| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Z-tests (known variance) |
| Type | Two-tail z-test |
| Difficulty | Standard +0.3 This is a straightforward application of a one-sample z-test with clearly stated parameters. Part (a) is a routine normal distribution calculation, part (b) follows a standard hypothesis testing procedure with all information provided, and part (c) asks for a p-value calculation. While it requires multiple steps and proper statistical reasoning, it involves no novel insight or complex problem-solving—just methodical application of standard Further Maths Statistics techniques. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| \([\text{Let } N = \text{height from region } A;\ P(N > 180) =]\ 0.24937\ldots\) awrt \(\mathbf{0.249}\) | B1 (1) | For awrt 0.249. |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0: \mu = 175.4 \quad H_1: \mu \neq 175.4\) | B1 | For both hypotheses correct in terms of \(\mu\). (See below for one-tail test.) |
| \([S = \text{height from region } B]\ \bar{S} \sim N\!\left(175.4,\ \dfrac{6.8^2}{52}\right)\), allow \(\sigma^2 =\) awrt 0.889 | M1 | For selecting the correct model, may be implied by standardisation using correct values or by a correct value in 1st A1. Condone use of \(S\) (or any other letter) instead of \(\bar{S}\). Condone use of \(\bar{S} \sim N\!\left(177.2,\ \frac{6.8^2}{52}\right)\) but this will lose 2nd A mark. |
| \([P(\bar{S} > 177.2)] = 0.02814\ldots\) | A1 | 1st A1 for probability awrt 0.028 (allow 0.03 if \(P(\bar{S} > 177.2)\) is seen). Condone \(1 - 0.02814\ldots = 0.9718\ldots\) (awrt 0..972) only if clearly compared with 0.975. Allow \(Z = 1.9(088\ldots)\) and comparison with 1.96. ALT CR of \([\bar{S}]\ldots 177.248\ldots\) (awrt 177.25). |
| \([0.028\ldots > 0.025, \text{ Not sig, do not reject } H_0]\) Insufficient evidence to support student's claim | A1 (4) | 2nd A1 (dep on 1st A1 and use of correct model) for conclusion using context. Do not allow 2nd A mark for contradictory statements. One-tail test [Max of 3/5 for (b) and (c)]: B0(hyp) M1 1st A1[prob or \(Z\) vs 1.6449] 2nd A1 for conclusion supporting claim or "heights of men from \(B\) is different from/greater than from \(A\)". |
| Answer | Marks | Guidance |
|---|---|---|
| \([p\text{-value} = 2 \times 0.02814\ldots =]\ 0.05628\ldots\) in range \(\mathbf{0.056}\)–\(\mathbf{0.06}\) or \(\mathbf{5.6(\%)}\)–\(\mathbf{6(\%)}\) | B1ft (1) | For answer in range 0.056–0.06 or 5.6%–6% (inclusive, condone missing %). Can ft their probability provided \(< 0.5\), from part (b) but not 0.025 leading to 5%. In (c) B0 for one-tail test. |
# Question 4:
## Part (a)
$[\text{Let } N = \text{height from region } A;\ P(N > 180) =]\ 0.24937\ldots$ awrt $\mathbf{0.249}$ | B1 **(1)** | For awrt 0.249.
## Part (b)
$H_0: \mu = 175.4 \quad H_1: \mu \neq 175.4$ | B1 | For both hypotheses correct in terms of $\mu$. **(See below for one-tail test.)**
$[S = \text{height from region } B]\ \bar{S} \sim N\!\left(175.4,\ \dfrac{6.8^2}{52}\right)$, allow $\sigma^2 =$ awrt 0.889 | M1 | For selecting the correct model, may be implied by standardisation using correct values or by a correct value in 1st A1. Condone use of $S$ (or any other letter) instead of $\bar{S}$. Condone use of $\bar{S} \sim N\!\left(177.2,\ \frac{6.8^2}{52}\right)$ **but this will lose 2nd A mark**.
$[P(\bar{S} > 177.2)] = 0.02814\ldots$ | A1 | 1st A1 for probability awrt 0.028 (allow 0.03 if $P(\bar{S} > 177.2)$ is seen). Condone $1 - 0.02814\ldots = 0.9718\ldots$ (awrt 0..972) **only if clearly compared with 0.975**. Allow $Z = 1.9(088\ldots)$ and comparison with 1.96. **ALT** CR of $[\bar{S}]\ldots 177.248\ldots$ (awrt 177.25).
$[0.028\ldots > 0.025, \text{ Not sig, do not reject } H_0]$ Insufficient evidence to support student's claim | A1 **(4)** | 2nd A1 (dep on 1st A1 and use of correct model) for conclusion using context. Do not allow 2nd A mark for contradictory statements. **One-tail test [Max of 3/5 for (b) and (c)]:** B0(hyp) M1 1st A1[prob or $Z$ vs 1.6449] 2nd A1 for conclusion supporting claim or "heights of men from $B$ is different from/greater than from $A$".
## Part (c)
$[p\text{-value} = 2 \times 0.02814\ldots =]\ 0.05628\ldots$ in range $\mathbf{0.056}$–$\mathbf{0.06}$ or $\mathbf{5.6(\%)}$–$\mathbf{6(\%)}$ | B1ft **(1)** | For answer in range 0.056–0.06 or 5.6%–6% (inclusive, condone missing %). Can ft their probability provided $< 0.5$, from part (b) but not 0.025 leading to 5%. In (c) B0 for one-tail test.
---\begin{enumerate}
\item A study was made of adult men from region $A$ of a country. It was found that their heights were normally distributed with a mean of 175.4 cm and standard deviation 6.8 cm .\\
(a) Find the proportion of these men that are taller than 180 cm .
\end{enumerate}
A student claimed that the mean height of adult men from region $B$ of this country was different from the mean height of adult men from region $A$.
A random sample of 52 adult men from region $B$ had a mean height of 177.2 cm\\
The student assumed that the standard deviation of heights of adult men was 6.8 cm both for region $A$ and region $B$.\\
(b) Use a suitable test to assess the student's claim.
You should
\begin{itemize}
\item state your hypotheses clearly
\item use a $5 \%$ level of significance\\
(c) Find the $p$-value for the test in part (b)
\end{itemize}
\hfill \mbox{\textit{Edexcel Paper 3 2023 Q4 [6]}}