| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2023 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Two-sample z-test large samples |
| Difficulty | Standard +0.3 This is a standard two-sample t-test with clearly provided summary statistics and straightforward hypotheses. Part (a) requires routine application of the test procedure (calculating pooled variance, test statistic, comparing to critical value), while part (b) involves simple arithmetic to compare profit per plant. The question is slightly easier than average because all necessary values are given, the context is clear, and no conceptual difficulties arise—it's a textbook application of S3 material. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.05c Hypothesis test: normal distribution for population mean |
| Mean | Standard deviation | Number of plants | |
| Fertiliser A | 1377 | 17.8 | 50 |
| Fertiliser B | 1368 | 18.4 | 40 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \mu_A = \mu_B\); \(H_1: \mu_A > \mu_B\) | B1 | Both hypotheses correct. Allow equivalent. Must be in terms of \(\mu\). If A and B not used, letter must be defined |
| \(se = \sqrt{\frac{17.8^2}{50}+\frac{18.4^2}{40}}\) | M1 | Correct attempt to find se or \(se^2\). Condone slip in sample sizes. May be implied by \(se =\) awrt 3.85 or \(se^2 =\) awrt 14.8. Allow for p-value of 0.0096 or awrt 0.0097 |
| \(z = \pm\frac{1377-1368}{\sqrt{\frac{17.8^2}{50}+\frac{18.4^2}{40}}}\) | M1 | Attempt to find \(z\) value. Allow slip in sample sizes and/or use of 17.8 and 18.4 rather than \(17.8^2\) and \(18.4^2\). Allow for p-value of 0.0096 or awrt 0.0097 |
| \(= \pm 2.339\ldots\) awrt \(\pm 2.34\) | A1 | awrt \(\pm2.34\). Allow for p-value of 0.0096 or awrt 0.0097 |
| One tailed c.v. \( | Z | = 2.3263\) or CR: \(Z \leqslant -2.3263\) or \(Z \geqslant 2.3263\) |
| In CR/Significant/Reject \(H_0\) | dM1 | Dep on previous dM1 awarded, ft their test statistic and CV only |
| Sufficient evidence to support that the mean yield from plants using fertiliser A is greater than the mean yield from plants using fertiliser B | A1ft | ft their \(z\) value and CR only. Correct contextual statement compatible with their test statistic and CV with context of yield and A and B |
| ALT (finding the CI): award B1M1M1A1B1M1A1 unless test statistic given. Award M1 for \(z=\pm\frac{D}{\sqrt{\frac{17.8^2}{50}+\frac{18.4^2}{40}}}\) dep on first M1 where \(2.3 \leqslant z \leqslant 2.4\). May be implied by \( | D | =8.949\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(A: 3\times1.377-\frac{75}{50}\) \(\quad\) \(B: 3\times1.368-\frac{50}{40}\) | M1 | Correct method for profit per \(n\) plants or \(m\) kg for either fertiliser. \(n(3\times1.377-\frac{75}{50})\) or \(n(3\times1.368-\frac{50}{40})\) or \(m(3-\frac{75}{50\times1.377})\) or \(m(3-\frac{50}{40\times1.368})\) where \(n\) and \(m\neq0\) |
| \(A\): £2.63(1) \(\quad\) \(B\): £2.85(4) | A1 | Must have 2 values which can be compared. Profit per \(n\) plant £2.63(1)\(n\) and £2.85(4)\(n\) or profit per \(m\) kg awrt £1.91\(m\) and awrt £2.09\(m\) (2dp) or cost per \(m\) kg awrt £1.09\(m\) and awrt £0.91\(m\) or number plants per £\(y\) awrt £0.38…\(y\) and awrt 0.35…\(y\) |
| Claire should use fertiliser \(B\) | dA1 | Dependent on 1st A1 being awarded. Correct statement |
# Question 5:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \mu_A = \mu_B$; $H_1: \mu_A > \mu_B$ | B1 | Both hypotheses correct. Allow equivalent. Must be in terms of $\mu$. If A and B not used, letter must be defined |
| $se = \sqrt{\frac{17.8^2}{50}+\frac{18.4^2}{40}}$ | M1 | Correct attempt to find se or $se^2$. Condone slip in sample sizes. May be implied by $se =$ awrt 3.85 or $se^2 =$ awrt 14.8. Allow for p-value of 0.0096 or awrt 0.0097 |
