| Exam Board | OCR MEI |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2009 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Standard 3×3 contingency table |
| Difficulty | Standard +0.3 This is a straightforward two-part question testing standard chi-squared independence test and one-tailed z-test procedures. Part (i) requires calculating expected frequencies, computing chi-squared statistic with contributions table, and comparing to critical value—all routine S2 content. Part (ii) is a textbook normal test with known variance. Both parts follow standard algorithms with no conceptual challenges or novel problem-solving required, making this slightly easier than average. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.06a Chi-squared: contingency tables |
| \multirow{2}{*}{} | Growth | \multirow[t]{2}{*}{Row totals} | |||
| Good | Average | Poor | |||
| \multirow{3}{*}{Type of plant} | Coriander | 12 | 28 | 15 | 55 |
| Aster | 7 | 18 | 23 | 48 | |
| Fennel | 14 | 22 | 11 | 47 | |
| Column totals | 33 | 68 | 49 | 150 | |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0\): no association between growth and type of plant; \(H_1\): some association between growth and type of plant | B1 | In context |
| Expected values: Coriander: 12.10, 24.93, 17.97; Aster: 10.56, 21.76, 15.68; Fennel: 10.34, 21.31, 15.35 | M1 A2 | For expected values to 2 dp; allow A1 for at least one row or column correct |
| Contributions \((O-E)^2/E\): Coriander: 0.0008, 0.3772, 0.4899; Aster: 1.2002, 0.6497, 3.4172; Fennel: 1.2955, 0.0226, 1.2344 | M1 | For valid attempt at \((O-E)^2/E\) |
| A1 | For all correct; NB these M1A1 marks cannot be implied by a correct final value of \(X^2\) | |
| \(X^2 = 8.69\) | M1 | For summation |
| A1 | For \(X^2\) CAO | |
| Refer to \(\chi^2_4\) | B1 | For 4 d.o.f. |
| Critical value at 5% level \(= 9.488\) | B1 CAO | For cv |
| Result is not significant; there is not enough evidence to suggest that there is some association between reported growth and type of plant | M1 A1 | NB if \(H_0\), \(H_1\) reversed, or 'correlation' mentioned, do not award first B1 or final A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Test statistic \(= \frac{49.2 - 47}{8.5/\sqrt{50}} = \frac{2.2}{1.202} = 1.830\) | M1 | Correct denominator |
| A1 | ||
| 1% level 1-tailed critical value of \(z = 2.326\) | B1 | For 2.326 |
| \(1.830 < 2.326\) so not significant; there is not sufficient evidence to reject \(H_0\) | M1 | Dep on first M1 for sensible comparison leading to a conclusion |
| There is insufficient evidence to conclude that the flowers are larger | A1 | For fully correct conclusion in words in context |
# Question 4:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0$: no association between growth and type of plant; $H_1$: some association between growth and type of plant | B1 | In context |
| Expected values: Coriander: 12.10, 24.93, 17.97; Aster: 10.56, 21.76, 15.68; Fennel: 10.34, 21.31, 15.35 | M1 A2 | For expected values to 2 dp; allow A1 for at least one row or column correct |
| Contributions $(O-E)^2/E$: Coriander: 0.0008, 0.3772, 0.4899; Aster: 1.2002, 0.6497, 3.4172; Fennel: 1.2955, 0.0226, 1.2344 | M1 | For valid attempt at $(O-E)^2/E$ |
| | A1 | For all correct; NB these M1A1 marks cannot be implied by a correct final value of $X^2$ |
| $X^2 = 8.69$ | M1 | For summation |
| | A1 | For $X^2$ CAO |
| Refer to $\chi^2_4$ | B1 | For 4 d.o.f. |
| Critical value at 5% level $= 9.488$ | B1 CAO | For cv |
| Result is not significant; there is not enough evidence to suggest that there is some association between reported growth and type of plant | M1 A1 | NB if $H_0$, $H_1$ reversed, or 'correlation' mentioned, do not award first B1 or final A1 |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Test statistic $= \frac{49.2 - 47}{8.5/\sqrt{50}} = \frac{2.2}{1.202} = 1.830$ | M1 | Correct denominator |
| | A1 | |
| 1% level 1-tailed critical value of $z = 2.326$ | B1 | For 2.326 |
| $1.830 < 2.326$ so not significant; there is not sufficient evidence to reject $H_0$ | M1 | Dep on first M1 for sensible comparison leading to a conclusion |
| There is insufficient evidence to conclude that the flowers are larger | A1 | For fully correct conclusion in words in context |4 A gardening research organisation is running a trial to examine the growth and the size of flowers of various plants.\\
(i) In the trial, seeds of three types of plant are sown. The growth of each plant is classified as good, average or poor. The results are shown in the table.
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|}
\hline
\multicolumn{2}{|c|}{\multirow{2}{*}{}} & \multicolumn{3}{|c|}{Growth} & \multirow[t]{2}{*}{Row totals} \\
\hline
& & Good & Average & Poor & \\
\hline
\multirow{3}{*}{Type of plant} & Coriander & 12 & 28 & 15 & 55 \\
\hline
& Aster & 7 & 18 & 23 & 48 \\
\hline
& Fennel & 14 & 22 & 11 & 47 \\
\hline
\multicolumn{2}{|c|}{Column totals} & 33 & 68 & 49 & 150 \\
\hline
\end{tabular}
\end{center}
Carry out a test at the $5 \%$ significance level to examine whether there is any association between growth and type of plant. State carefully your null and alternative hypotheses. Include a table of the contributions of each cell to the test statistic.\\
(ii) It is known that the diameter of marigold flowers is Normally distributed with mean 47 mm and standard deviation 8.5 mm . A certain fertiliser is expected to cause flowers to have a larger mean diameter, but without affecting the standard deviation. A large number of marigolds are grown using this fertiliser. The diameters of a random sample of 50 of the flowers are measured and the mean diameter is found to be 49.2 mm . Carry out a hypothesis test at the $1 \%$ significance level to check whether flowers grown with this fertiliser appear to be larger on average. Use hypotheses $\mathrm { H } _ { 0 } : \mu = 47 , \mathrm { H } _ { 1 } : \mu > 47$, where $\mu \mathrm { mm }$ represents the mean diameter of all marigold flowers grown with this fertiliser.
\hfill \mbox{\textit{OCR MEI S2 2009 Q4 [17]}}