| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2012 |
| Session | June |
| Marks | 11 |
| 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 standard one-sample t-test procedures with given summary statistics. Students must calculate sample mean and variance from summations, construct a confidence interval, and perform a one-tailed hypothesis test—all routine S3 techniques with no conceptual challenges or novel problem-solving required. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\bar{t} = 348/24\ (=14.5)\) | B1 | |
| \(s^2 = (5195.5 - 348^2/24)/23\ (=6.5)\) | B1 | |
| \(348/24 \pm t\, s/\sqrt{24}\) | M1 | Allow \(z=1.96\) |
| \(t = 2.069\) | B1 | |
| \((13.423,\ 15.557)\) | A1 | Rounding to \((13.4,\ 15.6)\) |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0: \mu = 15\ (\text{Or} \geq)\), \(H_1: \mu < 15\) | B1 B1 | |
| \(\alpha: (14.5-15)/\sqrt{"6.5"/24} = -0.961\) | M1 A1 | Must have /24. 15-14.5 etc M1A0 |
| \(> -1.319\), do not reject \(H_0\) | M1 | \(t\)-value. "\(0.961\)"\(<1.319\) M1, A1 possible |
| \(\beta: \bar{x} \leq 15 - 1.139\sqrt{6.5/24} = 14.3(1)\) | M1 A1 | \(t\)-value |
| \(< 14.5\) | M1 | |
| Insufficient evidence at the 5% SL that the machine is not working acceptably | A1ft | ft-TS, not over-assertive |
| [6] |
# Question 4:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\bar{t} = 348/24\ (=14.5)$ | B1 | |
| $s^2 = (5195.5 - 348^2/24)/23\ (=6.5)$ | B1 | |
| $348/24 \pm t\, s/\sqrt{24}$ | M1 | Allow $z=1.96$ |
| $t = 2.069$ | B1 | |
| $(13.423,\ 15.557)$ | A1 | Rounding to $(13.4,\ 15.6)$ |
| | **[5]** | |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \mu = 15\ (\text{Or} \geq)$, $H_1: \mu < 15$ | B1 B1 | |
| $\alpha: (14.5-15)/\sqrt{"6.5"/24} = -0.961$ | M1 A1 | Must have /24. 15-14.5 etc M1A0 |
| $> -1.319$, do not reject $H_0$ | M1 | $t$-value. "$0.961$"$<1.319$ M1, A1 possible |
| $\beta: \bar{x} \leq 15 - 1.139\sqrt{6.5/24} = 14.3(1)$ | M1 A1 | $t$-value |
| $< 14.5$ | M1 | |
| Insufficient evidence at the 5% SL that the machine is not working acceptably | A1ft | ft-TS, not over-assertive |
| | **[6]** | |
---4 The time interval, $T$ minutes, between consecutive stoppages of a particular grinding machine is regularly measured. $T$ is normally distributed with mean $\mu$.\\
24 randomly chosen values of $T$ are summarised by
$$\sum _ { i = 1 } ^ { 24 } t _ { i } = 348.0 \text { and } \sum _ { i = 1 } ^ { 24 } t _ { i } ^ { 2 } = 5195.5 .$$
(i) Calculate a symmetric $95 \%$ confidence interval for $\mu$.\\
(ii) For the machine to be working acceptably, $\mu$ should be at least 15.0 . Using a test at the 10\% significance level, decide whether the machine is working acceptably.
\hfill \mbox{\textit{OCR S3 2012 Q4 [11]}}