| Exam Board | OCR MEI |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2006 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Test statistic given, complete the test |
| Difficulty | Standard +0.3 This is a straightforward chi-squared test of independence with standard hypothesis writing, expected frequency calculation, and critical value comparison, plus a basic z-test. All procedures are routine textbook applications requiring no novel insight, though the multi-part structure and inclusion of two different tests makes it slightly above average difficulty. |
| Spec | 5.06a Chi-squared: contingency tables |
| Primary | Secondary | ||||
\multirow{3}{*}{
| Bus | 21 | 49 | ||
| \cline { 2 - 4 } | Car | 65 | 15 | ||
| \cline { 2 - 4 } | Cycle or Walk | 34 | 16 | ||
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0\): no association between method of travel and type of school | B1 | for both |
| \(H_1\): some association between method of travel and type of school | 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Expected frequency \(= \frac{120}{200} \times 70 = 42\) | M1 A1 | |
| Contribution \(= \frac{(21-42)^2}{42}\) | M1 | for valid attempt at \((O-E)^2/E\) |
| \(= 10.5\) | A1 FT | their 42 provided \(O = 21\) (at least 1 dp) 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(X^2 = 42.64\) | ||
| Refer to \(\chi^2_2\) | B1 | for 2 deg of f (seen) |
| Critical value at 5% level \(= 5.991\) | B1 | CAO for cv |
| Result is significant | B1 | for significant (FT their c.v. provided consistent with their d.o.f.) |
| There appears to be some association between method of travel and year group. NB if \(H_0\ H_1\) reversed, or 'correlation' mentioned, do not award first B1 or final E1 | E1 | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: \mu = 18.3\); \(H_1: \mu \neq 18.3\) | B1 | for both correct |
| Where \(\mu\) denotes the mean travel time by car for the whole population | B1 | for definition of \(\mu\) |
| Test statistic \(z = \frac{22.4 - 18.3}{8.0/\sqrt{20}} = \frac{4.1}{1.789}\) | M1 | standardizing sample mean |
| \(= 2.292\) | A1 | for test statistic |
| 10% level 2 tailed critical value of \(z\) is 1.645 | B1 | for 1.645 |
| \(2.292 > 1.645\) so significant | M1 | for comparison leading to a conclusion |
| There is evidence to reject \(H_0\); it is reasonable to conclude that the mean travel time by car is different from that by bus. | A1 | for conclusion in words and context 7 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Any two of the following: | E1, E1 | for any two valid comments |
| • The test suggests that students who travel by bus get to school more quickly. | ||
| • This may be due to their journeys being over a shorter distance. | ||
| • It may be due to bus lanes allowing buses to avoid congestion. | ||
| • It is possible that the test result was incorrect (ie implication of a Type I error). | ||
| • More investigation is needed before any firm conclusion can be reached. | 2 marks |
# Question 4:
## Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0$: no association between method of travel and type of school | B1 | for both |
| $H_1$: some association between method of travel and type of school | | **1 mark** |
## Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Expected frequency $= \frac{120}{200} \times 70 = 42$ | M1 A1 | |
| Contribution $= \frac{(21-42)^2}{42}$ | M1 | for valid attempt at $(O-E)^2/E$ |
| $= 10.5$ | A1 FT | their 42 provided $O = 21$ (at least 1 dp) **4 marks** |
## Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $X^2 = 42.64$ | | |
| Refer to $\chi^2_2$ | B1 | for 2 deg of f (seen) |
| Critical value at 5% level $= 5.991$ | B1 | CAO for cv |
| Result is significant | B1 | for significant (FT their c.v. provided consistent with their d.o.f.) |
| There appears to be some association between method of travel and year group. NB if $H_0\ H_1$ reversed, or 'correlation' mentioned, do not award first B1 or final E1 | E1 | **4 marks** |
## Part (iv):
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \mu = 18.3$; $H_1: \mu \neq 18.3$ | B1 | for both correct |
| Where $\mu$ denotes the mean travel time by car for the whole population | B1 | for definition of $\mu$ |
| Test statistic $z = \frac{22.4 - 18.3}{8.0/\sqrt{20}} = \frac{4.1}{1.789}$ | M1 | standardizing sample mean |
| $= 2.292$ | A1 | for test statistic |
| 10% level 2 tailed critical value of $z$ is 1.645 | B1 | for 1.645 |
| $2.292 > 1.645$ so significant | M1 | for comparison leading to a conclusion |
| There is evidence to reject $H_0$; it is reasonable to conclude that the mean travel time by car is different from that by bus. | A1 | for conclusion in words and context **7 marks** |
## Part (v):
| Answer | Mark | Guidance |
|--------|------|----------|
| Any two of the following: | E1, E1 | for any two valid comments |
| • The test suggests that students who travel by bus get to school more quickly. | | |
| • This may be due to their journeys being over a shorter distance. | | |
| • It may be due to bus lanes allowing buses to avoid congestion. | | |
| • It is possible that the test result was incorrect (ie implication of a Type I error). | | |
| • More investigation is needed before any firm conclusion can be reached. | | **2 marks** |4 The table summarises the usual method of travelling to school for 200 randomly selected pupils from primary and secondary schools in a city.
\begin{center}
\begin{tabular}{ | c | l | c | c | }
\hline
\multicolumn{2}{|c|}{} & Primary & Secondary \\
\hline
\multirow{3}{*}{\begin{tabular}{ c }
Method of \\
travel \\
\end{tabular}} & Bus & 21 & 49 \\
\cline { 2 - 4 }
& Car & 65 & 15 \\
\cline { 2 - 4 }
& Cycle or Walk & 34 & 16 \\
\hline
\end{tabular}
\end{center}
(i) Write down null and alternative hypotheses for a test to examine whether there is any association between method of travel and type of school.\\
(ii) Calculate the expected frequency for primary school bus users. Calculate also the corresponding contribution to the test statistic for the usual $\chi ^ { 2 }$ test.\\
(iii) Given that the value of the test statistic for the usual $\chi ^ { 2 }$ test is 42.64 , carry out the test at the $5 \%$ level of significance, stating your conclusion clearly.
The mean travel time for pupils who travel by bus is known to be 18.3 minutes. A survey is carried out to determine whether the mean travel time to school by car is different from 18.3 minutes. In the survey, 20 pupils who travel by car are selected at random. Their mean travel time is found to be 22.4 minutes.\\
(iv) Assuming that car travel times are Normally distributed with standard deviation 8.0 minutes, carry out a test at the $10 \%$ level, stating your hypotheses and conclusion clearly.\\
(v) Comment on the suggestion that pupils should use a bus if they want to get to school quickly.
\hfill \mbox{\textit{OCR MEI S2 2006 Q4 [18]}}