| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2007 |
| 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 question with straightforward calculations. Part (i) involves routine hypothesis stating, showing an expected frequency calculation (simple row×column/total), computing chi-squared contributions using the standard formula, and looking up a critical value. Part (ii) is a goodness-of-fit test with given ratios. All techniques are textbook exercises requiring only methodical application of formulas with no novel insight or complex reasoning. |
| Spec | 5.06a Chi-squared: contingency tables |
| Eye colour | |||||
| Blue | Brown | Other | Total | ||
| None | 12 | 17 | 10 | 39 | |
| Reaction | Mild | 31 | 21 | 11 | 63 |
| Strong | 22 | 4 | 12 | 38 | |
| Total | 65 | 42 | 33 | 140 | |
| Eye colour | ||||
| Blue | Brown | Other | ||
| None | 18.11 | 11.70 | 9.19 | |
| Reaction | Mild | 29.25 | 18.90 | 14.85 |
| Strong | 17.64 | 11.40 | 8.96 | |
| Answer | Marks | Guidance |
|---|---|---|
| (i)(a) \(H_0:\) Eye colour and reaction are not associated. \(H_1:\) Eye colour and reaction are associated | B1 | Or equivalent (independent, or unrelated) |
| (b) \(65 \times 39/140\); \(65 \times 39/140\) | B1, B1 | 2; 1 |
| (c) \(6.11^2/18.11 + 5.3^2/11.7 + 0.81^2/9.19\); \(2.061 + 2.401 + 0.071\); \(4.533\), \(4.53\) AG | M1, A1, A1 | Or equivalent; one correct; At least 3 dp here; But accept from 2 dp |
| (d) \(\nu = 4\); Use tables to obtain \(\alpha = 2\frac{1}{2}\) | B1, B1 | Stated or implied; 2 |
| (ii) \(H_0: p_{BE} = p_{BR} = 0.4\), \(p_O = 0.2\) (\(H_1:\) At least two prob. not as above) | B1 | Or in words, in terms of probs or proportions |
| E values 56, 56, 28; \(\chi^2 = 9^2/56 + 14^2/56 + 8^2/28 = 5.839\) | M1A1, M1 | |
| Compare correctly with 5.991; Accept that sample is consistent with hypothesis. | A1 | Accept 5.84; M1A0 if 5.991 seen and consistent conclusion but no explicit comparison |
| SR: If three tests for \(p\) then count only \(p_{BE} = 0.4\); \((42/140 - 0.4)/\sigma\); \(\sigma = \sqrt{(0.4 \times 0.6/140)}\); \(-2.415\); Compare with \(-1.96\); conclusion in context | M1, A1A1, M1A1 | the; Max 6/7 (with \(H_0\)) |
**(i)(a)** $H_0:$ Eye colour and reaction are not associated. $H_1:$ Eye colour and reaction are associated | B1 | Or equivalent (independent, or unrelated)
**(b)** $65 \times 39/140$; $65 \times 39/140$ | B1, B1 | 2; 1
**(c)** $6.11^2/18.11 + 5.3^2/11.7 + 0.81^2/9.19$; $2.061 + 2.401 + 0.071$; $4.533$, $4.53$ AG | M1, A1, A1 | Or equivalent; one correct; At least 3 dp here; But accept from 2 dp
**(d)** $\nu = 4$; Use tables to obtain $\alpha = 2\frac{1}{2}$ | B1, B1 | Stated or implied; 2
**(ii)** $H_0: p_{BE} = p_{BR} = 0.4$, $p_O = 0.2$ ($H_1:$ At least two prob. not as above) | B1 | Or in words, in terms of probs or proportions
E values 56, 56, 28; $\chi^2 = 9^2/56 + 14^2/56 + 8^2/28 = 5.839$ | M1A1, M1 |
Compare correctly with 5.991; Accept that sample is consistent with hypothesis. | A1 | Accept 5.84; M1A0 if 5.991 seen and consistent conclusion but no explicit comparison
SR: If three tests for $p$ then count only $p_{BE} = 0.4$; $(42/140 - 0.4)/\sigma$; $\sigma = \sqrt{(0.4 \times 0.6/140)}$; $-2.415$; Compare with $-1.96$; conclusion in context | M1, A1A1, M1A1 | the; Max 6/7 (with $H_0$)7 It is thought that a person's eye colour is related to the reaction of the person's skin to ultra-violet light. As part of a study, a random sample of 140 people were treated with a standard dose of ultra-violet light. The degree of reaction was classified as None, Mild or Strong. The results are given in Table 1. The corresponding expected frequencies for a $\chi ^ { 2 }$ test of association between eye colour and reaction are shown in Table 2.
\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Table 1\\
Observed frequencies}
\begin{tabular}{ | l l | c c c | c | }
\hline
& & \multicolumn{3}{|c|}{Eye colour} & \\
& & Blue & Brown & Other & Total \\
\hline
& None & 12 & 17 & 10 & 39 \\
Reaction & Mild & 31 & 21 & 11 & 63 \\
& Strong & 22 & 4 & 12 & 38 \\
\hline
& Total & 65 & 42 & 33 & 140 \\
\hline
\end{tabular}
\end{center}
\end{table}
\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Table 2\\
Expected frequencies}
\begin{tabular}{ | l l | r r r | }
\hline
& & \multicolumn{3}{|c|}{Eye colour} \\
& & Blue & Brown & Other \\
\hline
& None & 18.11 & 11.70 & 9.19 \\
Reaction & Mild & 29.25 & 18.90 & 14.85 \\
& Strong & 17.64 & 11.40 & 8.96 \\
\hline
\end{tabular}
\end{center}
\end{table}
\begin{enumerate}[label=(\roman*)]
\item (a) State suitable hypotheses for the test.\\
(b) Show how the expected frequency of 18.11 in Table 2 is obtained.\\
(c) Show that the three cells in the top row together contribute 4.53 to the calculated value of $\chi ^ { 2 }$, correct to 2 decimal places.\\
(d) You are given that the total calculated value of $\chi ^ { 2 }$ is 12.78 , correct to 2 decimal places. Give the smallest value of $\alpha$ obtained from the tables for which the null hypothesis would be rejected at the $\alpha \%$ significance level.
\item Test, at the $5 \%$ significance level, whether the proportions of people in the whole population with blue eyes, brown eyes and other colours are in the ratios $2 : 2 : 1$.
\end{enumerate}
\hfill \mbox{\textit{OCR S3 2007 Q7 [15]}}