| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2014 |
| Session | June |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Standard 2×2 contingency table |
| Difficulty | Standard +0.3 This is a standard two-part question covering routine hypothesis testing procedures. Part (a) is a straightforward paired t-test with clear hypotheses and data provided; part (b) is a textbook chi-squared goodness of fit test for Poisson distribution with the mean already given. Both require only standard application of learned procedures with no novel insight or complex multi-step reasoning. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.05c Hypothesis test: normal distribution for population mean |
| Trainee | Water | Beetroot juice |
| A | 75.1 | 72.9 |
| B | 86.2 | 79.9 |
| C | 77.3 | 71.6 |
| D | 89.1 | 90.2 |
| E | 67.9 | 68.2 |
| F | 101.5 | 95.2 |
| G | 82.5 | 76.5 |
| H | 83.3 | 80.2 |
| I | 102.5 | 99.1 |
| J | 91.3 | 82.2 |
| K | 92.5 | 90.1 |
| L | 77.2 | 77.9 |
| Number of birds | 0 | 1 | 2 | 3 | 4 | 5 | 6 | \(\geqslant 7\) |
| Frequency | 2 | 5 | 10 | 17 | 14 | 7 | 4 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0: \mu_D = 0 \quad H_1: \mu_D < 0\) | ||
| Where \(\mu_D\) is the population mean of the differences. MUST BE PAIRED COMPARISON \(t\) test. | M1 | |
| \(\bar{D} = -3.533 \quad s_D = 3.225\) | A1 | Do not allow \(s_n = 3.088\) |
| Test statistic is \(\dfrac{-3.533 - 0}{3.225/\sqrt{12}}\) | M1* | For method, allow their \(\bar{D}, s\). Allow confidence interval approach. |
| \(= -3.79\) | A1 | cao |
| Refer to \(t_{11}\) | M1 | No FT from here if wrong |
| Single-tailed 1% point is \(-2.718\) | A1 | No FT from here if wrong |
| \(-3.79 < -2.718\) so reject \(H_0\) | A1 dep | FT their \(-3.79\) if relevant M1 earned |
| Conclude mean time appears reduced | A1 dep | In context. FT their \(-3.79\) if relevant M1 earned |
| [8] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0\): The Poisson model fits the data | B1 | Both hypotheses. Must be the right way round. Do not accept "data fits model" or equivalent. |
| \(H_1\): The Poisson model does not fit the data | ||
| At least 3 probabilities to 3dp or 3 expected values to 3sf | M1 | |
| Multiply by 60 to obtain expected values | M1 | |
| All correct to 3sf or better | A1 | |
| Merge first 2 and last 2 cells | M1 | |
| \(X^2 = \dfrac{2.516^2}{9.516} + \dfrac{2.050^2}{12.050} + \dfrac{3.745^2}{13.255} + \dfrac{3.065^2}{10.935} + \dfrac{0.217^2}{7.217} + \dfrac{2.027^2}{7.027}\) | M1* | Calculation of \(X^2\) |
| \(= 3.522\) awrt 3.52 | A1 | cao |
| Refer to \(\chi^2_4\) | M1 | Allow correct df from wrongly grouped table |
| Upper 5% point is 9.49 | B1 | No FT from here if wrong |
| \(3.522 < 9.49\) cannot reject \(H_0\) | A1 dep | FT candidates 3.522 if relevant M1 earned |
| Poisson model appears to fit data. | A1 dep | FT candidates 3.522 if relevant M1 earned. Do not accept "data fits Poisson model" or equivalent. |
| [11] |
# Question 3:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \mu_D = 0 \quad H_1: \mu_D < 0$ | | |
| Where $\mu_D$ is the population mean of the differences. MUST BE PAIRED COMPARISON $t$ test. | M1 | |
| $\bar{D} = -3.533 \quad s_D = 3.225$ | A1 | Do not allow $s_n = 3.088$ |
| Test statistic is $\dfrac{-3.533 - 0}{3.225/\sqrt{12}}$ | M1* | For method, allow their $\bar{D}, s$. Allow confidence interval approach. |
| $= -3.79$ | A1 | cao |
| Refer to $t_{11}$ | M1 | No FT from here if wrong |
