| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2018 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | F-test and chi-squared for variance |
| Type | F-test then t-test sequential |
| Difficulty | Challenging +1.2 This is a multi-part S4 question requiring an F-test for variance comparison and a two-sample t-test for means. While it involves multiple hypothesis tests and careful interpretation of conditions, the procedures are standard for Further Maths Statistics 4. The calculations are straightforward applications of learned techniques with no novel problem-solving required, though the multi-stage decision process and algebraic manipulation in part (c) elevate it slightly above average difficulty. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Sample size \(( n )\) | Sample mean \(( \bar { x } )\) | Standard deviation \(( s )\) | |
| Tackfast \(( T )\) | 6 | 5.27 | 0.31 |
| Goglue \(( G )\) | 5 | 10.12 | 0.66 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: \sigma_G^2 = \sigma_T^2\) against \(H_1: \sigma_G^2 > \sigma_T^2\) | B1 | Both hypotheses; allow \(H_0: \sigma_T = \sigma_G\); must use \(\sigma\) or \(\sigma^2\) |
| Test stat: \(F_{4,5} = \frac{0.66^2}{0.31^2} = 4.53\) \(\left(\frac{1}{F_{4,5}} = \frac{0.31^2}{0.66^2} = 0.221\right)\) | M1A1 | Allow use of 0.31 and 0.66 rather than \(0.31^2\) and \(0.66^2\) if formula written down |
| Critical value: \(F_{4,5} = 5.19\) (0.1927) | B1 | Correct CV or correct comparison if use \(p\) |
| Not in critical region, therefore no evidence to reject \(H_0\); No evidence of difference in standard deviation (allow variance) | A1cso (5) | All previous marks must be awarded |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(s_p^2 = \frac{5\times0.31^2 + 4\times0.66^2}{5+4}\) | M1 | Allow use of 0.31 and 0.66 |
| \(s_p^2 = 0.24698...\) or \(s_p = 0.49697...\) awrt 0.247 or 0.497 | A1 | |
| \(H_0: \mu_G = \mu_T + 4\), \(H_1: \mu_G > \mu_T + 4\) | B1 | Both hypotheses using \(\mu\); do not allow \(\geq\) instead of \(>\) |
| Critical value CR: \(t_9(0.05) > \pm1.833\) | B1 | Correct CV; must match \(t\)-value or correct comparison if use \(p\) |
| \(t = \pm\frac{10.12 - 5.27 - 4}{\sqrt{0.24698\left(\frac{1}{5}+\frac{1}{6}\right)}} = \pm2.8245...\) or \(p =\) awrt \(0.0099549\), awrt 2.82, 2.825 | M1, A1 | Use of correct formula with their \(s_p\); condone missing 4 |
| There is evidence to reject \(H_0\); \(\mu_G\) is greater than \(\mu_T + 4\). The suppliers claim is supported. | A1 (7) | Correct statement; must include word force and mean and more than 4 greater; no contradicting statements |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{\bar{X}_G - \bar{X}_T - 4}{\sqrt{0.24698\left(\frac{1}{5}+\frac{1}{6}\right)}} > 1.833\) | M1 | Correct LHS with 1.833 (or CV from (b)) |
| \(\bar{X}_G - \bar{X}_T > 1.833 \times \sqrt{0.24698\left(\frac{1}{5}+\frac{1}{6}\right)} + 4\) | ||
| \(\bar{X}_G - \bar{X}_T > 4.55\) | A1 (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| No change to standard deviation | B1 | |
| \(\bar{X}_G - \bar{X}_T = 4.35\) | M1 | |
| Previously they would have changed to *Goglue*, now they will remain with *Tackfast*; or they will no longer change, or they would have changed but now they will not | A1 (3) |
## Question 4:
### Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \sigma_G^2 = \sigma_T^2$ against $H_1: \sigma_G^2 > \sigma_T^2$ | B1 | Both hypotheses; allow $H_0: \sigma_T = \sigma_G$; must use $\sigma$ or $\sigma^2$ |
| Test stat: $F_{4,5} = \frac{0.66^2}{0.31^2} = 4.53$ $\left(\frac{1}{F_{4,5}} = \frac{0.31^2}{0.66^2} = 0.221\right)$ | M1A1 | Allow use of 0.31 and 0.66 rather than $0.31^2$ and $0.66^2$ if formula written down |
| Critical value: $F_{4,5} = 5.19$ (0.1927) | B1 | Correct CV or correct comparison if use $p$ |
| Not in critical region, therefore no evidence to reject $H_0$; No evidence of difference in **standard deviation** (allow variance) | A1cso (5) | All previous marks must be awarded |
### Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $s_p^2 = \frac{5\times0.31^2 + 4\times0.66^2}{5+4}$ | M1 | Allow use of 0.31 and 0.66 |
| $s_p^2 = 0.24698...$ or $s_p = 0.49697...$ awrt 0.247 or 0.497 | A1 | |
| $H_0: \mu_G = \mu_T + 4$, $H_1: \mu_G > \mu_T + 4$ | B1 | Both hypotheses using $\mu$; do not allow $\geq$ instead of $>$ |
| Critical value CR: $t_9(0.05) > \pm1.833$ | B1 | Correct CV; must match $t$-value or correct comparison if use $p$ |
| $t = \pm\frac{10.12 - 5.27 - 4}{\sqrt{0.24698\left(\frac{1}{5}+\frac{1}{6}\right)}} = \pm2.8245...$ or $p =$ awrt $0.0099549$, awrt 2.82, 2.825 | M1, A1 | Use of correct formula with their $s_p$; condone missing 4 |
| There is evidence to reject $H_0$; $\mu_G$ is greater than $\mu_T + 4$. The suppliers claim is supported. | A1 (7) | Correct statement; must include word **force** and **mean** and **more than 4 greater**; no contradicting statements |
### Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{\bar{X}_G - \bar{X}_T - 4}{\sqrt{0.24698\left(\frac{1}{5}+\frac{1}{6}\right)}} > 1.833$ | M1 | Correct LHS with 1.833 (or CV from (b)) |
| $\bar{X}_G - \bar{X}_T > 1.833 \times \sqrt{0.24698\left(\frac{1}{5}+\frac{1}{6}\right)} + 4$ | | |
| $\bar{X}_G - \bar{X}_T > 4.55$ | A1 (2) | |
### Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| No change to standard deviation | B1 | |
| $\bar{X}_G - \bar{X}_T = 4.35$ | M1 | |
| Previously they would have changed to *Goglue*, now they will remain with *Tackfast*; or they will no longer change, or they would have changed but now they will not | A1 (3) | |
**Total: 17**
---\begin{enumerate}
\item A glue supplier claims that Goglue is stronger than Tackfast. A company is presently using Tackfast but agrees to change to Goglue if, at the 5\% significance level,
\end{enumerate}
\begin{itemize}
\item the standard deviation of the force required for Goglue to fail is not greater than the standard deviation of the force required for Tackfast to fail and
\item the mean force required for Goglue to fail is more than 4 newtons greater than the mean force for Tackfast to fail.
\end{itemize}
A series of trials is carried out, using Goglue and Tackfast, and the glues are tested to destruction. The force, $x$ newtons, at which each glue fails is recorded.
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\hline
& Sample size $( n )$ & Sample mean $( \bar { x } )$ & Standard deviation $( s )$ \\
\hline
Tackfast $( T )$ & 6 & 5.27 & 0.31 \\
\hline
Goglue $( G )$ & 5 & 10.12 & 0.66 \\
\hline
\end{tabular}
\end{center}
It can be assumed that the force at which each glue fails is normally distributed.\\
(a) Test, at the $5 \%$ level of significance, whether or not there is evidence that the standard deviation of the force required for Goglue to fail is greater than the standard deviation of the force required for the Tackfast to fail. State your hypotheses clearly.
The supplier claims that the mean force required for its Goglue to fail is more than 4 newtons greater than the mean force required for Tackfast to fail.\\
(b) Stating your hypotheses clearly and using a $5 \%$ level of significance, test the supplier's claim.\\
(c) Show that, at the $5 \%$ level of significance, the supplier's claim will be accepted if $\bar { X } _ { G } - \bar { X } _ { T } > 4.55$, where $\bar { X } _ { G }$ and $\bar { X } _ { T }$ are the mean forces required for Goglue to fail and Tackfast to fail respectively.
Later, it was found that an error had been made when recording the results for Goglue. This resulted in all the forces recorded for Goglue being 0.5 newtons more than they should have been. The results for Tackfast were correct.\\
(d) Explain whether or not this information affects the decision about which glue the supplier decides to use.
\hfill \mbox{\textit{Edexcel S4 2018 Q4 [17]}}