| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2011 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Single sample confidence interval t-distribution |
| Difficulty | Standard +0.3 This is a straightforward application of a one-sample t-test with standard parts: justification for using t-test, carrying out the test with given data (requiring calculation of sample mean and standard deviation), interpreting significance level, and constructing a confidence interval. All steps are routine S3 procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(t\) test used because population variance is unknown | E1 | Allow "sample is small" as alternative |
| Background population is Normal | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: \mu = 15.3\), \(H_1: \mu < 15.3\) | B1 | Both hypotheses must include "population". Do NOT allow \(\bar{X} = \ldots\) unless clearly stated as population mean |
| where \(\mu\) is the mean of Gerry's times | B1 | For adequate verbal definition. Allow absence of "population" if correct notation \(\mu\) used |
| \(\bar{x} = 14.987\), \(s_{n-1} = 0.4567(5)\) | B1 | \(s_n = 0.4333\) but do NOT allow this here or in construction of test statistic |
| Test statistic is \(\dfrac{14.987 - 15.3}{\dfrac{0.45675}{\sqrt{10}}}\) | M1 | Allow c's \(\bar{x}\) and/or \(s_{n-1}\). Alternative: \(15.3 + (\text{c's} -1.833) \times \dfrac{0.45675}{\sqrt{10}}\) |
| \(= -2.167(0)\) | A1 | c.a.o. but ft from here if wrong. Use of \(\mu - \bar{x}\) scores M1A0 |
| Refer to \(t_9\) | M1 | No ft from here if wrong |
| Single-tailed 5% point is \(-1.833\) | A1 | Must be minus 1.833 unless absolute values compared. \(P(t < -2.167(0)) = 0.0292\) |
| Significant | A1 | ft only c's test statistic |
| Gerry's times have been reduced on average | A1 | ft only c's test statistic. Conclusion in context to include "average" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| 5% significance level means probability of rejecting \(H_0\) given it is true is 0.05 | E1 | |
| Decreasing significance level would make it less likely a true \(H_0\) would be rejected | E1 | |
| Evidence for rejecting \(H_0\) would need to be stronger | E1 | Allow answers relating to context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| CI given by \(14.987 \pm\) | M1 | ZERO/4 if not same distribution as test. Recovery to \(t_9\) is OK |
| \(2.262\) | B1 | |
| \(\times \dfrac{0.45675}{\sqrt{10}}\) | M1 | |
| \(= 14.987 \pm 0.3267 = (14.66(0),\ 15.31(3))\) | A1 | c.a.o. Must be expressed as an interval |
# Question 1:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $t$ test used because population variance is unknown | E1 | Allow "sample is small" as alternative |
| Background population is Normal | E1 | |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \mu = 15.3$, $H_1: \mu < 15.3$ | B1 | Both hypotheses must include "population". Do NOT allow $\bar{X} = \ldots$ unless clearly stated as population mean |
| where $\mu$ is the mean of Gerry's times | B1 | For adequate verbal definition. Allow absence of "population" if correct notation $\mu$ used |
| $\bar{x} = 14.987$, $s_{n-1} = 0.4567(5)$ | B1 | $s_n = 0.4333$ but do NOT allow this here or in construction of test statistic |
| Test statistic is $\dfrac{14.987 - 15.3}{\dfrac{0.45675}{\sqrt{10}}}$ | M1 | Allow c's $\bar{x}$ and/or $s_{n-1}$. Alternative: $15.3 + (\text{c's} -1.833) \times \dfrac{0.45675}{\sqrt{10}}$ |
| $= -2.167(0)$ | A1 | c.a.o. but ft from here if wrong. Use of $\mu - \bar{x}$ scores M1A0 |
| Refer to $t_9$ | M1 | No ft from here if wrong |
| Single-tailed 5% point is $-1.833$ | A1 | Must be minus 1.833 unless absolute values compared. $P(t < -2.167(0)) = 0.0292$ |
| Significant | A1 | ft only c's test statistic |
| Gerry's times have been reduced on average | A1 | ft only c's test statistic. Conclusion in context to include "average" |
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| 5% significance level means probability of rejecting $H_0$ given it is true is 0.05 | E1 | |
| Decreasing significance level would make it less likely a true $H_0$ would be rejected | E1 | |
| Evidence for rejecting $H_0$ would need to be stronger | E1 | Allow answers relating to context |
## Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| CI given by $14.987 \pm$ | M1 | ZERO/4 if not same distribution as test. Recovery to $t_9$ is OK |
| $2.262$ | B1 | |
| $\times \dfrac{0.45675}{\sqrt{10}}$ | M1 | |
| $= 14.987 \pm 0.3267 = (14.66(0),\ 15.31(3))$ | A1 | c.a.o. Must be expressed as an interval |
---1 Gerry runs 5000 -metre races for his local athletics club. His coach has been monitoring his practice times for several months and he believes that they can be modelled using a Normal distribution with mean 15.3 minutes. The coach suggests that Gerry should try running with a pacemaker in order to see if this can improve his times. Subsequently a random sample of 10 of Gerry's times with the pacemaker is collected to see if any reduction has been achieved. The sample of times (in minutes) is as follows.
$$\begin{array} { l l l l l l l l l l }
14.86 & 15.00 & 15.62 & 14.44 & 15.27 & 15.64 & 14.58 & 14.30 & 15.08 & 15.08
\end{array}$$
(i) Why might a $t$ test be used for these data?\\
(ii) Using a $5 \%$ significance level, carry out the test to see whether, on average, Gerry's times have been reduced.\\
(iii) What is meant by 'a $5 \%$ significance level'? What would be the consequence of decreasing the significance level?\\
(iv) Find a $95 \%$ confidence interval for the true mean of Gerry's times using a pacemaker.
\hfill \mbox{\textit{OCR MEI S3 2011 Q1 [18]}}