| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2008 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Single sample t-test |
| Difficulty | Standard +0.3 This is a straightforward application of a one-sample t-test with standard parts: justifying the test, performing calculations with given data, constructing a confidence interval, and discussing confidence level trade-offs. All components are routine textbook exercises requiring no novel insight, though the multi-part structure and need to calculate sample statistics from raw data makes it slightly above the most basic recall questions. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| 374.0 | 378.1 | 363.0 | 357.0 | 377.9 | 388.4 |
| 379.6 | 372.4 | 362.4 | 377.3 | 385.2 | 370.6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| A \(t\) test should be used because the sample is small | E1 | |
| the population variance is unknown | E1 | |
| the background population is Normal | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: \mu = 380\); \(H_1: \mu < 380\) | B1 | Both hypotheses. Hypotheses in words only must include "population". |
| where \(\mu\) is the mean temperature in the chamber | B1 | For adequate verbal definition. Allow absence of "population" if correct notation \(\mu\) is used, but do NOT allow "\(\bar{X} = \ldots\)" or similar unless \(\bar{X}\) is clearly and explicitly stated to be a population mean. |
| \(\bar{x} = 373.825\), \(s_{n-1} = 9.368\) | B1 | \(s_n = 8.969\) but do NOT allow this here or in construction of test statistic, but FT from there. |
| Test statistic is \(\frac{373.825 - 380}{\dfrac{9.368}{\sqrt{12}}}\) | M1 | Allow c's \(\bar{x}\) and/or \(s_{n-1}\). Allow alternative: \(380 + (\text{c's} - 1.796) \times \frac{9.368}{\sqrt{12}}\ (= 375.143)\) for subsequent comparison with \(\bar{x}\). |
| \(= -2.283(359)\) | A1 | c.a.o. but ft from here in any case if wrong. Use of \(380 - \bar{x}\) scores M1A0, but ft. |
| Refer to \(t_{11}\). Single-tailed 5% point is \(-1.796\). | M1, A1 | No ft from here if wrong. Must be minus 1.796 unless absolute values are being compared. No ft from here if wrong. |
| Significant. Seems mean temperature in the chamber has fallen. | A1, A1 | ft only c's test statistic. ft only c's test statistic. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| CI is given by \(373.825 \pm 2.201 \times \frac{9.368}{\sqrt{12}}\) | M1, B1, M1 | |
| \(= 373.825 \pm 5.952 = (367.87(3),\ 379.77(7))\) | A1 | c.a.o. Must be expressed as an interval. ZERO/4 if not same distribution as test. Same wrong distribution scores maximum M1B0M1A0. Recovery to \(t_{11}\) is OK. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Advantage: greater certainty. | E1 | Or equivalents. |
| Disadvantage: less precision. | E1 |
# Question 3:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| A $t$ test should be used because the sample is small | E1 | |
| the population variance is unknown | E1 | |
| the background population is Normal | E1 | |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \mu = 380$; $H_1: \mu < 380$ | B1 | Both hypotheses. Hypotheses in words only must include "population". |
| where $\mu$ is the mean temperature in the chamber | B1 | For adequate verbal definition. Allow absence of "population" if correct notation $\mu$ is used, but do NOT allow "$\bar{X} = \ldots$" or similar unless $\bar{X}$ is clearly and explicitly stated to be a population mean. |
| $\bar{x} = 373.825$, $s_{n-1} = 9.368$ | B1 | $s_n = 8.969$ but do NOT allow this here or in construction of test statistic, but FT from there. |
| Test statistic is $\frac{373.825 - 380}{\dfrac{9.368}{\sqrt{12}}}$ | M1 | Allow c's $\bar{x}$ and/or $s_{n-1}$. Allow alternative: $380 + (\text{c's} - 1.796) \times \frac{9.368}{\sqrt{12}}\ (= 375.143)$ for subsequent comparison with $\bar{x}$. |
| $= -2.283(359)$ | A1 | c.a.o. but ft from here in any case if wrong. Use of $380 - \bar{x}$ scores M1A0, but ft. |
| Refer to $t_{11}$. Single-tailed 5% point is $-1.796$. | M1, A1 | No ft from here if wrong. Must be minus 1.796 unless absolute values are being compared. No ft from here if wrong. |
| Significant. Seems mean temperature in the chamber has fallen. | A1, A1 | ft only c's test statistic. ft only c's test statistic. |
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| CI is given by $373.825 \pm 2.201 \times \frac{9.368}{\sqrt{12}}$ | M1, B1, M1 | |
| $= 373.825 \pm 5.952 = (367.87(3),\ 379.77(7))$ | A1 | c.a.o. Must be expressed as an interval. ZERO/4 if not same distribution as test. Same wrong distribution scores maximum M1B0M1A0. Recovery to $t_{11}$ is OK. |
## Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| Advantage: greater certainty. | E1 | Or equivalents. |
| Disadvantage: less precision. | E1 | |
---3 Engineers in charge of a chemical plant need to monitor the temperature inside a reaction chamber. Past experience has shown that when functioning correctly the temperature inside the chamber can be modelled by a Normal distribution with mean $380 ^ { \circ } \mathrm { C }$. The engineers are concerned that the mean operating temperature may have fallen. They decide to test the mean using the following random sample of 12 recent temperature readings.
\begin{center}
\begin{tabular}{ l l l l l l }
374.0 & 378.1 & 363.0 & 357.0 & 377.9 & 388.4 \\
379.6 & 372.4 & 362.4 & 377.3 & 385.2 & 370.6 \\
\end{tabular}
\end{center}
(i) Give three reasons why a $t$ test would be appropriate.\\
(ii) Carry out the test using a $5 \%$ significance level. State your hypotheses and conclusion carefully.\\
(iii) Find a 95\% confidence interval for the true mean temperature in the reaction chamber.\\
(iv) Describe briefly one advantage and one disadvantage of having a 99\% confidence interval instead of a 95\% confidence interval.
\hfill \mbox{\textit{OCR MEI S3 2008 Q3 [18]}}