| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2008 |
| Session | January |
| Marks | 15 |
| 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 standard chi-squared test of independence with straightforward calculations. Part (i) is routine expected value calculation, part (ii) follows the standard test procedure with given expected values, and part (iii) requires a simple goodness-of-fit test. All steps are textbook applications with no novel insight required, making it slightly easier than average. |
| Spec | 5.06a Chi-squared: contingency tables |
| Shade | ||||
| \cline { 2 - 5 } | \(G _ { 1 }\) | \(G _ { 2 }\) | \(G _ { 3 }\) | |
| \cline { 2 - 5 } Gender | Male | 11 | 24 | 23 |
| Female | 18 | 13 | 31 | |
| \cline { 2 - 5 } | ||||
| \cline { 2 - 5 } | ||||
| Shade | ||||
| \cline { 2 - 5 } | \(G _ { 1 }\) | \(G _ { 2 }\) | \(G _ { 3 }\) | |
| \cline { 2 - 5 } Gender | Male | 14.02 | 17.88 | 26.10 |
| Female | 14.98 | 19.12 | 27.90 | |
| \cline { 2 - 5 } | ||||
| \cline { 2 - 5 } | ||||
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| M1 A1 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| M1 A1 | ||
| A1 B1 M1 A1N 7 | Ft \(\chi^2\). Can be assertive. |
| Answer | Marks | Guidance |
|---|---|---|
| G₁ | G₂ | G₃ |
| O | 29 | 37 |
| E | 40 | 40 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| M1 A1 M1 A1 M1 | ||
| A1 6 (15) | For combining |
## Part (i)
$\frac{37 \times 58}{120}$
$17.883..., 17.88$ AG
| Answer | Marks | Guidance |
|--------|-------|----------|
| | M1 A1 **2** | |
## Part (ii)
$H_0$: Gender and shade are independent
$H_1$: are not independent
$3.02^2(14.02^{-1}+14.98^{-1}) + 6.12^2(17.88^{-1}+19.12^{-1}) + 3.1^2(26.1^{-1}+27.9^{-1})$
$= 6.03$
EITHER: CV 5.991
$6.03 > 5.991$, reject $H_0$ and accept that gender and shade are not independent
OR: $P(\chi^2 > 6.03) = 0.049 < 0.05$, reject $H_0$ and accept that gender and shade are not independent
| Answer | Marks | Guidance |
|--------|-------|----------|
| | M1 A1 | |
| | A1 B1 M1 A1N **7** | Ft $\chi^2$. Can be assertive. |
## Part (iii)
| | G₁ | G₂ | G₃ |
|---|----|----|-----|
| O | 29 | 37 | 54 |
| E | 40 | 40 | 40 |
$\frac{121}{40} + \frac{9}{40}+\frac{196}{40} = 8.15$
Using df = 2
2.5% tables, 1.7% calculator
| Answer | Marks | Guidance |
|--------|-------|----------|
| | M1 A1 M1 A1 M1 | |
| | A1 **6 (15)** | For combining |
---
6 The Research and Development department of a paint manufacturer has produced paint of three different shades of grey, $G _ { 1 } , G _ { 2 }$ and $G _ { 3 }$. In order to find the reaction of the public to these shades, each of a random sample of 120 people was asked to state which shade they preferred. The results, classified by gender, are shown in Table 1.
\begin{table}[h]
\begin{center}
\begin{tabular}{ l | l | c | c | c | }
\multicolumn{5}{c}{Shade} \\
\cline { 2 - 5 }
& & $G _ { 1 }$ & $G _ { 2 }$ & $G _ { 3 }$ \\
\cline { 2 - 5 }
Gender & Male & 11 & 24 & 23 \\
Female & 18 & 13 & 31 & \\
\cline { 2 - 5 }
& & & & \\
\cline { 2 - 5 }
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Table 1}
\end{center}
\end{table}
Table 2 shows the corresponding expected values, correct to 2 decimal places, for a test of independence.
\begin{table}[h]
\begin{center}
\begin{tabular}{ l | l | c | c | c | }
\multicolumn{5}{c}{Shade} \\
\cline { 2 - 5 }
& & $G _ { 1 }$ & $G _ { 2 }$ & $G _ { 3 }$ \\
\cline { 2 - 5 }
Gender & Male & 14.02 & 17.88 & 26.10 \\
& Female & 14.98 & 19.12 & 27.90 \\
\cline { 2 - 5 }
& & & & \\
\cline { 2 - 5 }
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Table 2}
\end{center}
\end{table}
(i) Show how the value 17.88 for Male, $G _ { 2 }$ was obtained.\\
(ii) Test, at the $5 \%$ significance level, whether gender and preferred shade are independent.\\
(iii) Determine the smallest significance level obtained from tables or calculator for which there is evidence that not all shades are equally preferred by people in general, irrespective of gender.
\hfill \mbox{\textit{OCR S3 2008 Q6 [15]}}