| Exam Board | AQA |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| 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 standard one-tailed binomial hypothesis test with straightforward setup (H₀: p=0.12, H₁: p<0.12) and calculation of P(X≤4) using tables or calculator. Part (b) requires stating standard assumptions (independence, constant probability). While it requires proper statistical method and interpretation, it follows a well-practiced template with no conceptual surprises, making it slightly easier than average. |
| Spec | 2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail |
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| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: p = 0.12\), \(H_1: p < 0.12\) (one-tailed test, population proportion \(= 0.12\)) | B1 | AO 2.5 |
| States \(X \sim B(60, 0.12)\) under null hypothesis | M1 | AO 3.3 |
| Calculates \(P(X \leq 4) = 0.139\) or \(P(X \leq 3) = 0.060\); do not accept \(P(X=4)\) or \(P(X=3)\) | M1 | AO 1.1a |
| Correct value \(P(X \leq 4) = 0.139\) (accept 0.14) | A1 | AO 1.1b |
| Evaluates Binomial model by comparing \(0.139\) (accept \(0.14\)) or \(0.060\) with \(0.10\); must be clear comparison | M1 | AO 3.5a |
| Infers \(H_0\) accepted or \(H_1\) rejected (condone 'do not reject') | A1 | AO 2.2b |
| There is insufficient evidence to suggest that the proportion of faulty chargers has reduced | R1 | AO 3.2a — only award for full complete solution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Calculates \(P(X \leq 4)\) and \(P(X \leq 3)\) but not \(P(X=4)\) and \(P(X=3)\) | M1 | AO 1.1a |
| Identifies correct critical region \(X \leq 3\); must have considered both \(P(X \leq 4)\) and \(P(X \leq 3)\) | A1 | AO 1.1b |
| Compares \(X = 4\) or \(X = 3\) with critical region; must be clear comparison | M1 | AO 3.5a |
| As 4 does not lie in the critical region, accept \(H_0\) | A1 | AO 2.2b |
| There is insufficient evidence to suggest that the proportion of faulty chargers has reduced | R1 | AO 3.2a — only award for full complete solution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The probability of a faulty charger is fixed | E1 | AO 3.5b — must include 'faulty'; do not accept "number of trials is fixed at 60" or "charger is either faulty or not faulty" |
| A charger being faulty is independent of any other charger being faulty; also accept: the sample of chargers was randomly selected | E1 | AO 3.5b — must include 'faulty' if assumption refers to probability or independence |
## Question 16:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: p = 0.12$, $H_1: p < 0.12$ (one-tailed test, population proportion $= 0.12$) | B1 | AO 2.5 |
| States $X \sim B(60, 0.12)$ under null hypothesis | M1 | AO 3.3 |
| Calculates $P(X \leq 4) = 0.139$ or $P(X \leq 3) = 0.060$; do **not** accept $P(X=4)$ or $P(X=3)$ | M1 | AO 1.1a |
| Correct value $P(X \leq 4) = 0.139$ (accept 0.14) | A1 | AO 1.1b |
| Evaluates Binomial model by comparing $0.139$ (accept $0.14$) or $0.060$ with $0.10$; must be clear comparison | M1 | AO 3.5a |
| Infers $H_0$ accepted or $H_1$ rejected (condone 'do not reject') | A1 | AO 2.2b |
| There is insufficient evidence to suggest that the proportion of faulty chargers has reduced | R1 | AO 3.2a — only award for full complete solution |
**Alternative (critical region method):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Calculates $P(X \leq 4)$ **and** $P(X \leq 3)$ but **not** $P(X=4)$ and $P(X=3)$ | M1 | AO 1.1a |
| Identifies correct critical region $X \leq 3$; must have considered both $P(X \leq 4)$ **and** $P(X \leq 3)$ | A1 | AO 1.1b |
| Compares $X = 4$ or $X = 3$ with critical region; must be clear comparison | M1 | AO 3.5a |
| As 4 does not lie in the critical region, accept $H_0$ | A1 | AO 2.2b |
| There is insufficient evidence to suggest that the proportion of faulty chargers has reduced | R1 | AO 3.2a — only award for full complete solution |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| The probability of a faulty charger is fixed | E1 | AO 3.5b — must include 'faulty'; do **not** accept "number of trials is fixed at 60" or "charger is either faulty or not faulty" |
| A charger being faulty is independent of any other charger being faulty; also accept: the sample of chargers was randomly selected | E1 | AO 3.5b — must include 'faulty' if assumption refers to probability or independence |\begin{center}
\begin{tabular}{|c|l|}
\hline
16 \\
16
\begin{enumerate}[label=(\alph*)]
\item & \begin{tabular}{l}
Andrea is the manager of a company which makes mobile phone chargers. \\
In the past, she had found that $12 \%$ of all chargers are faulty. \\
Andrea decides to move the manufacture of chargers to a different factory. \\
Andrea tests 60 of the new chargers and finds that 4 chargers are faulty. \\
Investigate, at the $10 \%$ level of significance, whether the proportion of faulty chargers has reduced. \\[0pt]
[7 marks] $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ $\_\_\_\_$ \\
\end{tabular} \\
\hline
\end{tabular}
\end{center}
16
\item State, in context, two assumptions that are necessary for the distribution that you have used in part (a) to be valid.
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 2 2019 Q16 [9]}}