| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2018 |
| Session | June |
| Marks | 13 |
| 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 S4 confidence interval question requiring standard calculations: t-interval for mean and chi-squared interval for standard deviation, followed by simple comparison to given criteria. All techniques are routine for Further Maths Statistics students with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(\bar{x} = \dfrac{468}{9} = 52\), \(s^2 = \dfrac{9}{8}\!\left(\dfrac{24560}{9} - 52^2\right) = 28\) | M1A1 | M1: attempting \(s\) or \(s^2\); A1: 28 only |
| (i) \(t_8 = 2.306\) | B1 | CV awrt 2.306 |
| \(95\%\ \text{CI} = 52 \pm 2.306 \times \dfrac{\sqrt{``28''}}{\sqrt{9}}\) | M1 | \(\bar{x} \pm t\text{-value} \times \dfrac{\sqrt{\text{their Var}}}{\sqrt{9}}\) |
| \(= (47.93\ldots,\ 56.06\ldots)\) | A1 | awrt 47.9 and awrt 56.1 |
| (ii) \(\dfrac{8 \times 28}{17.535} < \sigma^2 < \dfrac{8 \times 28}{2.180}\) | M1 B1 | M1: \(\dfrac{8(\text{their } s)^2}{\chi^2}\); B1: awrt 17.535 and awrt 2.18 |
| \(12.77 < \sigma^2 < 102.75\) | — | |
| \(3.57 < \sigma < 10.14\) | M1d A1 (9) | M1d: dep on previous M; rearranging to interval for \(\sigma\) — must square root; A1: awrt 3.57 and 10.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(38 \times 1.2 = 45.6\) or \(26\%\) or \(1.26\) | B1 | 45.6 seen or awrt 26% or awrt 1.26 |
| 45.6 is below the CI for *Fruity* therefore there is evidence that the mean for *Zesty* is more than 20% higher than his Fruity | M1 | Correct reason for mean ft their CI |
| 5.5 is in the CI therefore there is no evidence that the standard deviation is less than 5.5 | M1 | Correct reason for sd ft their CI |
| He should not change to *Zesty* | A1cso (4) | Correct conclusion; not ft on incorrect intervals |
## Question 2(a):
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $\bar{x} = \dfrac{468}{9} = 52$, $s^2 = \dfrac{9}{8}\!\left(\dfrac{24560}{9} - 52^2\right) = 28$ | M1A1 | M1: attempting $s$ or $s^2$; A1: 28 only |
| **(i)** $t_8 = 2.306$ | B1 | CV awrt 2.306 |
| $95\%\ \text{CI} = 52 \pm 2.306 \times \dfrac{\sqrt{``28''}}{\sqrt{9}}$ | M1 | $\bar{x} \pm t\text{-value} \times \dfrac{\sqrt{\text{their Var}}}{\sqrt{9}}$ |
| $= (47.93\ldots,\ 56.06\ldots)$ | A1 | awrt 47.9 and awrt 56.1 |
| **(ii)** $\dfrac{8 \times 28}{17.535} < \sigma^2 < \dfrac{8 \times 28}{2.180}$ | M1 B1 | M1: $\dfrac{8(\text{their } s)^2}{\chi^2}$; B1: awrt 17.535 and awrt 2.18 |
| $12.77 < \sigma^2 < 102.75$ | — | |
| $3.57 < \sigma < 10.14$ | M1d A1 (9) | M1d: dep on previous M; rearranging to interval for $\sigma$ — must square root; A1: awrt 3.57 and 10.1 |
---
## Question 2(b):
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $38 \times 1.2 = 45.6$ or $26\%$ or $1.26$ | B1 | 45.6 seen or awrt 26% or awrt 1.26 |
| 45.6 is **below the CI** for *Fruity* therefore there is evidence that the **mean** for *Zesty* is more than **20% higher** than his Fruity | M1 | Correct reason for mean ft their CI |
| 5.5 is **in the CI** therefore there is no evidence that the **standard deviation** is less than 5.5 | M1 | Correct reason for sd ft their CI |
| He should not change to *Zesty* | A1cso (4) | Correct conclusion; not ft on incorrect intervals |\begin{enumerate}
\item Jeremiah currently uses a Fruity model of juicer. He agrees to trial a new model of juicer, Zesty. The amounts of juice extracted, $x \mathrm { ml }$, from each of 9 randomly selected oranges, using the Zesty are summarised as
\end{enumerate}
$$\sum x = 468 \quad \sum x ^ { 2 } = 24560$$
Given that the amounts of juice extracted follow a normal distribution,\\
(a) calculate a 95\% confidence interval for\\
(i) the mean amount of juice extracted from an orange using the Zesty,\\
(ii) the standard deviation of the amount of juice extracted from an orange using the Zesty.
Jeremiah knows that, for his Fruity, the mean amount of juice extracted from an orange is 38 ml and the standard deviation of juice extracted from an orange is 5 ml .
He decides that he will replace his Fruity with a Zesty if both
\begin{itemize}
\item the mean for the Zesty is more than $20 \%$ higher than the mean for his Fruity and
\item the standard deviation for the Zesty is less than 5.5 ml .\\
(b) Using your answers to part (a), explain whether or not Jeremiah should replace his Fruity with the Zesty.\\
\end{itemize}
\hfill \mbox{\textit{Edexcel S4 2018 Q2 [13]}}