| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2015 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Single sample confidence interval t-distribution |
| Difficulty | Standard +0.8 This S4 question requires confidence intervals for standard deviation using chi-squared distribution, then a t-test for mean with unknown variance. While mechanically straightforward for Further Maths students, it involves multiple statistical procedures, careful distribution selection, and interpretation across two contexts. The chi-squared CI for SD is less routine than mean tests, placing this above average difficulty. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\bar{x} = \frac{843}{8} = 105.375\) | B1 | Correct mean |
| \(s^2 = \frac{1}{7}\left(89327 - \frac{843^2}{8}\right) = \frac{1}{7}(89327 - 88830.125) = \frac{496.875}{7} = 70.982...\) so \(s = 8.425...\) | M1 A1 | Correct method for \(s^2\); correct value |
| \(\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}\) | M1 | Correct structure of CI |
| \(\chi^2_7(5\%) = 14.067\), \(\chi^2_7(95\%) = 2.167\) | B1 | Both critical values correct |
| \(\sqrt{\frac{7 \times 70.982}{14.067}} \leq \sigma \leq \sqrt{\frac{7 \times 70.982}{2.167}}\) | ||
| \(5.96 \leq \sigma \leq 15.2\) (or equivalent) | A1 | Correct interval for \(\sigma\) (allow equivalent in \(\sigma^2\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| The weights of the blocks of cheese are normally distributed | B1 | Must state normality of weights |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Since 5 is within the confidence interval (5.96, 15.2) [or lower bound > 5], Fred may not be achieving the target / there is no evidence that Fred is not meeting the target | B1ft | Follow through from part (a); comment must be in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: \mu = 100\), \(H_1: \mu \neq 100\) | B1 | Both hypotheses stated correctly |
| \(t = \frac{102.6 - 100}{\sqrt{\frac{19.4}{20}}} = \frac{2.6}{\sqrt{0.97}} = \frac{2.6}{0.9849...} = 2.6396...\) | M1 A1 | Correct method; correct value |
| \(t_{19}(5\%) = 1.729\) (two-tailed 10%) | B1 | Correct critical value |
| \(2.640 > 1.729\), reject \(H_0\) | M1 | Correct comparison |
| There is evidence at the 10% significance level that the mean weight of blocks cut by Olga is not 100 g | A1 | Correct conclusion in context |
# Question 2:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\bar{x} = \frac{843}{8} = 105.375$ | B1 | Correct mean |
| $s^2 = \frac{1}{7}\left(89327 - \frac{843^2}{8}\right) = \frac{1}{7}(89327 - 88830.125) = \frac{496.875}{7} = 70.982...$ so $s = 8.425...$ | M1 A1 | Correct method for $s^2$; correct value |
| $\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$ | M1 | Correct structure of CI |
| $\chi^2_7(5\%) = 14.067$, $\chi^2_7(95\%) = 2.167$ | B1 | Both critical values correct |
| $\sqrt{\frac{7 \times 70.982}{14.067}} \leq \sigma \leq \sqrt{\frac{7 \times 70.982}{2.167}}$ | | |
| $5.96 \leq \sigma \leq 15.2$ (or equivalent) | A1 | Correct interval for $\sigma$ (allow equivalent in $\sigma^2$) |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| The weights of the blocks of cheese are normally distributed | B1 | Must state normality of weights |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| Since 5 is within the confidence interval (5.96, 15.2) [or lower bound > 5], Fred may not be achieving the target / there is no evidence that Fred is not meeting the target | B1ft | Follow through from part (a); comment must be in context |
## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \mu = 100$, $H_1: \mu \neq 100$ | B1 | Both hypotheses stated correctly |
| $t = \frac{102.6 - 100}{\sqrt{\frac{19.4}{20}}} = \frac{2.6}{\sqrt{0.97}} = \frac{2.6}{0.9849...} = 2.6396...$ | M1 A1 | Correct method; correct value |
| $t_{19}(5\%) = 1.729$ (two-tailed 10%) | B1 | Correct critical value |
| $2.640 > 1.729$, reject $H_0$ | M1 | Correct comparison |
| There is evidence at the 10% significance level that the mean weight of blocks cut by Olga is not 100 g | A1 | Correct conclusion in context |\begin{enumerate}
\item Fred is a new employee in a delicatessen. He is asked to cut cheese into 100 g blocks. A random sample of 8 of these blocks of cheese is selected. The weight, in grams, of each block of cheese is given below
\end{enumerate}
$$94 , \quad 106 , \quad 115 , \quad 98 , \quad 111 , \quad 104 , \quad 113 , \quad 102$$
(a) Calculate a $90 \%$ confidence interval for the standard deviation of the weights of the blocks of cheese cut by Fred.
Given that the weights of the blocks of cheese are independent,\\
(b) state what further assumption is necessary for this confidence interval to be valid.
The delicatessen manager expects the standard deviation of the weights of the blocks of cheese cut by an employee to be less than 5 g. Any employee who does not achieve this target is given training.\\
(c) Use your answer from part (a) to comment on Fred's results.
A second employee, Olga, has just been given training. Olga is asked to cut cheese into 100 g blocks. A random sample of 20 of these blocks of cheese is selected. The weight of each block of cheese, $x$ grams, is recorded and the results are summarised below.
$$\bar { x } = 102.6 \quad s ^ { 2 } = 19.4$$
Given that the assumption in part (b) is also valid in this case,\\
(d) test, at a $10 \%$ level of significance, whether or not the mean weight of the blocks of cheese cut by Olga after her training is 100 g . State your hypotheses clearly.\\
(6)
\hfill \mbox{\textit{Edexcel S4 2015 Q2 [14]}}