| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2007 |
| Session | January |
| Marks | 10 |
| 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 summary statistics provided. Students must calculate sample mean and standard deviation from given sums, then apply textbook formulas for confidence intervals and hypothesis testing. While it requires multiple steps and careful calculation, it involves no conceptual challenges or novel problem-solving—just routine application of S3 techniques with clearly stated hypotheses and significance level. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(s^2 = \frac{1}{n-1}(2604.4 - 177.6^2/12) = 1.0836...\) | M1, A1 | aef |
| Use \(\bar{x} \pm t\sqrt{\frac{s^2}{12}}\) | M1 | |
| \(t = 2.201\); \(\bar{x} = 177.6/12 = 14.8\); \((14.14, 15.46)\), \((14.1, 15.5)\) | B1, A1, A1 | 6 |
| (ii) EITHER: \((14.8 - 15.4)/(\sqrt{s^2/12}) = -1.997\) | M1, A1 | With their variance |
| Compare correctly with \(-1.796\) | M1 | |
| Reject \(H_0\) and accept that there is evidence that the mean is less than 15.4 | A1√ | In context; Ft \(-1.997\) |
| OR: \(\bar{x} - 15.4 \le -ks^2; \bar{x} \le 14.86\) | M1A1 | Allow \(<\) or \(=\) |
| \(14.8 < 14.86\), reject \(H_0\) etc | M1A1√ | 4 Or equivalent; Ft 14.86 |
**(i)** $s^2 = \frac{1}{n-1}(2604.4 - 177.6^2/12) = 1.0836...$ | M1, A1 | aef
Use $\bar{x} \pm t\sqrt{\frac{s^2}{12}}$ | M1 |
$t = 2.201$; $\bar{x} = 177.6/12 = 14.8$; $(14.14, 15.46)$, $(14.1, 15.5)$ | B1, A1, A1 | 6
**(ii)** EITHER: $(14.8 - 15.4)/(\sqrt{s^2/12}) = -1.997$ | M1, A1 | With their variance
Compare correctly with $-1.796$ | M1 |
Reject $H_0$ and accept that there is evidence that the mean is less than 15.4 | A1√ | In context; Ft $-1.997$
OR: $\bar{x} - 15.4 \le -ks^2; \bar{x} \le 14.86$ | M1A1 | Allow $<$ or $=$
$14.8 < 14.86$, reject $H_0$ etc | M1A1√ | 4 Or equivalent; Ft 14.864 A machine is set to produce metal discs with mean diameter 15.4 mm . In order to test the correctness of the setting, a random sample of 12 discs was selected and the diameters, $x \mathrm {~mm}$, were measured. The results are summarised by $\Sigma x = 177.6$ and $\Sigma x ^ { 2 } = 2640.40$. Diameters may be assumed to be normally distributed with mean $\mu \mathrm { mm }$.\\
(i) Find a $95 \%$ confidence interval for $\mu$.\\
(ii) Test, at the $5 \%$ significance level, the null hypothesis $\mu = 15.4$ against the alternative hypothesis $\mu < 15.4$.
\hfill \mbox{\textit{OCR S3 2007 Q4 [10]}}