
Student smokes & 21 & 27 \\
\hline
Student does not smoke & 17 & 55 \\
\hline
\end{tabular}
\end{center}
\item The manufacturer of a particular brand of cigarette claims that the nicotine content of these cigarettes is Normally distributed with mean 0.87 mg . A researcher suspects that the mean nicotine content of this brand is higher than the value claimed by the manufacturer. The nicotine content, $x \mathrm { mg }$, is measured for a random sample of 100 cigarettes. The data are summarised as follows.
$$\sum x = 88.20 \quad \sum x ^ { 2 } = 78.68$$
Carry out a test at the $1 \%$ significance level to investigate the researcher's belief.
\section*{END OF QUESTION PAPER}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S2 2015 Q4 [20]}}