| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2006 |
| 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 straightforward paired t-test question followed by a standard confidence interval calculation. Both parts require routine application of well-practiced S3 techniques with clear data and standard hypotheses. The paired design is explicitly suggested by the temperature blocking, and all calculations follow textbook procedures without requiring novel insight or complex interpretation. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Ambient temperature | \(T_1\) | \(T_2\) | \(T_3\) | \(T_4\) | \(T_5\) | \(T_6\) | \(T_7\) | \(T_8\) | \(T_9\) | \(T_{10}\) |
| Amount delivered by machine A | 246.2 | 251.6 | 252.0 | 246.6 | 258.4 | 251.0 | 247.5 | 247.1 | 248.1 | 253.4 |
| Amount delivered by machine B | 248.3 | 252.6 | 252.8 | 247.2 | 258.8 | 250.0 | 247.2 | 247.9 | 249.0 | 254.5 |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0: \mu_D = 0\) (or \(\mu_A = \mu_B\)) | B1 | Hypotheses in words only must include "population". Or "<" for A−B. |
| \(H_1: \mu_D > 0\) (or \(\mu_B > \mu_A\)) where \(\mu_D\) is "mean for B − mean for A" | B1, B1 | For adequate verbal definition. Allow absence of "population" if correct notation \(\mu\) is used, but do NOT allow "\(\bar{x} = \bar{x}_a\)" or similar unless \(\bar{x}\) is clearly and explicitly stated to be a population mean. |
| Normality of differences is required MUST be PAIRED COMPARISON \(t\) test. | B1 | |
| Differences are: 2.1 1.0 0.8 0.6 0.4 −1.0 −0.3 0.8 0.9 1.1 | ||
| \(\bar{d} = 0.64\), \(s_{n-1} = 0.8316\) | B1 | \(s_n = 0.7889\) but do NOT allow this here or in construction of test statistic, but FT from there. Allow c's \(\bar{d}\) and/or \(s_{n-1}\). Allow alternative: \(0 + (c's \, 1.833) \times \frac{0.8316}{\sqrt{10}} (= 0.4821)\) for subsequent comparison with \(\bar{d}\). (Or \(\bar{d} - (c's \, 1.833) \times \frac{0.8316}{\sqrt{10}} (= 0.1579)\) for comparison with 0.) |
| Test statistic is \(\frac{0.64 - 0}{0.8316/\sqrt{10}}\) | M | |
| \(= 2.43(37)\) | A1 | c.a.o. but ft from here in any case if wrong. Use of \(0 - \bar{d}\) scores M1A0, but ft. No ft from here if wrong. |
| Refer to \(t_9\) | M | |
| Single-tailed 5% point is 1.833. Significant. Seems mean amount delivered by B is greater that that by A | A1, E1, E1 | No ft from here if wrong. ft only c's test statistic. ft only c's test statistic. Special case: (\(t_{10}\) and 1.812) can score 1 of these last 2 marks if either form of conclusion is given. |
| Answer | Marks | Guidance |
|---|---|---|
| We now require Normality for the amounts delivered by machine A. | B1 | |
| For machine A, \(\bar{x} = 250.19\), \(s_{n-1} = 3.8527\) | B1 | \(s_n = 3.6549(83)\) but do NOT allow this here or in construction of CI. |
| CI is given by \(250.19 \pm 2.262 \frac{3.8527}{\sqrt{10}}\) | M | ft c's \(\bar{x} \pm\). 2.262. ft c's \(s_{n1}\). |
| \(= 250.19 \pm 2.75(6) = (247.43(4), 252.94(6))\) | 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 \(t_9\) is OK. |
| 250 is in the CI, so would accept \(H_0: \mu = 250\), so no evidence that machine is not working correctly in this respect. | E1 |
## (a)
$H_0: \mu_D = 0$ (or $\mu_A = \mu_B$) | B1 | Hypotheses in words only must include "population". Or "<" for A−B. | 11 marks
$H_1: \mu_D > 0$ (or $\mu_B > \mu_A$) where $\mu_D$ is "mean for B − mean for A" | B1, B1 | For adequate verbal definition. Allow absence of "population" if correct notation $\mu$ is used, but do NOT allow "$\bar{x} = \bar{x}_a$" or similar unless $\bar{x}$ is clearly and explicitly stated to be a population mean. |
