| Exam Board | OCR MEI |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| Session | June |
| Marks | 17 |
| 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 straightforward chi-squared test of independence with most calculations provided. Students only need to calculate one expected frequency using the standard formula (row total × column total / grand total), verify a given contribution using (O-E)²/E, look up critical values, and state a conclusion. The conceptual demand is minimal—it's a standard textbook exercise with no novel problem-solving required. |
| Spec | 5.06a Chi-squared: contingency tables |
| \multirow{2}{*}{} | Classification of person | \multirow{2}{*}{Row totals} | ||
| Child | Adult | |||
| \multirow{4}{*}{Type of cake} | Chocolate | 34 | 23 | 57 |
| Carrot | 16 | 18 | 34 | |
| Lemon | 4 | 18 | 22 | |
| Ginger | 13 | 24 | 37 | |
| Column totals | 67 | 83 | 150 | |
| Classification of person | ||||||
| \cline { 3 - 4 } \multicolumn{2}{|c|}{} | Child | Adult | ||||
\multirow{3}{*}{
| Chocolate | 2.8646 | 2.3124 | |||
| \cline { 2 - 4 } | Carrot | 0.0436 | 0.0352 | |||
| \cline { 2 - 4 } | Lemon | 3.4549 | 2.7889 | |||
| \cline { 2 - 4 } | Ginger | 0.7526 | 0.6075 | |||
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(5 - 1.645 \times 0.0072 \div \sqrt{8}\) | M1\*B1\* | |
| \(= 4.9958\ldots\) | A1 | |
| \(4.995 < 4.99581\ldots\) | M1dep\* | Sensible comparison |
| Correct conclusion in words & context | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(4.995 + 1.645 \times 0.0072 \div \sqrt{8}\) | M1\*B1\* | |
| \(= 4.9991\ldots\) | A1 | Final M1dep\* A1 available only if 1.645 used |
| \(5 > 4.9991\ldots\) | M1 | |
| Correct conclusion in words & context | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Finding \(P(\text{sample mean} < 4.995) = 0.0248\) | M1\*A1B1 | |
| \(0.0248 < \mathbf{0.05}\) | M1dep\* | For sensible comparison if a conclusion is made |
| Correct conclusion in words & context | A1 | Condone \(P(\text{sample mean} > 4.995) = 0.9752\) for M1 but only allow A1B1 if later compared with 0.95 |
| Answer | Marks |
|---|---|
| Guidance | |
| Final answers correct to 5 or more significant figures will be penalised. Candidates may lose no more than 2 marks per question and no more than 4 marks in total. Exception: Question 3 part (iv) — see guidance note. |
## Question 4(b):
**Critical Value Method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $5 - 1.645 \times 0.0072 \div \sqrt{8}$ | M1\*B1\* | |
| $= 4.9958\ldots$ | A1 | |
| $4.995 < 4.99581\ldots$ | M1dep\* | Sensible comparison |
| Correct conclusion in words & context | A1 | |
**"Confidence Interval" Method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $4.995 + 1.645 \times 0.0072 \div \sqrt{8}$ | M1\*B1\* | |
| $= 4.9991\ldots$ | A1 | Final M1dep\* A1 available only if 1.645 used |
| $5 > 4.9991\ldots$ | M1 | |
| Correct conclusion in words & context | A1 | |
**Probability Method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| Finding $P(\text{sample mean} < 4.995) = 0.0248$ | M1\*A1B1 | |
| $0.0248 < \mathbf{0.05}$ | M1dep\* | For sensible comparison if a conclusion is made |
| Correct conclusion in words & context | A1 | Condone $P(\text{sample mean} > 4.995) = 0.9752$ for M1 but only allow A1B1 if later compared with 0.95 |
---
## Over-specification Note:
| Guidance |
|----------|
| Final answers correct to 5 or more significant figures will be penalised. Candidates may lose no more than 2 marks per question and no more than 4 marks in total. Exception: Question 3 part (iv) — see guidance note. |4
\begin{enumerate}[label=(\alph*)]
\item Mary is opening a cake shop. As part of her market research, she carries out a survey into which type of cake people like best. She offers people 4 types of cake to taste: chocolate, carrot, lemon and ginger. She selects a random sample of 150 people and she classifies the people as children and adults. The results are as follows.
\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
\multicolumn{2}{|c|}{\multirow{2}{*}{}} & \multicolumn{2}{|c|}{Classification of person} & \multirow{2}{*}{Row totals} \\
\hline
& & Child & Adult & \\
\hline
\multirow{4}{*}{Type of cake} & Chocolate & 34 & 23 & 57 \\
\hline
& Carrot & 16 & 18 & 34 \\
\hline
& Lemon & 4 & 18 & 22 \\
\hline
& Ginger & 13 & 24 & 37 \\
\hline
\multicolumn{2}{|r|}{Column totals} & 67 & 83 & 150 \\
\hline
\end{tabular}
\end{center}
The contributions to the test statistic for the usual $\chi ^ { 2 }$ test are shown in the table below.
\begin{center}
\begin{tabular}{ | c | l | c | c | }
\hline
\multicolumn{2}{|c|}{} & \multicolumn{2}{c|}{Classification of person} \\
\cline { 3 - 4 }
\multicolumn{2}{|c|}{} & Child & Adult \\
\hline
\multirow{3}{*}{\begin{tabular}{ c }
Type \\
of \\
cake \\
\end{tabular}} & Chocolate & 2.8646 & 2.3124 \\
\cline { 2 - 4 }
& Carrot & 0.0436 & 0.0352 \\
\cline { 2 - 4 }
& Lemon & 3.4549 & 2.7889 \\
\cline { 2 - 4 }
& Ginger & 0.7526 & 0.6075 \\
\hline
\end{tabular}
\end{center}
The sum of these contributions, correct to 2 decimal places, is 12.86 .
\begin{enumerate}[label=(\roman*)]
\item Calculate the expected frequency for children preferring chocolate cake. Verify the corresponding contribution, 2.8646, to the test statistic.
\item Carry out the test at the $1 \%$ level of significance.
\end{enumerate}\item Mary buys flour in bags which are labelled as containing 5 kg . She suspects that the average contents of these bags may be less than 5 kg . In order to test this, she selects a random sample of 8 bags and weighs their contents. Assuming that weights are Normally distributed with standard deviation 0.0072 kg , carry out a test at the $5 \%$ level, given that the weights of the 8 bags in kg are as follows.\\
4.992\\
4.981\\
4.982\\
4.996\\
4.991\\
5.006\\
5.009\\
5.003\\[0pt]
[9]
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\end{enumerate}
\hfill \mbox{\textit{OCR MEI S2 2012 Q4 [17]}}