| Exam Board | Edexcel |
|---|---|
| Module | FS1 (Further Statistics 1) |
| Year | 2022 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Z-tests (known variance) |
| Type | Two-tail z-test |
| Difficulty | Challenging +1.2 This is a standard hypothesis testing question with three routine parts: (a) performing a two-tailed z-test with known variance, (b) finding a critical value expression for a one-tailed test, and (c) using Type II error probability to find sample size. Part (c) requires solving simultaneous equations involving normal distributions, which adds some computational complexity beyond typical FS1 questions, but the conceptual framework is entirely standard for Further Statistics 1. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\bar{X} \sim N(1000, 90)\) | M1 | May be implied by correct prob or \(z\) value; \(\mu=1000, \sigma^2=90\) or \(\sigma=\sqrt{90}\) or awrt \(9.49\) |
| \(P(\bar{X} > 1020) = 0.0175...\) or \(z = 2.108\) | A1 | Using model to find correct \(z\) value or \(P(\bar{X}>1020)=\) awrt \(0.0175\) |
| \(0.0175... < 0.025\) or \(z=2.108...>1.96\), therefore reject \(H_0\) | M1 | Correct comparison or non-contextual conclusion; dep on \(P(\bar{X}>1020)\); M0 if contradictory statements |
| Evidence mean weight of flour in bag is not \(1000\text{ g}\), or evidence of change in mean weight | A1 (cso) | Dep on M1A1M1; correct conclusion in context with underlined words; do NOT accept "mean weight has increased" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\left[\bar{Y} \sim N\!\left(1000,\frac{900}{n}\right)\right] \Rightarrow \frac{c-1000}{30/\sqrt{n}}=1.6449\) | M1 | \(\frac{c-1000}{30/\sqrt{n}}=z\) where \( |
| \(c = 1000 + \frac{49.347}{\sqrt{n}}\) | A1 | Correct equation in form \(c=\ldots\); use of awrt \(1.6449\) (implied by awrt \(49.3[4]\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\dfrac{\text{"1000}+\frac{49.347}{\sqrt{n}}\text{"}-1020}{30/\sqrt{n}} = -2.5758\) | M1, A1ft | Standardising using their \(c\) and equating to \(z\) (\( |
| \(\frac{126.621}{\sqrt{n}}=20\) or \(\frac{49.34...}{c-1000}=\frac{-77.274}{c-1020}\) | dM1 | Dep 1st M1; isolating/eliminating \(\sqrt{n}\) or \(n\) or eliminating \(c\) |
| \(n = \mathbf{40}\) | A1 | For 40 (allow 41); must be integer with correct working |
| \(c = 1007.8...\) awrt \(\mathbf{1010}\) | A1 | For awrt 1010 from correct working |
# Question 7:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\bar{X} \sim N(1000, 90)$ | M1 | May be implied by correct prob or $z$ value; $\mu=1000, \sigma^2=90$ or $\sigma=\sqrt{90}$ or awrt $9.49$ |
| $P(\bar{X} > 1020) = 0.0175...$ or $z = 2.108$ | A1 | Using model to find correct $z$ value or $P(\bar{X}>1020)=$ awrt $0.0175$ |
| $0.0175... < 0.025$ or $z=2.108...>1.96$, therefore reject $H_0$ | M1 | Correct comparison or non-contextual conclusion; dep on $P(\bar{X}>1020)$; M0 if contradictory statements |
| Evidence mean weight of flour in bag is not $1000\text{ g}$, or evidence of change in mean weight | A1 (cso) | Dep on M1A1M1; correct conclusion in context with underlined words; do NOT accept "mean weight has increased" |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\left[\bar{Y} \sim N\!\left(1000,\frac{900}{n}\right)\right] \Rightarrow \frac{c-1000}{30/\sqrt{n}}=1.6449$ | M1 | $\frac{c-1000}{30/\sqrt{n}}=z$ where $|z|>1.5$; allow any inequality or $=$ for M1; condone $\bar{X}$ used for $c$ |
| $c = 1000 + \frac{49.347}{\sqrt{n}}$ | A1 | Correct equation in form $c=\ldots$; use of awrt $1.6449$ (implied by awrt $49.3[4]$) |
## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\dfrac{\text{"1000}+\frac{49.347}{\sqrt{n}}\text{"}-1020}{30/\sqrt{n}} = -2.5758$ | M1, A1ft | Standardising using their $c$ and equating to $z$ ($|z|>2$); ft their $\sigma$ used in (b) |
| $\frac{126.621}{\sqrt{n}}=20$ or $\frac{49.34...}{c-1000}=\frac{-77.274}{c-1020}$ | dM1 | Dep 1st M1; isolating/eliminating $\sqrt{n}$ or $n$ or eliminating $c$ |
| $n = \mathbf{40}$ | A1 | For 40 (allow 41); must be integer with correct working |
| $c = 1007.8...$ awrt $\mathbf{1010}$ | A1 | For awrt 1010 from correct working |\begin{enumerate}
\item A machine fills bags with flour. The weight of flour delivered by the machine into a bag, $X$ grams, is normally distributed with mean $\mu$ grams and standard deviation 30 grams. To check if there is any change to the mean weight of flour delivered by the machine into each bag, Olaf takes a random sample of 10 bags. The weight of flour, $x$ grams, in each bag is recorded and $\bar { x } = 1020$\\
(a) Test, at the $5 \%$ level of significance, $\mathrm { H } _ { 0 } : \mu = 1000$ against $\mathrm { H } _ { 1 } : \mu \neq 1000$
\end{enumerate}
Olaf decides to alter the test so that the hypotheses are $\mathrm { H } _ { 0 } : \mu = 1000$ and $\mathrm { H } _ { 1 } : \mu > 1000$ but keeps the level of significance at 5\%
He takes a second sample of size $n$ and finds the critical region, $\bar { X } > c$\\
(b) Find an equation for $c$ in terms of $n$
When the true value of $\mu$ is 1020 grams, the probability of making a Type II error is 0.0050 , to 2 significant figures.\\
(c) Calculate the value of $n$ and the value of $c$
\hfill \mbox{\textit{Edexcel FS1 2022 Q7 [11]}}