| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2012 |
| 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 one-sample t-test with standard steps: calculate sample mean and standard deviation, perform hypothesis test at 5% level, then construct a confidence interval. While it requires knowledge of t-distribution procedures and distributional assumptions, it follows a routine template with no conceptual challenges or novel problem-solving required. Slightly above average difficulty due to being Further Maths content and requiring careful calculation, but still a standard textbook exercise. |
| Spec | 2.01a Population and sample: terminology5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Simple random sampling is when every sample of the required size has an equal chance of being chosen. | E1, E1 | SC: Allow E1 for "Every member of the population has an equal chance of being chosen" |
| e.g. One needs access to the entire population in order to establish the sampling frame. | E2 | E2, 1, 0. Reward any sensible point. SC1 "Sample may not be representative of the population" |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0: \mu = 70.3\), \(H_1: \mu < 70.3\) | B1 | Both. Hypotheses in words only must include "population". Do NOT allow "\(\bar{X} = ...\)" unless \(\bar{X}\) is clearly stated to be a population mean |
| Where \(\mu\) is the (population) mean drying rate. | B1 | For adequate verbal definition. Allow absence of "population" if correct notation \(\mu\) is used |
| \(\bar{x} = 67.675\), \(s_{n-1} = 4.129\) \((s_{n-1}^2 = 17.049)\) | B1 | Do not allow \(s_n = 3.953\) \((s_n^2 = 15.629)\) |
| Test statistic is \(\dfrac{67.675 - 70.3}{\left(\dfrac{4.129}{\sqrt{12}}\right)}\) | M1 | Allow c's \(\bar{x}\) and/or \(s_{n-1}\). Allow alternative: \(70.3 + (\text{c's} -1.796) \times \dfrac{4.129}{\sqrt{12}}\) (= 68.15(9)) for subsequent comparison with \(\bar{x}\) |
| \(= -2.202(2)\) | A1 | cao but ft from here in any case if wrong. Use of \(70.3 - \bar{x}\) scores M1A0, but ft |
| Refer to \(t_{11}\). Lower 5% point is \(-1.796\). | M1, A1 | No ft from here if wrong. Allow any \(t_{11}\) value from tables. Must be \(-1.796\) unless absolute values are being used. \(P(t < -2.202) = 0.0249\) |
| \(-2.202 < -1.796\), \(\therefore\) Result is significant. Seems mean drying rate has reduced. | A1, A1 | ft only c's test statistic. "Non-assertive" conclusion in context to include "on average" |
| Underlying population is Normal. | B1 | |
| [10] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| CI is given by \(67.675 \pm\) | M1 | ZERO if not same distribution as test. ft c's \(\bar{x} \pm\) |
| \(2.201\) | B1 | |
| \(\times \dfrac{4.129}{\sqrt{12}}\) | M1 | ft c's \(s_{n-1}\) |
| \(= 67.675 \pm 2.623 = (65.05(2),\ 70.29(8))\) | A1 | cao. Must be expressed as an interval. (\(t = 1.796\) gives \((65.534, 69.816)\) (M1B0M1A0)) |
| [4] |
# Question 1:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Simple random sampling is when every sample of the required size has an equal chance of being chosen. | E1, E1 | SC: Allow E1 for "Every member of the population has an equal chance of being chosen" |
| e.g. One needs access to the entire population in order to establish the sampling frame. | E2 | E2, 1, 0. Reward any sensible point. SC1 "Sample may not be representative of the population" |
| **[4]** | | |
## Part (b)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \mu = 70.3$, $H_1: \mu < 70.3$ | B1 | Both. Hypotheses in words only must include "population". Do NOT allow "$\bar{X} = ...$" unless $\bar{X}$ is clearly stated to be a population mean |
| Where $\mu$ is the (population) mean drying rate. | B1 | For adequate verbal definition. Allow absence of "population" if correct notation $\mu$ is used |
| $\bar{x} = 67.675$, $s_{n-1} = 4.129$ $(s_{n-1}^2 = 17.049)$ | B1 | Do not allow $s_n = 3.953$ $(s_n^2 = 15.629)$ |
| Test statistic is $\dfrac{67.675 - 70.3}{\left(\dfrac{4.129}{\sqrt{12}}\right)}$ | M1 | Allow c's $\bar{x}$ and/or $s_{n-1}$. Allow alternative: $70.3 + (\text{c's} -1.796) \times \dfrac{4.129}{\sqrt{12}}$ (= 68.15(9)) for subsequent comparison with $\bar{x}$ |
| $= -2.202(2)$ | A1 | cao but ft from here in any case if wrong. Use of $70.3 - \bar{x}$ scores M1A0, but ft |
| Refer to $t_{11}$. Lower 5% point is $-1.796$. | M1, A1 | No ft from here if wrong. Allow any $t_{11}$ value from tables. Must be $-1.796$ unless absolute values are being used. $P(t < -2.202) = 0.0249$ |
| $-2.202 < -1.796$, $\therefore$ Result is significant. Seems mean drying rate has reduced. | A1, A1 | ft only c's test statistic. "Non-assertive" conclusion in context to include "on average" |
| Underlying population is Normal. | B1 | |
| **[10]** | | |
## Part (b)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| CI is given by $67.675 \pm$ | M1 | ZERO if not same distribution as test. ft c's $\bar{x} \pm$ |
| $2.201$ | B1 | |
| $\times \dfrac{4.129}{\sqrt{12}}$ | M1 | ft c's $s_{n-1}$ |
| $= 67.675 \pm 2.623 = (65.05(2),\ 70.29(8))$ | A1 | cao. Must be expressed as an interval. ($t = 1.796$ gives $(65.534, 69.816)$ (M1B0M1A0)) |
| **[4]** | | |
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\begin{enumerate}[label=(\alph*)]
\item Define simple random sampling. Describe briefly one difficulty associated with simple random sampling.
\item Freeze-drying is an economically important process used in the production of coffee. It improves the retention of the volatile aroma compounds. In order to maintain the quality of the coffee, technologists need to monitor the drying rate, measured in suitable units, at regular intervals. It is known that, for best results, the mean drying rate should be 70.3 units and anything substantially less than this would be detrimental to the coffee. Recently, a random sample of 12 observations of the drying rate was as follows.
$$\begin{array} { l l l l l l l l l l l l }
66.0 & 66.1 & 59.8 & 64.0 & 70.9 & 71.4 & 66.9 & 76.2 & 65.2 & 67.9 & 69.2 & 68.5
\end{array}$$
\begin{enumerate}[label=(\roman*)]
\item Carry out a test to investigate at the $5 \%$ level of significance whether the mean drying rate appears to be less than 70.3. State the distributional assumption that is required for this test.
\item Find a 95\% confidence interval for the true mean drying rate.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI S3 2012 Q1 [18]}}