Edexcel FS2 2023 June — Question 2 12 marks

Exam BoardEdexcel
ModuleFS2 (Further Statistics 2)
Year2023
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicZ-tests (known variance)
TypeKnown variance (z-distribution)
DifficultyStandard +0.3 This is a standard Further Statistics 2 question testing routine confidence intervals and hypothesis testing with known variance. Parts (a), (b), and (e) are direct applications of formulas. Part (c) requires algebraic manipulation to find n, which is slightly more challenging but still a standard exercise. Part (d) tests understanding of CLT. All techniques are textbook procedures with no novel insight required, making it slightly easier than average even for Further Maths.
Spec5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution

  1. Camilo grows two types of apple, green apples and red apples.
The standard deviation of the weights of green apples is known to be 3.5 grams.
A random sample of 80 green apples has a mean weight of 128 grams.
  1. Find a 98\% confidence interval for the mean weight of the population of green apples. Show your working clearly and give the confidence interval limits to 2 decimal places. Camilo believes that the mean weight of the population of green apples is more than 10 grams greater than the mean weight of the population of red apples. A random sample of \(n\) red apples has a mean weight of 117 grams.
    The standard deviation of the weights of the red apples is known to be 4 grams.
    A test of Camilo's belief is carried out at the 5\% level of significance.
  2. State the null and alternative hypotheses for this test.
  3. Find the smallest value of \(n\) for which the null hypothesis will be rejected.
  4. Explain the relevance of the Central Limit Theorem in parts (a) and (c).
  5. Given that \(n = 85\), state the conclusion of the hypothesis test.

Question 2:
Part 2(a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(z = 2.3263\)B1 2.3263 or better
98% CI for \(\mu\): \(128 \pm 2.3263 \times \frac{3.5}{\sqrt{80}}\)M1 For \(128 \pm \sqrt{\frac{3.5^2}{80}} \times\) "z value", \(
\(= (127.0896..., 128.9103...) = \text{awrt } (127.09, 128.91)\)A1 awrt 127.09 and awrt 128.91
Part 2(b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(H_0: \mu_{g(reen)} - \mu_{r(ed)} = 10\), \(H_1: \mu_{g(reen)} - \mu_{r(ed)} > 10\)B1 Both hypotheses correct with correct notation (if using \(\mu_x\) and \(\mu_y\) these must be defined)
Part 2(c):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Normal model: \(\bar{G} - \bar{R} \sim \text{N}\left(10, \frac{3.5^2}{80} + \frac{4^2}{n}\right)\)M1 Setting up normal model with correct mean or variance. May be shown in standardisation with correct mean or variance and \(
A1Correct variance and mean, may be seen in standardisation
\(z = \frac{(128-117)-10}{\sqrt{\frac{3.5^2}{80} + \frac{4^2}{n}}} > 1.6449\)M1 Setting up equation or inequality by standardising with their mean and standard deviation and \(
B1Correct critical value 1.6449 or better, allow 1.64 or better
\(\frac{1}{1.6449} > \sqrt{\frac{3.5^2}{80} + \frac{4^2}{n}} \rightarrow \left(\frac{1}{1.6449}\right)^2 - \frac{12.25}{80} > \frac{16}{n} \rightarrow n > 73.9...\)M1 Rearranging to solve for \(n\) (may be implied by 73.9...)
\(n = 74\)A1 cao
Part 2(d):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Sample sizes are large so CLT means even though we do not know the distributions of \(G\) and \(R\), the means \(\bar{G}\) and \(\bar{G}-\bar{R}\) will be (approximately) normally distributed.B1 Correct explanation mentioning large sample size and the means being distributed normally
Part 2(e):
AnswerMarks Guidance
Answer/WorkingMark Guidance
[Since \(85 > 73.9...\)] there is significant evidence to support Camilo's belief/the mean weight of green apples is more than 10 grams greater than red apples is supported.B1ft Drawing a correct inference in context, must be consistent with their value of \(n\)
# Question 2:

## Part 2(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $z = 2.3263$ | B1 | 2.3263 or better |
| 98% CI for $\mu$: $128 \pm 2.3263 \times \frac{3.5}{\sqrt{80}}$ | M1 | For $128 \pm \sqrt{\frac{3.5^2}{80}} \times$ "z value", $|z|>1$. May be implied by correct CI |
| $= (127.0896..., 128.9103...) = \text{awrt } (127.09, 128.91)$ | A1 | awrt **127.09** and awrt **128.91** |

## Part 2(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \mu_{g(reen)} - \mu_{r(ed)} = 10$, $H_1: \mu_{g(reen)} - \mu_{r(ed)} > 10$ | B1 | Both hypotheses correct with correct notation (if using $\mu_x$ and $\mu_y$ these must be defined) |

## Part 2(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Normal model: $\bar{G} - \bar{R} \sim \text{N}\left(10, \frac{3.5^2}{80} + \frac{4^2}{n}\right)$ | M1 | Setting up normal model with correct mean or variance. May be shown in standardisation with correct mean or variance and $|z|>1$ |
| | A1 | Correct variance and mean, may be seen in standardisation |
| $z = \frac{(128-117)-10}{\sqrt{\frac{3.5^2}{80} + \frac{4^2}{n}}} > 1.6449$ | M1 | Setting up equation or inequality by standardising with their mean and standard deviation and $|z|>1$ |
| | B1 | Correct critical value 1.6449 or better, allow **1.64 or better** |
| $\frac{1}{1.6449} > \sqrt{\frac{3.5^2}{80} + \frac{4^2}{n}} \rightarrow \left(\frac{1}{1.6449}\right)^2 - \frac{12.25}{80} > \frac{16}{n} \rightarrow n > 73.9...$ | M1 | Rearranging to solve for $n$ (may be implied by 73.9...) |
| $n = 74$ | A1 | cao |

## Part 2(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sample sizes are large so CLT means even though we do not know the distributions of $G$ and $R$, the means $\bar{G}$ and $\bar{G}-\bar{R}$ will be (approximately) normally distributed. | B1 | Correct explanation mentioning large sample size and the means being distributed normally |

## Part 2(e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| [Since $85 > 73.9...$] there is significant evidence to support Camilo's belief/the mean weight of green apples is more than 10 grams greater than red apples is supported. | B1ft | Drawing a correct inference in context, must be consistent with their value of $n$ |

---
\begin{enumerate}
  \item Camilo grows two types of apple, green apples and red apples.
\end{enumerate}

The standard deviation of the weights of green apples is known to be 3.5 grams.\\
A random sample of 80 green apples has a mean weight of 128 grams.\\
(a) Find a 98\% confidence interval for the mean weight of the population of green apples. Show your working clearly and give the confidence interval limits to 2 decimal places.

Camilo believes that the mean weight of the population of green apples is more than 10 grams greater than the mean weight of the population of red apples.

A random sample of $n$ red apples has a mean weight of 117 grams.\\
The standard deviation of the weights of the red apples is known to be 4 grams.\\
A test of Camilo's belief is carried out at the 5\% level of significance.\\
(b) State the null and alternative hypotheses for this test.\\
(c) Find the smallest value of $n$ for which the null hypothesis will be rejected.\\
(d) Explain the relevance of the Central Limit Theorem in parts (a) and (c).\\
(e) Given that $n = 85$, state the conclusion of the hypothesis test.

\hfill \mbox{\textit{Edexcel FS2 2023 Q2 [12]}}