| Single-tailed 1% point is $-2.718$ | A1 | No FT from here if wrong |
| $-3.79 < -2.718$ so reject $H_0$ | A1 dep | FT their $-3.79$ if relevant M1 earned |
| Conclude mean time appears reduced | A1 dep | In context. FT their $-3.79$ if relevant M1 earned |
| **[8]** | | |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0$: The Poisson model fits the data | B1 | Both hypotheses. Must be the right way round. Do not accept "data fits model" or equivalent. |
| $H_1$: The Poisson model does not fit the data | | |
| At least 3 probabilities to 3dp or 3 expected values to 3sf | M1 | |
| Multiply by 60 to obtain expected values | M1 | |
| All correct to 3sf or better | A1 | |
| Merge first 2 and last 2 cells | M1 | |
| $X^2 = \dfrac{2.516^2}{9.516} + \dfrac{2.050^2}{12.050} + \dfrac{3.745^2}{13.255} + \dfrac{3.065^2}{10.935} + \dfrac{0.217^2}{7.217} + \dfrac{2.027^2}{7.027}$ | M1* | Calculation of $X^2$ |
| $= 3.522$ awrt 3.52 | A1 | cao |
| Refer to $\chi^2_4$ | M1 | Allow correct df from wrongly grouped table |
| Upper 5% point is 9.49 | B1 | No FT from here if wrong |
| $3.522 < 9.49$ cannot reject $H_0$ | A1 dep | FT candidates 3.522 if relevant M1 earned |
| Poisson model appears to fit data. | A1 dep | FT candidates 3.522 if relevant M1 earned. Do not accept "data fits Poisson model" or equivalent. |
| **[11]** | | |3
\begin{enumerate}[label=(\alph*)]
\item A personal trainer believes that drinking a glass of beetroot juice an hour before exercising enables endurance tests to be completed more quickly. To test his belief he takes a random sample of 12 of his trainees and, on two occasions, asks them to carry out 100 repetitions of a particular exercise as quickly as possible. Each trainee drinks a glass of water on one occasion and a glass of beetroot juice on the other occasion.
The times in seconds taken by the trainees are given in the table.
\begin{center}
\begin{tabular}{|l|l|l|}
\hline
Trainee & Water & Beetroot juice \\
\hline
A & 75.1 & 72.9 \\
\hline
B & 86.2 & 79.9 \\
\hline
C & 77.3 & 71.6 \\
\hline
D & 89.1 & 90.2 \\
\hline
E & 67.9 & 68.2 \\
\hline
F & 101.5 & 95.2 \\
\hline
G & 82.5 & 76.5 \\
\hline
H & 83.3 & 80.2 \\
\hline
I & 102.5 & 99.1 \\
\hline
J & 91.3 & 82.2 \\
\hline
K & 92.5 & 90.1 \\
\hline
L & 77.2 & 77.9 \\
\hline
\end{tabular}
\end{center}
The trainer wishes to test his belief using a paired $t$ test at the $1 \%$ level of significance. Assuming any necessary assumptions are valid, carry out a test of the hypotheses $\mathrm { H } _ { 0 } : \mu _ { D } = 0 , \mathrm { H } _ { 1 } : \mu _ { D } < 0$, where $\mu _ { D }$ is the population mean difference in times (time with beetroot juice minus time with water).
\item An ornithologist believes that the number of birds landing on the bird feeding station in her garden in a given interval of time during the morning should follow a Poisson distribution. In order to test her belief, she makes the following observations in 60 randomly chosen minutes one morning.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | c | }
\hline
Number of birds & 0 & 1 & 2 & 3 & 4 & 5 & 6 & $\geqslant 7$ \\
\hline
Frequency & 2 & 5 & 10 & 17 & 14 & 7 & 4 & 1 \\
\hline
\end{tabular}
\end{center}
Given that the data in the table have a mean value of 3.3, use a goodness of fit test, with a significance level of $5 \%$, to investigate whether the ornithologist is justified in her belief.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S3 2014 Q3 [19]}}