| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2014 |
| Session | June |
| Marks | 19 |
| 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 one-sample t-tests with clear hypotheses and standard procedures. Part (i) is definitional recall, part (ii) is a routine one-tailed test, and part (iii) adds minimal complexity by requiring recognition that symmetry allows the t-test despite non-normality. The calculations are standard with small samples, making this slightly easier than average for A-level statistics. |
| Spec | 2.01a Population and sample: terminology5.05c Hypothesis test: normal distribution for population mean5.07b Sign test: and Wilcoxon signed-rank |
| 1.055 | 1.064 | 1.063 | 1.043 | 1.062 | 1.070 | 1.059 | 1.044 | 1.054 |
| 1.053 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| A simple random sample is one where every sample of the same size has an equal probability of being selected. | E2,1,0 | Allow E1 for every item has the same probability of being selected |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0: \mu = 1.02 \quad H_1: \mu < 1.02\) | B1 | Both hypotheses. Must include "population". Do NOT allow "\(\bar{X} = ...\)" unless clearly stated to be a population mean. |
| Where \(\mu\) is the population mean volume | B1 | For adequate verbal definition. Allow absence of "population" if correct notation \(\mu\) is used. |
| \(\bar{x} = 1.0131 \quad s = 0.009146\) | B1 | Do not allow \(s_n = 0.00868\) here or in construction of test statistic |
| Test statistic is \(\dfrac{1.0131 - 1.02}{0.009146/\sqrt{10}}\) | M1* | For method, allow candidates \(\bar{x}, s\). Allow confidence interval approach. |
| \(= -2.3857\), value between \(-2.38\) and \(-2.39\) | A1 | cao |
| Refer to \(t_9\) | M1 | For \(t_9\). No FT from here |
| 5% point is \(\pm 1.833\) | A1 | For matching 1.833 seen. No FT from here |
| \(-2.3857 < -1.833\) reject \(H_0\) | M1 dep | For rejection. Must compare test statistic with matching 1.833 unless absolute values are being compared. |
| Conclude mean appears to be below 1.02 and so machine will be cleaned. | E1 dep | Needs context |
| [9] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0\): median of population volumes \(= 1.05\) | B1 | Both. Allow location parameter or mean (if explanation is given) |
| \(H_1\): median of population volumes \(\neq 1.05\) | ||
| Differences from 1.05, ranked: | M1 | For differences from 1.05. ZERO from this point if differences not used. |
| M1 | For ranks | |
| A1 | ft from here if ranks wrong | |
| \(W_- = 4 + 5 = 9\) | B1 | Or \(W_+ = 1+2+3+6+7+8+9+10 = 46\) |
| Refer to tables of Wilcoxon paired/single sample statistic for \(n = 10\) | M1 | No FT from here if wrong |
| Critical value \(= 8\) (47 for upper) | A1 | No FT from here if wrong (i.e. 1 tail) |
| Cannot reject \(H_0\) and conclude median volume appears to be 1.05 | E1 | Must have context |
| [8] |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| A simple random sample is one where every sample of the same size has an equal probability of being selected. | E2,1,0 | Allow E1 for every item has the same probability of being selected |
| **[2]** | | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \mu = 1.02 \quad H_1: \mu < 1.02$ | B1 | Both hypotheses. Must include "population". Do NOT allow "$\bar{X} = ...$" unless clearly stated to be a population mean. |
| Where $\mu$ is the population mean volume | B1 | For adequate verbal definition. Allow absence of "population" if correct notation $\mu$ is used. |
| $\bar{x} = 1.0131 \quad s = 0.009146$ | B1 | Do not allow $s_n = 0.00868$ here or in construction of test statistic |
| Test statistic is $\dfrac{1.0131 - 1.02}{0.009146/\sqrt{10}}$ | M1* | For method, allow candidates $\bar{x}, s$. Allow confidence interval approach. |
| $= -2.3857$, value between $-2.38$ and $-2.39$ | A1 | cao |
| Refer to $t_9$ | M1 | For $t_9$. No FT from here |
| 5% point is $\pm 1.833$ | A1 | For matching 1.833 seen. No FT from here |
| $-2.3857 < -1.833$ reject $H_0$ | M1 dep | For rejection. Must compare test statistic with matching 1.833 unless absolute values are being compared. |
| Conclude mean appears to be below 1.02 and so machine will be cleaned. | E1 dep | Needs context |
| **[9]** | | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0$: median of population volumes $= 1.05$ | B1 | Both. Allow location parameter or mean (if explanation is given) |
| $H_1$: median of population volumes $\neq 1.05$ | | |
| Differences from 1.05, ranked: | M1 | For differences from 1.05. ZERO from this point if differences not used. |
| | M1 | For ranks |
| | A1 | ft from here if ranks wrong |
| $W_- = 4 + 5 = 9$ | B1 | Or $W_+ = 1+2+3+6+7+8+9+10 = 46$ |
| Refer to tables of Wilcoxon paired/single sample statistic for $n = 10$ | M1 | No FT from here if wrong |
| Critical value $= 8$ (47 for upper) | A1 | No FT from here if wrong (i.e. 1 tail) |
| Cannot reject $H_0$ and conclude median volume appears to be 1.05 | E1 | Must have context |
| **[8]** | | |
---2 (i) Explain what is meant by a simple random sample.
A manufacturer produces tins of paint which nominally contain 1 litre. The quantity of paint delivered by the machine that fills the tins can be assumed to be a Normally distributed random variable.
The machine is designed to deliver an average of 1.05 litres to each tin. However, over time paint builds up in the delivery nozzle of the machine, reducing the quantity of paint delivered. Random samples of 10 tins are taken regularly from the production process. If a significance test, carried out at the $5 \%$ level, suggests that the average quantity of paint delivered is less than 1.02 litres, the machine is cleaned.\\
(ii) By carrying out an appropriate test, determine whether or not the sample below leads to the machine being cleaned.
$$\begin{array} { l l l l l l l l l l }
0.994 & 1.010 & 1.021 & 1.015 & 1.016 & 1.022 & 1.009 & 1.007 & 1.011 & 1.026
\end{array}$$
Each time the machine has been cleaned, a random sample of 10 tins is taken to determine whether or not the average quantity of paint delivered has returned to 1.05 litres.\\
(iii) On one occasion after the machine has been cleaned, the quality control manager thinks that the distribution of the quantity of paint is symmetrical but not necessarily Normal. The sample on this occasion is as follows.
\begin{center}
\begin{tabular}{ l l l l l l l l l }
1.055 & 1.064 & 1.063 & 1.043 & 1.062 & 1.070 & 1.059 & 1.044 & 1.054 \\
1.053 & & & & & & & & \\
\end{tabular}
\end{center}
By carrying out an appropriate test at the $5 \%$ level of significance, determine whether or not this sample supports the conclusion that the average quantity of paint delivered is 1.05 litres.
\hfill \mbox{\textit{OCR MEI S3 2014 Q2 [19]}}