| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2009 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Paired sample t-test |
| Difficulty | Standard +0.3 This is a standard paired t-test question requiring routine application of the procedure: calculating differences, finding mean and standard deviation, performing the test, and working backwards from a confidence interval. While it involves multiple parts and some calculation, all steps follow textbook methodology with no novel insight required. The confidence interval reverse-calculation in part (iv) adds slight challenge but remains algorithmic. Slightly above average difficulty due to the computational load and four-part structure, but well within typical S3 expectations. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Immediately | 15.21 | 13.36 | 15.97 | 21.07 | 12.82 | 10.80 | 11.50 | 12.05 |
| After freezing | 15.96 | 10.65 | 13.38 | 15.00 | 12.11 | 12.65 | 12.48 | 8.49 |
| Immediately | 10.90 | 18.48 | 13.43 | 13.16 | 16.62 | 14.91 | 17.08 |
| After freezing | 9.13 | 15.53 | 11.84 | 8.99 | 16.24 | 14.03 | 16.13 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| A paired test is appropriate because there are clearly differences between specimens which the pairing eliminates. | E1, E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: \mu_D = 0\) | B1 | Both. Accept alternatives e.g. \(\mu_D < 0\) for \(H_1\), or \(\mu_A - \mu_B\) etc provided adequately defined. Hypotheses must include "population". |
| \(H_1: \mu_D > 0\) | ||
| Where \(\mu_D\) is the (population) mean reduction in hormone concentration. | B1 | For adequate verbal definition. Allow absence of "population" if correct notation \(\mu\) is used. Do NOT allow "\(\bar{X} = ...\)" unless \(\bar{X}\) is clearly stated to be a population mean. |
| Sample is random | B1 | |
| Normality of differences | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| MUST be PAIRED COMPARISON \(t\) test. Differences (reductions) (before – after): \(-0.75\ 2.71\ 2.59\ 6.07\ 0.71\ -1.85\ -0.98\ 3.56\ 1.77\ 2.95\ 1.59\ 4.17\ 0.38\ 0.88\ 0.95\) | Allow "after – before" if consistent with alternatives above. | |
| \(\bar{x} = 1.65,\ s_{n-1} = 2.100(3)\ (s_{n-1}^2 = 4.4112)\) | B1 | Do not allow \(s_n = 2.0291\ (s_n^2 = 4.1171)\) |
| Test statistic \(= \frac{1.65-0}{\frac{2.100}{\sqrt{15}}} = 3.043\) | M1, A1 | Allow c's \(\bar{x}\) and/or \(s_{n-1}\). Use of \(0 - \bar{x}\) scores M1A0, but ft. |
| Refer to \(t_{14}\) | M1 | No ft from here if wrong. \(P(t > 3.043) = 0.00438\). |
| Single-tailed 1% point is 2.624. | A1 | No ft from here if wrong. |
| Significant. | A1 | ft only c's test statistic. |
| Seems mean concentration of hormone has fallen. | A1 | ft only c's test statistic. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| CI is \(1.65 \pm k \times \frac{2.100}{\sqrt{15}}\) | M1, M1 | ft c's \(\bar{x} \pm\); ft c's \(s_{n1}\). |
| \(= (0.4869,\ 2.8131)\) | A1 | A correct equation in \(k\) using either end of the interval or the width of the interval. |
| \(\therefore k = 2.145\); By reference to \(t_{14}\) tables this is a 95% CI. | A1, A1 | Allow ft c's \(\bar{x}\) and \(s_{n1}\). c.a.o. |
# Question 3:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| A paired test is appropriate because there are clearly differences between specimens which the pairing eliminates. | E1, E1 | |
**Total: 2 marks**
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \mu_D = 0$ | B1 | Both. Accept alternatives e.g. $\mu_D < 0$ for $H_1$, or $\mu_A - \mu_B$ etc provided adequately defined. Hypotheses must include "population". |
