OCR S3 2008 January — Question 6 15 marks

Exam BoardOCR
ModuleS3 (Statistics 3)
Year2008
SessionJanuary
Marks15
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicChi-squared test of independence
TypeExpected frequencies partially provided
DifficultyStandard +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.
Spec5.06a Chi-squared: contingency tables

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]
Shade
\cline { 2 - 5 }\(G _ { 1 }\)\(G _ { 2 }\)\(G _ { 3 }\)
\cline { 2 - 5 } GenderMale112423
Female181331
\cline { 2 - 5 }
\cline { 2 - 5 }
\captionsetup{labelformat=empty} \caption{Table 1}
\end{table} Table 2 shows the corresponding expected values, correct to 2 decimal places, for a test of independence. \begin{table}[h]
Shade
\cline { 2 - 5 }\(G _ { 1 }\)\(G _ { 2 }\)\(G _ { 3 }\)
\cline { 2 - 5 } GenderMale14.0217.8826.10
Female14.9819.1227.90
\cline { 2 - 5 }
\cline { 2 - 5 }
\captionsetup{labelformat=empty} \caption{Table 2}
\end{table}
  1. Show how the value 17.88 for Male, \(G _ { 2 }\) was obtained.
  2. Test, at the \(5 \%\) significance level, whether gender and preferred shade are independent.
  3. 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.

Part (i)
\(\frac{37 \times 58}{120}\)
\(17.883..., 17.88\) AG
AnswerMarks Guidance
AnswerMarks 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
AnswerMarks Guidance
AnswerMarks Guidance
M1 A1
A1 B1 M1 A1N 7Ft \(\chi^2\). Can be assertive.
Part (iii)
AnswerMarks Guidance
G₁G₂ G₃
O29 37
E40 40
\(\frac{121}{40} + \frac{9}{40}+\frac{196}{40} = 8.15\)
Using df = 2
2.5% tables, 1.7% calculator
AnswerMarks Guidance
AnswerMarks 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 |

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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]}}