Normality of differences is required MUST be PAIRED COMPARISON $t$ test. | B1 | |
Differences are: 2.1 1.0 0.8 0.6 0.4 −1.0 −0.3 0.8 0.9 1.1 | | |
$\bar{d} = 0.64$, $s_{n-1} = 0.8316$ | B1 | $s_n = 0.7889$ but do NOT allow this here or in construction of test statistic, but FT from there. Allow c's $\bar{d}$ and/or $s_{n-1}$. Allow alternative: $0 + (c's \, 1.833) \times \frac{0.8316}{\sqrt{10}} (= 0.4821)$ for subsequent comparison with $\bar{d}$. (Or $\bar{d} - (c's \, 1.833) \times \frac{0.8316}{\sqrt{10}} (= 0.1579)$ for comparison with 0.) |
Test statistic is $\frac{0.64 - 0}{0.8316/\sqrt{10}}$ | M | |
$= 2.43(37)$ | A1 | c.a.o. but ft from here in any case if wrong. Use of $0 - \bar{d}$ scores M1A0, but ft. No ft from here if wrong. |
Refer to $t_9$ | M | |
Single-tailed 5% point is 1.833. Significant. Seems mean amount delivered by B is greater that that by A | A1, E1, E1 | No ft from here if wrong. ft only c's test statistic. ft only c's test statistic. Special case: ($t_{10}$ and 1.812) can score 1 of these last 2 marks if either form of conclusion is given. |
## (b)
We now require Normality for the amounts delivered by machine A. | B1 | | 18 marks
For machine A, $\bar{x} = 250.19$, $s_{n-1} = 3.8527$ | B1 | $s_n = 3.6549(83)$ but do NOT allow this here or in construction of CI. | 7 marks
CI is given by $250.19 \pm 2.262 \frac{3.8527}{\sqrt{10}}$ | M | ft c's $\bar{x} \pm$. 2.262. ft c's $s_{n1}$. |
$= 250.19 \pm 2.75(6) = (247.43(4), 252.94(6))$ | 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 $t_9$ is OK. |
250 is in the CI, so would accept $H_0: \mu = 250$, so no evidence that machine is not working correctly in this respect. | E1 | |
---A production line has two machines, A and B, for delivering liquid soap into bottles. Each machine is set to deliver a nominal amount of 250 ml, but it is not expected that they will work to a high level of accuracy. In particular, it is known that the ambient temperature affects the rate of flow of the liquid and leads to variation in the amounts delivered.
The operators think that machine B tends to deliver a somewhat greater amount than machine A, no matter what the ambient temperature. This is being investigated by an experiment. A random sample of 10 results from the experiment is shown below. Each column of data is for a different ambient temperature.
\begin{tabular}{|l|c|c|c|c|c|c|c|c|c|c|}
\hline
Ambient temperature & $T_1$ & $T_2$ & $T_3$ & $T_4$ & $T_5$ & $T_6$ & $T_7$ & $T_8$ & $T_9$ & $T_{10}$ \\
\hline
Amount delivered by machine A & 246.2 & 251.6 & 252.0 & 246.6 & 258.4 & 251.0 & 247.5 & 247.1 & 248.1 & 253.4 \\
\hline
Amount delivered by machine B & 248.3 & 252.6 & 252.8 & 247.2 & 258.8 & 250.0 & 247.2 & 247.9 & 249.0 & 254.5 \\
\hline
\end{tabular}
\begin{enumerate}[label=(\alph*)]
\item Use an appropriate $t$ test to examine, at the 5\% level of significance, whether the mean amount delivered by machine B may be taken as being greater than that delivered by machine A, stating carefully your null and alternative hypotheses and the required distributional assumption. [11]
\item Using the data for machine A in the table above, provide a two-sided 95\% confidence interval for the mean amount delivered by this machine, stating the required distributional assumption. Explain whether you would conclude that the machine appears to be working correctly in terms of the nominal amount as set. [7]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S3 2006 Q3 [18]}}