| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2013 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | F-test and chi-squared for variance |
| Type | F-test two variances hypothesis |
| Difficulty | Challenging +1.2 This is a multi-part S4 question requiring F-test and t-test procedures with some data reconstruction. While it involves several steps and two hypothesis tests, the techniques are standard for Further Maths Statistics 4: calculating sample variances, performing an F-test for equality of variances, and a two-sample t-test. The 'missing data' element adds mild complexity but follows directly from given summary statistics. This is moderately harder than average A-level due to being Further Maths content and requiring careful execution of multiple procedures, but remains a textbook application without novel insight. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Ardo data: 13.1, 13.6, 13.2, 13.8, 12.8, 13.5, 13.8; \(\sum x = 93.8\), \(\bar{x}_A = 13.4\) | B1 | May be implied |
| \(S_{xx} = 13.1^2 + 13.6^2 + ... + 13.8^2 - \frac{93.8^2}{7}\) | M1 | Correct method for \(S_{xx}\) |
| \(= 1258.18 - 1257.0057... = 1.1743...\) | A1 | Correct \(S_{xx}\) (accept awrt 1.174) |
| \(s_A^2 = \frac{1.1743}{6} = 0.1957...\) awrt 0.196 | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Pooled estimate \(= \frac{(n_A-1)s_A^2 + (n_B-1)s_B^2}{n_A+n_B-2} = 0.261\) | M1 | Correct pooled formula used |
| \(\frac{6 \times 0.1957 + 8 \times s_B^2}{14} = 0.261\) | M1 | Substituting values |
| \(s_B^2 = \frac{14 \times 0.261 - 6 \times 0.1957}{8}\) awrt 0.310 | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: \sigma_A^2 = \sigma_B^2\); \(H_1: \sigma_A^2 \neq \sigma_B^2\) | B1 | Both hypotheses correct |
| \(F = \frac{s_B^2}{s_A^2} = \frac{0.310}{0.196}\) | M1 | Larger variance divided by smaller |
| \(F \approx 1.58\) | A1 | Correct test statistic |
| Critical value: \(F_{8,6}(0.05) = 4.15\) | B1 | Correct critical value (two-tailed 10%) |
| \(1.58 < 4.15\), do not reject \(H_0\); no significant difference in variability | A1 | Correct conclusion in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: \mu_B - \mu_A = 0.9\); \(H_1: \mu_B - \mu_A > 0.9\) | B1 | Both hypotheses correct |
| \(t = \frac{(\bar{x}_B - \bar{x}_A) - 0.9}{\sqrt{s_p^2\left(\frac{1}{n_A}+\frac{1}{n_B}\right)}}\) | M1 | Correct structure of test statistic |
| \(= \frac{(14.8 - 13.4) - 0.9}{\sqrt{0.261\left(\frac{1}{7}+\frac{1}{9}\right)}}\) | M1 | Correct substitution |
| \(= \frac{0.5}{\sqrt{0.261 \times 0.2540}} = \frac{0.5}{0.2574...}\) | A1 | Correct numerator and denominator |
| \(t \approx 1.94\) | A1 | Correct test statistic |
| Critical value: \(t_{14}(0.05) = 1.761\); \(1.94 > 1.761\), reject \(H_0\) | B1 | Correct CV and conclusion in context |
# Question 4:
## Part (a)(i) - Finding $s_A^2$
| Answer | Mark | Guidance |
|--------|------|----------|
| Ardo data: 13.1, 13.6, 13.2, 13.8, 12.8, 13.5, 13.8; $\sum x = 93.8$, $\bar{x}_A = 13.4$ | B1 | May be implied |
| $S_{xx} = 13.1^2 + 13.6^2 + ... + 13.8^2 - \frac{93.8^2}{7}$ | M1 | Correct method for $S_{xx}$ |
| $= 1258.18 - 1257.0057... = 1.1743...$ | A1 | Correct $S_{xx}$ (accept awrt 1.174) |
| $s_A^2 = \frac{1.1743}{6} = 0.1957...$ awrt **0.196** | A1 | Correct answer |
## Part (a)(ii) - Finding $s_B^2$
| Answer | Mark | Guidance |
|--------|------|----------|
| Pooled estimate $= \frac{(n_A-1)s_A^2 + (n_B-1)s_B^2}{n_A+n_B-2} = 0.261$ | M1 | Correct pooled formula used |
| $\frac{6 \times 0.1957 + 8 \times s_B^2}{14} = 0.261$ | M1 | Substituting values |
| $s_B^2 = \frac{14 \times 0.261 - 6 \times 0.1957}{8}$ awrt **0.310** | A1 | Correct answer |
## Part (b) - F-test for equal variance
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \sigma_A^2 = \sigma_B^2$; $H_1: \sigma_A^2 \neq \sigma_B^2$ | B1 | Both hypotheses correct |
| $F = \frac{s_B^2}{s_A^2} = \frac{0.310}{0.196}$ | M1 | Larger variance divided by smaller |
| $F \approx 1.58$ | A1 | Correct test statistic |
| Critical value: $F_{8,6}(0.05) = 4.15$ | B1 | Correct critical value (two-tailed 10%) |
| $1.58 < 4.15$, do not reject $H_0$; no significant difference in variability | A1 | Correct conclusion in context |
## Part (c) - Two-sample t-test
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \mu_B - \mu_A = 0.9$; $H_1: \mu_B - \mu_A > 0.9$ | B1 | Both hypotheses correct |
| $t = \frac{(\bar{x}_B - \bar{x}_A) - 0.9}{\sqrt{s_p^2\left(\frac{1}{n_A}+\frac{1}{n_B}\right)}}$ | M1 | Correct structure of test statistic |
| $= \frac{(14.8 - 13.4) - 0.9}{\sqrt{0.261\left(\frac{1}{7}+\frac{1}{9}\right)}}$ | M1 | Correct substitution |
| $= \frac{0.5}{\sqrt{0.261 \times 0.2540}} = \frac{0.5}{0.2574...}$ | A1 | Correct numerator and denominator |
| $t \approx 1.94$ | A1 | Correct test statistic |
| Critical value: $t_{14}(0.05) = 1.761$; $1.94 > 1.761$, reject $H_0$ | B1 | Correct CV and conclusion in context |
---4. A company carries out an investigation into the strengths of rods from two different suppliers, Ardo and Bards. Independent random samples of rods were taken from each supplier and the force, $x \mathrm { kN }$, needed to break each rod was recorded. The company wrote the results on a piece of paper but unfortunately spilt ink on it so some of the results can not be seen.\\
The paper with the results on is shown below.\\
\includegraphics[max width=\textwidth, alt={}, center]{4f096806-33da-453f-a4c1-12be20d1a96d-08_435_1522_541_244}
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Use the data from Ardo to calculate an unbiased estimate, $s _ { A } ^ { 2 }$, of the variance.
\item Hence find an unbiased estimate, $s _ { B } ^ { 2 }$, of the variance for the sample of 9 values from Bards.
\end{enumerate}\item Stating your hypotheses clearly, test at the $10 \%$ level of significance whether or not there is a difference in variability of strength between the rods from Ardo and the rods from Bards.\\
(You may assume the two samples come from independent normal distributions.)
\item Use a $5 \%$ level of significance to test whether the mean strength of rods from Bards is more than 0.9 kN greater than the mean strength of rods from Ardo.\\
(6)
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2013 Q4 [15]}}