| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Cell combining required |
| Difficulty | Standard +0.3 This is a standard chi-squared test of independence with routine row combination. All steps are textbook procedures: stating hypotheses, checking expected frequencies (rule of 5), calculating a single cell contribution using the standard formula, and completing the test with given test statistic. The only mild challenge is explaining why combination is needed, but this is a well-rehearsed concept in S3. Slightly easier than average due to most calculations being provided. |
| Spec | 5.06a Chi-squared: contingency tables |
| Treatment \(A\) | Treatment \(B\) | Treatment \(C\) | |
| Much improved | 12 | 16 | 4 |
| Improved | 13 | 12 | 6 |
| Slightly improved | 7 | 6 | 7 |
| No change | 5 | 3 | 9 |
| Treatment \(A\) | Treatment \(B\) | Treatment \(C\) | |
| Much improved | 12 | 16 | 4 |
| Improved | 13 | 12 | 6 |
| Slightly improved/ No change | 12 | 9 | 16 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0\): no assoc between level of improvement and treatment; \(H_1\): there is an assoc between level of improvement and treatment. | B1 | oe |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{26 \times 17}{100} = 4.42\) | M1, A1 | |
| Expected value for NC, tr C \(< 5\) | A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{26 \times 37}{100}\ (= 9.62)\) | M1 | |
| \(\frac{(16 - \text{"9.62"})^2}{\text{"9.62"}}\) | M1 | |
| \(= 4.231\) AG | A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(10.51 > 9.488\), reject \(H_0\) | M1 | \(p < 0.05\) and reject \(H_0\) |
| There is sufficient evidence that there is an assoc between level of improvement and treatment. | A1 | Contextualised. \(p = 0.03266\) and conclusion. |
| [2] |
## Question 7:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0$: no assoc between level of improvement and treatment; $H_1$: there is an assoc between level of improvement and treatment. | B1 | oe |
| **[1]** | | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{26 \times 17}{100} = 4.42$ | M1, A1 | |
| Expected value for NC, tr C $< 5$ | A1 | |
| **[3]** | | |
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{26 \times 37}{100}\ (= 9.62)$ | M1 | |
| $\frac{(16 - \text{"9.62"})^2}{\text{"9.62"}}$ | M1 | |
| $= 4.231$ AG | A1 | |
| **[3]** | | |
### Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $10.51 > 9.488$, reject $H_0$ | M1 | $p < 0.05$ and reject $H_0$ |
| There is sufficient evidence that there is an assoc between level of improvement and treatment. | A1 | Contextualised. $p = 0.03266$ and conclusion. |
| **[2]** | | |
---7 A random sample of 100 adults with a chronic disease was chosen. Each adult was randomly assigned to one of three different treatments. After six months of treatment, each adult was then assessed and classified as 'much improved', 'improved', 'slightly improved' or 'no change'. The results are summarised in Table 1.
\begin{table}[h]
\begin{center}
\begin{tabular}{|l|l|l|l|}
\hline
& Treatment $A$ & Treatment $B$ & Treatment $C$ \\
\hline
Much improved & 12 & 16 & 4 \\
\hline
Improved & 13 & 12 & 6 \\
\hline
Slightly improved & 7 & 6 & 7 \\
\hline
No change & 5 & 3 & 9 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Table 1}
\end{center}
\end{table}
A $\chi ^ { 2 }$ test, at the $5 \%$ significance level, is to be carried out.\\
(i) State suitable hypotheses.
Combining the last two rows of Table 1 gives Table 2.
\begin{table}[h]
\begin{center}
\begin{tabular}{|l|l|l|l|}
\hline
& Treatment $A$ & Treatment $B$ & Treatment $C$ \\
\hline
Much improved & 12 & 16 & 4 \\
\hline
Improved & 13 & 12 & 6 \\
\hline
Slightly improved/ No change & 12 & 9 & 16 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Table 2}
\end{center}
\end{table}
(ii) By considering the expected frequencies for Treatment $C$ in Table 1, explain why it was necessary to combine rows.\\
(iii) Show that the contribution to the $\chi ^ { 2 }$ value for the cell 'slightly improved/no change, Treatment $C$ ' is 4.231 , correct to 3 decimal places.
You are given that the $\chi ^ { 2 }$ test statistic is 10.51 , correct to 2 decimal places.\\
(iv) Carry out the test.
\hfill \mbox{\textit{OCR S3 2014 Q7 [9]}}