| $z = \pm\frac{1377-1368}{\sqrt{\frac{17.8^2}{50}+\frac{18.4^2}{40}}}$ | M1 | Attempt to find $z$ value. Allow slip in sample sizes and/or use of 17.8 and 18.4 rather than $17.8^2$ and $18.4^2$. Allow for p-value of 0.0096 or awrt 0.0097 |
| $= \pm 2.339\ldots$ awrt $\pm 2.34$ | A1 | awrt $\pm2.34$. Allow for p-value of 0.0096 or awrt 0.0097 |
| One tailed c.v. $|Z| = 2.3263$ or CR: $Z \leqslant -2.3263$ or $Z \geqslant 2.3263$ | B1 | $\pm2.3263$ or better seen. Must be compatible with their test statistic |
| In CR/Significant/Reject $H_0$ | dM1 | Dep on previous dM1 awarded, ft their test statistic and CV only |
| Sufficient evidence to support that the mean **yield** from plants using fertiliser **A** is greater than the mean **yield** from plants using fertiliser **B** | A1ft | ft their $z$ value and CR only. Correct contextual statement compatible with their test statistic and CV with context of yield and A and B |
**ALT (finding the CI):** award B1M1M1A1B1M1A1 unless test statistic given. Award M1 for $z=\pm\frac{D}{\sqrt{\frac{17.8^2}{50}+\frac{18.4^2}{40}}}$ dep on first M1 where $2.3 \leqslant z \leqslant 2.4$. May be implied by $|D|=8.949$
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $A: 3\times1.377-\frac{75}{50}$ $\quad$ $B: 3\times1.368-\frac{50}{40}$ | M1 | Correct method for profit per $n$ plants or $m$ kg for either fertiliser. $n(3\times1.377-\frac{75}{50})$ or $n(3\times1.368-\frac{50}{40})$ or $m(3-\frac{75}{50\times1.377})$ or $m(3-\frac{50}{40\times1.368})$ where $n$ and $m\neq0$ |
| $A$: £2.63(1) $\quad$ $B$: £2.85(4) | A1 | Must have 2 values which can be compared. Profit per $n$ plant £2.63(1)$n$ and £2.85(4)$n$ or profit per $m$ kg awrt £1.91$m$ and awrt £2.09$m$ (2dp) or cost per $m$ kg awrt £1.09$m$ and awrt £0.91$m$ or number plants per £$y$ awrt £0.38…$y$ and awrt 0.35…$y$ |
| Claire should use fertiliser $B$ | dA1 | Dependent on 1st A1 being awarded. Correct statement |
---5 Claire grows strawberries on her farm. She wants to compare two brands of fertiliser, brand $A$ and brand $B$.
She grows two sets of plants of the same variety of strawberries under the same conditions, fertilising one set with brand $A$ and the other with brand $B$.
The yields per plant, in grams, from each set of plants are summarised below.
\begin{center}
\begin{tabular}{|l|l|l|l|}
\hline
& Mean & Standard deviation & Number of plants \\
\hline
Fertiliser A & 1377 & 17.8 & 50 \\
\hline
Fertiliser B & 1368 & 18.4 & 40 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Stating your hypotheses clearly, carry out a suitable test to assess whether the mean yield from plants using fertiliser $A$ is greater than the mean yield from plants using fertiliser $B$.\\
Use a 1\% level of significance and state your test statistic and critical value.
The total cost of fertiliser $A$ for Claire's 50 plants was $\pounds 75$\\
The total cost of fertiliser $B$ for Claire's 40 plants was $\pounds 50$\\
Claire sells all her strawberries at $\pounds 3$ per kilogram.
\item Use this information, together with your answer in part (a), to advise Claire on which of the two brands of fertiliser she should use next year in order to maximise her expected profit per plant, giving a reason for your answer.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2023 Q5 [10]}}