| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2010 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Wilcoxon tests |
| Type | Paired t-test |
| Difficulty | Standard +0.3 This is a straightforward paired t-test application with clear data and standard procedures. Part (i) requires stating normality assumption, calculating differences, and performing a one-tailed t-test—all routine S3 techniques. Part (ii) involves working backwards from a confidence interval to find sample statistics using the standard formula, which is slightly less routine but still mechanical. The context is accessible and the calculations are standard, making this slightly easier than average for S3. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Patient | A | B | C | D | E | F | G | H | I | J | K | L |
| Before | 5.7 | 5.7 | 4.0 | 6.8 | 7.4 | 5.5 | 6.7 | 6.4 | 7.2 | 7.2 | 7.1 | 4.4 |
| After | 5.8 | 4.0 | 5.2 | 5.7 | 6.0 | 5.0 | 5.8 | 4.2 | 7.3 | 5.2 | 6.4 | 4.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Must assume: Normality of population; of differences | B1, B1 | |
| \(H_0: \mu_D = 0\); \(H_1: \mu_D > 0\) | B1 | Both. Accept alternatives e.g. \(\mu_D < 0\) for \(H_1\), or \(\mu_B - \mu_A\) etc provided adequately defined. Hypotheses in words must include "population" |
| Where \(\mu_D\) is the (population) mean reduction/difference in cholesterol level | B1 | For adequate verbal definition |
| MUST be PAIRED COMPARISON \(t\) test | ||
| Differences: \(-0.1,\ 1.7,\ -1.2,\ 1.1,\ 1.4,\ 0.5,\ 0.9,\ 2.2,\ -0.1,\ 2.0,\ 0.7,\ 0.3\) | ||
| \(\bar{x} = 0.7833\), \(s_{n-1} = 0.9833(46)\) | B1 | Do not allow \(s_n = 0.9415\) |
| Test statistic \(= \frac{0.7833 - 0}{\frac{0.9833}{\sqrt{12}}} = 2.7595\) | M1, A1 | Allow candidate's \(\bar{x}\) and/or \(s_{n-1}\). c.a.o. but ft from here |
| Refer to \(t_{11}\) | M1 | No ft from here if wrong. \(P(t > 2.7595) = 0.009286\) |
| Single-tailed 1% point is 2.718 | A1 | No ft from here if wrong |
| Significant | A1 | ft only candidate's test statistic |
| Seems mean cholesterol level has fallen | A1 | ft only candidate's test statistic [11] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| CI is \(\bar{x} \pm 2.201 \times \frac{s}{\sqrt{12}}\) | M1, B1 | Overall structure seen or implied. From \(t_{11}\) seen or implied |
| \(= (-0.5380,\ 1.4046)\) | A1 | Fully correct pair of equations using the given interval |
| \(\bar{x} = \frac{1}{2}(1.4046 - 0.5380) = 0.4333\) | B1 | |
| \(s = (1.4046 - 0.4333) \times \frac{\sqrt{12}}{2.201} = 1.5287\) | M1, A1 | Substitute \(\bar{x}\) and rearrange to find \(s\). c.a.o. |
| Using this interval the doctor might conclude that the mean cholesterol level did not seem to have been reduced | E1 | Accept any sensible comment or interpretation of this interval [7] |
# Question 3:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Must assume: Normality of population; of differences | B1, B1 | |
| $H_0: \mu_D = 0$; $H_1: \mu_D > 0$ | B1 | Both. Accept alternatives e.g. $\mu_D < 0$ for $H_1$, or $\mu_B - \mu_A$ etc provided adequately defined. Hypotheses in words must include "population" |
| Where $\mu_D$ is the (population) mean reduction/difference in cholesterol level | B1 | For adequate verbal definition |
| MUST be PAIRED COMPARISON $t$ test | | |
| Differences: $-0.1,\ 1.7,\ -1.2,\ 1.1,\ 1.4,\ 0.5,\ 0.9,\ 2.2,\ -0.1,\ 2.0,\ 0.7,\ 0.3$ | | |
| $\bar{x} = 0.7833$, $s_{n-1} = 0.9833(46)$ | B1 | Do not allow $s_n = 0.9415$ |
| Test statistic $= \frac{0.7833 - 0}{\frac{0.9833}{\sqrt{12}}} = 2.7595$ | M1, A1 | Allow candidate's $\bar{x}$ and/or $s_{n-1}$. c.a.o. but ft from here |
| Refer to $t_{11}$ | M1 | No ft from here if wrong. $P(t > 2.7595) = 0.009286$ |
| Single-tailed 1% point is 2.718 | A1 | No ft from here if wrong |
| Significant | A1 | ft only candidate's test statistic |
| Seems mean cholesterol level has fallen | A1 | ft only candidate's test statistic **[11]** |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| CI is $\bar{x} \pm 2.201 \times \frac{s}{\sqrt{12}}$ | M1, B1 | Overall structure seen or implied. From $t_{11}$ seen or implied |
| $= (-0.5380,\ 1.4046)$ | A1 | Fully correct pair of equations using the given interval |
| $\bar{x} = \frac{1}{2}(1.4046 - 0.5380) = 0.4333$ | B1 | |
| $s = (1.4046 - 0.4333) \times \frac{\sqrt{12}}{2.201} = 1.5287$ | M1, A1 | Substitute $\bar{x}$ and rearrange to find $s$. c.a.o. |
| Using this interval the doctor might conclude that the mean cholesterol level did not seem to have been reduced | E1 | Accept any sensible comment or interpretation of this interval **[7]** |
---3 Cholesterol is a lipid (fat) which is manufactured by the liver from the fatty foods that we eat. It plays a vital part in allowing the body to function normally. However, when high levels of cholesterol are present in the blood there is a risk of arterial disease. Among the factors believed to assist with achieving and maintaining low cholesterol levels are weight loss and exercise.
A doctor wishes to test the effectiveness of exercise in lowering cholesterol levels. For a random sample of 12 of her patients, she measures their cholesterol levels before and after they have followed a programme of exercise. The measurements obtained are as follows.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | c | c | c | c | }
\hline
Patient & A & B & C & D & E & F & G & H & I & J & K & L \\
\hline
Before & 5.7 & 5.7 & 4.0 & 6.8 & 7.4 & 5.5 & 6.7 & 6.4 & 7.2 & 7.2 & 7.1 & 4.4 \\
\hline
After & 5.8 & 4.0 & 5.2 & 5.7 & 6.0 & 5.0 & 5.8 & 4.2 & 7.3 & 5.2 & 6.4 & 4.1 \\
\hline
\end{tabular}
\end{center}
(i) A $t$ test is to be used in order to see if, on average, the exercise programme seems to be effective in lowering cholesterol levels. State the distributional assumption necessary for the test, and carry out the test using a $1 \%$ significance level.\\
(ii) A second random sample of 12 patients gives a $95 \%$ confidence interval of $( - 0.5380,1.4046 )$ for the true mean reduction (before - after) in cholesterol level. Find the mean and standard deviation for this sample. How might the doctor interpret this interval in relation to the exercise programme?
\hfill \mbox{\textit{OCR MEI S3 2010 Q3 [18]}}