| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2013 |
| Session | June |
| Marks | 12 |
| 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: showing an expected frequency calculation (row total × column total / grand total), stating degrees of freedom (r-1)(c-1), and comparing a given test statistic to critical values. All steps are routine S3 procedures requiring only recall and basic application, making it slightly easier than average. |
| Spec | 5.06a Chi-squared: contingency tables |
| \multirow{2}{*}{} | Hair colour | |||||
| Black | Brown | Red | Blonde | Total | ||
| \multirow{5}{*}{Eye colour} | Brown | 45 | 125 | 15 | 58 | 243 |
| Blue | 34 | 90 | 10 | 58 | 192 | |
| Hazel | 20 | 38 | 16 | 26 | 100 | |
| Green | 6 | 29 | 7 | 23 | 65 | |
| Total | 105 | 282 | 48 | 165 | 600 | |
| \multirow{2}{*}{} | Hair colour | ||||
| Black | Brown | Red | Blonde | ||
| \multirow{4}{*}{Eye colour} | Brown | 42.5 | 114.2 | 19.4 | 66.8 |
| Blue | 33.6 | 90.2 | 15.4 | 52.8 | |
| Hazel | 17.5 | 47 | 8 | 27.5 | |
| Green | 11.4 | 30.6 | 5.2 | 17.9 | |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\frac{282 \times 100}{600}\) | B1 | (Do not accept \(282 - 114.2 - 90.2 - 30.6\) (o.e.)) |
| (b) 9 | B1 | (1) |
| (c) 2.5 or better | B1 | (Do not accept 0.025) |
| (d) \(H_0\): hair colour occurs in the ratio 2:6:1:3 | B1 | |
| \(H_1\): hair colour does not occur in the ratio 2:6:1:3 | ||
| black | brown | red |
| observed | 105 | 282 |
| expected | 100 | 300 |
| \((O_i - E_i)^2\) | 0.25 | 1.08 |
| \(E_i\) | ||
| B1 expected M1 | ||
| \(O_i^2\) | 110.25 | 265.08 |
| \(E_i\) | ||
| A1 | ||
| \(\sum \frac{(O_i - E_i)^2}{E_i} = 2.91\) or \(\sum \frac{O_i^2}{E_i} - 600 = 602.91 - 600 = 2.91\) | A1 | |
| (awrt 2.91) | ||
| \(\nu = 3\) | B1 | |
| cv is 7.815 | B1 | |
| [2.91 < 7.815] so insufficient evidence to reject \(H_0\) or not significant | dM1 | |
| There is evidence to suggest that hair colour does occur in the given ratio. | A1 | (9) |
**(a)** $\frac{282 \times 100}{600}$ | B1 | (Do not accept $282 - 114.2 - 90.2 - 30.6$ (o.e.))
**(b)** 9 | B1 | (1)
**(c)** 2.5 or better | B1 | (Do not accept 0.025) | (1)
**(d)** $H_0$: hair colour occurs in the ratio 2:6:1:3 | B1 |
$H_1$: hair colour does not occur in the ratio 2:6:1:3 | |
| | black | brown | red | blonde |
|---|---|---|---|---|
| observed | 105 | 282 | 48 | 165 |
| expected | 100 | 300 | 50 | 150 |
| $(O_i - E_i)^2$ | 0.25 | 1.08 | 0.08 | 1.5 |
| $E_i$ | | | | |
| B1 expected M1 |
| $O_i^2$ | 110.25 | 265.08 | 46.08 | 181.5 |
| $E_i$ | | | | |
| A1 |
$\sum \frac{(O_i - E_i)^2}{E_i} = 2.91$ or $\sum \frac{O_i^2}{E_i} - 600 = 602.91 - 600 = 2.91$ | A1 |
(awrt **2.91**) | |
$\nu = 3$ | B1 |
cv is 7.815 | B1 |
[2.91 < 7.815] so insufficient evidence to reject $H_0$ or not significant | dM1 |
There is evidence to suggest that hair colour does occur in the given ratio. | A1 | (9)
**[Total 12]**
**Notes:**
**(d)** 1st B1 for both hypotheses. Must mention hair colour and ratio
e.g. "hair colour in the given ratio" Allow use of ditto
2nd B1 for all 4 correct expected frequencies
1st M1 for at least 2 correct calculations from 3rd or 4th row
1st A1 for all correct calculations to at least 3sf if row 4. If awrt 2.91 is seen with no incorrect working award B1M1A1A1
2nd dM1 Dep on 1st M1 for a correct statement linking their test statistic and their cv (cv > 3.5)
3rd A1 for a correct comment in context - must mention "hair colour" and "ratios" or "model". e.g. "There is evidence of to support the given model" No follow through. If hypotheses are the wrong way round score A0.
---\begin{enumerate}
\item John thinks that a person's eye colour is related to their hair colour. He takes a random sample of 600 people and records their eye and hair colours. The results are shown in Table 1.
\end{enumerate}
\begin{table}[h]
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline
\multicolumn{2}{|c|}{\multirow{2}{*}{}} & \multicolumn{5}{|c|}{Hair colour} \\
\hline
& & Black & Brown & Red & Blonde & Total \\
\hline
\multirow{5}{*}{Eye colour} & Brown & 45 & 125 & 15 & 58 & 243 \\
\hline
& Blue & 34 & 90 & 10 & 58 & 192 \\
\hline
& Hazel & 20 & 38 & 16 & 26 & 100 \\
\hline
& Green & 6 & 29 & 7 & 23 & 65 \\
\hline
& Total & 105 & 282 & 48 & 165 & 600 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Table 1}
\end{center}
\end{table}
John carries out a $\chi ^ { 2 }$ test in order to test whether eye colour and hair colour are related. He calculates the expected frequencies shown in Table 2.
\begin{table}[h]
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|}
\hline
\multicolumn{2}{|c|}{\multirow{2}{*}{}} & \multicolumn{4}{|c|}{Hair colour} \\
\hline
& & Black & Brown & Red & Blonde \\
\hline
\multirow{4}{*}{Eye colour} & Brown & 42.5 & 114.2 & 19.4 & 66.8 \\
\hline
& Blue & 33.6 & 90.2 & 15.4 & 52.8 \\
\hline
& Hazel & 17.5 & 47 & 8 & 27.5 \\
\hline
& Green & 11.4 & 30.6 & 5.2 & 17.9 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Table 2}
\end{center}
\end{table}
(a) Show how the value 47 in Table 2 has been calculated.\\
(b) Write down the number of degrees of freedom John should use in this $\chi ^ { 2 }$ test.
Given that the value of the $\chi ^ { 2 }$ statistic is 20.6 , to 3 significant figures,\\
(c) find the smallest value of $\alpha$ for which the null hypothesis will be rejected at the $\alpha \%$ level of significance.\\
(d) Use the data from Table 1 to test at the $5 \%$ level of significance whether or not the proportions of people in the population with black, brown, red and blonde hair are in the ratio 2:6:1:3 State your hypotheses clearly.
\hfill \mbox{\textit{Edexcel S3 2013 Q4 [12]}}