| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | One-tailed hypothesis test (lower tail, H₁: p < p₀) |
| Difficulty | Moderate -0.3 This is a straightforward one-tailed binomial hypothesis test with standard structure. Part (a) involves routine binomial probability calculations using tables or calculator. Part (b) requires setting up H₀: p=0.12 vs H₁: p<0.12, finding the critical region or p-value for X≤3 with n=60, and making a conclusion—all standard AS-level procedure with no conceptual challenges. Slightly easier than average due to clear context and small test statistic making calculations manageable. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0.153588\ldots\) awrt \(\mathbf{0.154}\) | B1 | For awrt 0.154 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X \leqslant 14) - P(X \leqslant 11)\) with at least one value; \(P(X \leqslant 14) = 0.97707\ldots\); \(P(X \leqslant 11) = 0.797603\ldots\) | M1 | For correct expression with at least one correct probability substituted (2sf truncated or rounded). Alt: \(P(X=12)+P(X=13)+P(X=14)\) with at least one correct term |
| \(= 0.17947\ldots\) awrt \(\mathbf{0.179}\) | A1 | Allow 0.1795. Correct answers score 2 out of 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: p = 0.12 \quad H_1: p < 0.12\) | B1 | For both hypotheses correct in terms of \(p\) or \(\pi\) |
| \([D = \text{no. of defective items in sample}]\ D \sim B(60, 0.12)\) | M1 | For sight or correct use of \(B(60, 0.12)\). Implied by awrt 0.0601 or awrt 0.0405 or awrt 0.0196 |
| \([P(D \leqslant 3)] = 0.06013\ldots\) awrt \(\mathbf{0.060}\) or \([P(D \leqslant 2)] =\) awrt 0.0196 with reference to CR | A1 | 1st A1 for final answer awrt 0.060 (allow 0.06 if \(P(D,,3)\) is seen with \(B(60,0.12)\)). NB: \(P(D,,2) =\) awrt 0.0196 on its own scores A0 here |
| \([0.06\ldots > 5\%\) not significant, do not reject \(H_0]\) Insufficient evidence that proportion of defective items has decreased | A1 | 2nd A1 (dep on M1A1 but independent of hypotheses). Must NOT reject \(H_0\) (if stated) and mention underlined words o.e. Allow e.g. 'no' for insufficient; allow proportion/probability/percentage but not number |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| "\(0.06\ldots\)" | B1ft | For 0.06 or better. Allow as percentage. ft their final (\(p\)-value) answer from part (b) to 1sf. NB: using a critical region approach in (b) scores B0ft if they state their CR probability as the \(p\)-value |
## Question 4:
**Part (a)(i):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $0.153588\ldots$ awrt $\mathbf{0.154}$ | B1 | For awrt 0.154 |
**Part (a)(ii):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X \leqslant 14) - P(X \leqslant 11)$ with at least one value; $P(X \leqslant 14) = 0.97707\ldots$; $P(X \leqslant 11) = 0.797603\ldots$ | M1 | For correct expression with at least one correct probability substituted (2sf truncated or rounded). Alt: $P(X=12)+P(X=13)+P(X=14)$ with at least one correct term |
| $= 0.17947\ldots$ awrt $\mathbf{0.179}$ | A1 | Allow 0.1795. Correct answers score 2 out of 2 |
**Part (b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: p = 0.12 \quad H_1: p < 0.12$ | B1 | For both hypotheses correct in terms of $p$ or $\pi$ |
| $[D = \text{no. of defective items in sample}]\ D \sim B(60, 0.12)$ | M1 | For sight or correct use of $B(60, 0.12)$. Implied by awrt 0.0601 or awrt 0.0405 or awrt 0.0196 |
| $[P(D \leqslant 3)] = 0.06013\ldots$ awrt $\mathbf{0.060}$ **or** $[P(D \leqslant 2)] =$ awrt 0.0196 with reference to CR | A1 | 1st A1 for **final answer** awrt 0.060 (allow 0.06 if $P(D,,3)$ is seen with $B(60,0.12)$). NB: $P(D,,2) =$ awrt 0.0196 **on its own** scores A0 here |
| $[0.06\ldots > 5\%$ not significant, do not reject $H_0]$ Insufficient evidence that proportion of defective items has decreased | A1 | 2nd A1 (dep on M1A1 but independent of hypotheses). Must NOT reject $H_0$ (if stated) **and** mention underlined words o.e. Allow e.g. 'no' for insufficient; allow proportion/probability/percentage but **not** number |
**Part (c):**
| Answer | Mark | Guidance |
|--------|------|----------|
| "$0.06\ldots$" | B1ft | For 0.06 or better. Allow as percentage. ft their **final** ($p$-value) **answer** from part (b) to 1sf. NB: using a **critical region** approach in (b) scores B0ft if they state their CR probability as the $p$-value |
---4. The random variable $X \sim \mathrm {~B} ( 27,0.35 )$
\begin{enumerate}[label=(\alph*)]
\item Find
\begin{enumerate}[label=(\roman*)]
\item $\mathrm { P } ( X = 10 )$
\item $\mathrm { P } ( 12 \leqslant X < 15 )$
Historical records show that the proportion of defective items produced by a machine is 0.12
Following a maintenance service of the machine, a random sample of 60 items is taken and 3 defective items are found.
\end{enumerate}\item Carry out a suitable test to determine whether the proportion of defective items produced by the machine has decreased following the maintenance service. You should state your hypotheses clearly and use a $5 \%$ level of significance.
\item Write down the $p$-value for your test in part (b)
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 2 2024 Q4 [8]}}