| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2006 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Expected frequencies partially provided |
| Difficulty | Moderate -0.3 This is a straightforward chi-squared test of independence with all calculations provided. Students only need to verify one expected frequency calculation using the standard formula, check validity conditions (all expected frequencies > 5), state standard hypotheses, compare the given test statistic to critical value, and make a simple interpretation from the data table. No complex reasoning or novel problem-solving required—slightly easier than average due to the scaffolded structure. |
| Spec | 5.06a Chi-squared: contingency tables |
| Day | ||||
| Monday | Wednesday | Friday | ||
| \multirow{3}{*}{Volume} | Low | 15 | 13 | 2 |
| Medium | 23 | 26 | 23 | |
| High | 12 | 9 | 27 | |
| Day | ||||
| Monday | Wednesday | Friday | ||
| Low | 10.00 | 9.60 | 10.40 | |
| Volume | Medium | 24.00 | 23.04 | 24.96 |
| High | 16.00 | 15.36 | 16.64 | |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((48 \times 72/150)\) or \((48/150)(72/150) \times 150\) | M1 | Multiply and divide relevant values |
| A1 | 2 | All correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| No, no expected value less than 5 | B1 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_0\): Volume and day are independent | Attributes specified | |
| (\(H_1\): Volume and day are not independent) | B1 | |
| Critical value for 4 df \(= 13.28\) | B1 | |
| Test statistic \(> 13.28\), reject \(H_0\) | M1 | |
| Accept that volume and day are not independent | A1 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Choose Friday | B1 | |
| Highest volume | B1 | 2 |
# Question 5(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(48 \times 72/150)$ or $(48/150)(72/150) \times 150$ | M1 | Multiply and divide relevant values |
| | A1 | **2** | All correct |
---
# Question 5(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| No, no expected value less than 5 | B1 | **1** | |
---
# Question 5(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0$: Volume and day are independent | | Attributes specified |
| ($H_1$: Volume and day are not independent) | B1 | |
| Critical value for 4 df $= 13.28$ | B1 | |
| Test statistic $> 13.28$, reject $H_0$ | M1 | |
| Accept that volume and day are not independent | A1 | **4** | |
---
# Question 5(iv):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Choose Friday | B1 | |
| Highest volume | B1 | **2** | Not reference to E values |
---5 Gloria is a market trader who sells jeans. She trades on Mondays, Wednesdays and Fridays. Wishing to investigate whether the volume of trade depends on the day of the week, Gloria analysed a random sample of 150 days' sales and classified them by day and volume (low, medium and high). The results are given in the table below.
\begin{center}
\begin{tabular}{ | c l | c c c | }
\hline
& & \multicolumn{3}{|c|}{Day} \\
& & Monday & Wednesday & Friday \\
\hline
\multirow{3}{*}{Volume} & Low & 15 & 13 & 2 \\
& Medium & 23 & 26 & 23 \\
& High & 12 & 9 & 27 \\
\hline
\end{tabular}
\end{center}
Gloria asked a statistician to perform a suitable test of independence and, as part of this test, expected frequencies were calculated. These are shown in the table below.
\begin{center}
\begin{tabular}{ | c l | c c c | }
\hline
& & \multicolumn{3}{|c|}{Day} \\
& & Monday & Wednesday & Friday \\
\hline
& Low & 10.00 & 9.60 & 10.40 \\
Volume & Medium & 24.00 & 23.04 & 24.96 \\
& High & 16.00 & 15.36 & 16.64 \\
\hline
\end{tabular}
\end{center}
(i) Show how the value 23.04 for medium volume on Wednesday has been obtained.\\
(ii) State, giving a reason, if it is necessary to combine any rows or columns in order to carry out the test.
The value of the test statistic is found to be 21.15, correct to 2 decimal places.\\
(iii) Stating suitable hypotheses for the test, give its conclusion using a $1 \%$ significance level.
Gloria wishes to hold a sale and asks the statistician to advise her on which day to hold it in order to sell as much as possible.\\
(iv) State the day that the statistician should advise and give a reason for the choice.
\hfill \mbox{\textit{OCR S3 2006 Q5 [9]}}