| $H_1: \mu_D > 0$ | | |
| Where $\mu_D$ is the (population) mean reduction in hormone concentration. | B1 | For adequate verbal definition. Allow absence of "population" if correct notation $\mu$ is used. Do NOT allow "$\bar{X} = ...$" unless $\bar{X}$ is clearly stated to be a population mean. |
| Sample is random | B1 | |
| Normality of differences | B1 | |
**Total: 4 marks**
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| MUST be PAIRED COMPARISON $t$ test. Differences (reductions) (before – after): $-0.75\ 2.71\ 2.59\ 6.07\ 0.71\ -1.85\ -0.98\ 3.56\ 1.77\ 2.95\ 1.59\ 4.17\ 0.38\ 0.88\ 0.95$ | | Allow "after – before" if consistent with alternatives above. |
| $\bar{x} = 1.65,\ s_{n-1} = 2.100(3)\ (s_{n-1}^2 = 4.4112)$ | B1 | Do not allow $s_n = 2.0291\ (s_n^2 = 4.1171)$ |
| Test statistic $= \frac{1.65-0}{\frac{2.100}{\sqrt{15}}} = 3.043$ | M1, A1 | Allow c's $\bar{x}$ and/or $s_{n-1}$. Use of $0 - \bar{x}$ scores M1A0, but ft. |
| Refer to $t_{14}$ | M1 | No ft from here if wrong. $P(t > 3.043) = 0.00438$. |
| Single-tailed 1% point is 2.624. | A1 | No ft from here if wrong. |
| Significant. | A1 | ft only c's test statistic. |
| Seems mean concentration of hormone has fallen. | A1 | ft only c's test statistic. |
**Total: 7 marks**
## Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| CI is $1.65 \pm k \times \frac{2.100}{\sqrt{15}}$ | M1, M1 | ft c's $\bar{x} \pm$; ft c's $s_{n1}$. |
| $= (0.4869,\ 2.8131)$ | A1 | A correct equation in $k$ using either end of the interval or the width of the interval. |
| $\therefore k = 2.145$; By reference to $t_{14}$ tables this is a 95% CI. | A1, A1 | Allow ft c's $\bar{x}$ and $s_{n1}$. c.a.o. |
**Total: 5 marks**
---3 Pathology departments in hospitals routinely analyse blood specimens. Ideally the analysis should be done while the specimens are fresh to avoid any deterioration, but this is not always possible. A researcher decides to study the effect of freezing specimens for later analysis by measuring the concentrations of a particular hormone before and after freezing. He collects and divides a sample of 15 specimens. One half of each specimen is analysed immediately, the other half is frozen and analysed a month later. The concentrations of the particular hormone (in suitable units) are as follows.
\begin{center}
\begin{tabular}{ | l | r | r | r | r | r | r | r | r | }
\hline
Immediately & 15.21 & 13.36 & 15.97 & 21.07 & 12.82 & 10.80 & 11.50 & 12.05 \\
\hline
After freezing & 15.96 & 10.65 & 13.38 & 15.00 & 12.11 & 12.65 & 12.48 & 8.49 \\
\hline
\end{tabular}
\end{center}
\begin{center}
\begin{tabular}{ | l | r | r | r | r | r | r | r | }
\hline
Immediately & 10.90 & 18.48 & 13.43 & 13.16 & 16.62 & 14.91 & 17.08 \\
\hline
After freezing & 9.13 & 15.53 & 11.84 & 8.99 & 16.24 & 14.03 & 16.13 \\
\hline
\end{tabular}
\end{center}
A $t$ test is to be used in order to see if, on average, there is a reduction in hormone concentration as a result of being frozen.\\
(i) Explain why a paired test is appropriate in this situation.\\
(ii) State the hypotheses that should be used, together with any necessary assumptions.\\
(iii) Carry out the test using a $1 \%$ significance level.\\
(iv) A $p \%$ confidence interval for the true mean reduction in hormone concentration is found to be ( $0.4869,2.8131$ ). Determine the value of $p$.
\hfill \mbox{\textit{OCR MEI S3 2009 Q3 [18]}}