| Exam Board | Edexcel |
|---|---|
| Module | FS1 (Further Statistics 1) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | State hypotheses only |
| Difficulty | Easy -1.2 Part (a) asks only to state hypotheses for a one-tailed binomial test, which is straightforward recall requiring no calculation or problem-solving. This is a routine bookwork question testing basic understanding of hypothesis notation (H₀: p = 0.03, H₁: p > 0.03), significantly easier than average A-level questions that typically require multi-step calculations or application of techniques. |
| Spec | 5.02b Expectation and variance: discrete random variables5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(H_0: p = 0.03\) \(\quad H_1: p > 0.03\) | B1 |
| (b) | \(D_{100} \sim B(100, 0.03)\) \([P(D_{100} \geq 5) = 1 - P(D_{100} \leq 4)]\) | M1 |
| \(= 0.18214\ldots\) awrt \(\mathbf{0.182}\) | A1 | 1.1b |
| (c) | \(P(D_{80} \geq 5) + P(D_{80} = 4) \times P(D_{80} > 1)\) | M1 |
| \(= 0.09279\ldots + 0.12654\ldots = 0.20826\ldots\) awrt \(\mathbf{0.208}\) | A1, A1 | 1.1b, 1.1b |
| (d)(i) | Test A: \([X \sim B(100, 0.06) P(X \geq 5) =] 0.72322\ldots\) awrt \(\mathbf{0.723}\) | B1 |
| (d)(ii) | Expected number \(= 80 + 80 \times P(Y = 4)\) with \(Y \sim B(80, 0.06) = 94.87\ldots = \mathbf{95}\) | M1, A1 |
| (e) | Tests of similar size and power but Test B involves sampling fewer components so would advise to use test B. | B1 |
(a) | $H_0: p = 0.03$ $\quad H_1: p > 0.03$ | B1 | 2.5 |
(b) | $D_{100} \sim B(100, 0.03)$ $[P(D_{100} \geq 5) = 1 - P(D_{100} \leq 4)]$ | M1 | 3.3 |
| $= 0.18214\ldots$ awrt $\mathbf{0.182}$ | A1 | 1.1b |
(c) | $P(D_{80} \geq 5) + P(D_{80} = 4) \times P(D_{80} > 1)$ | M1 | 2.1 |
| $= 0.09279\ldots + 0.12654\ldots = 0.20826\ldots$ awrt $\mathbf{0.208}$ | A1, A1 | 1.1b, 1.1b |
(d)(i) | Test A: $[X \sim B(100, 0.06) P(X \geq 5) =] 0.72322\ldots$ awrt $\mathbf{0.723}$ | B1 | 1.2 |
(d)(ii) | Expected number $= 80 + 80 \times P(Y = 4)$ with $Y \sim B(80, 0.06) = 94.87\ldots = \mathbf{95}$ | M1, A1 | 3.4, 1.1b |
(e) | Tests of similar size and power but Test B involves sampling fewer components so would advise to use test B. | B1 | 2.4 |
**Notes:**
- (a) B1 for both hypotheses in terms of $p$ or $\pi$
- (b) M1 for sight or use of the correct model. Allow for $1 - 0.81785\ldots$ or $1 - 0.91916\ldots$
- (b) A1 for awrt 0.182
- (c) M1 for a correct expression for required probability. May be implied by 1st A1
- (c) 1st A1 for correct numerical expression – values to 2 s.f. or better
- (c) 2nd A1 for awrt 0.208
- (d)(i) B1 for power of test A = awrt 0.723
- (d)(ii) M1 Correct expression with $Y \sim B(80, 0.06)$ or for use of B(80, 0.06) to obtain a probability of 0.814 or 0.186
- (d)(ii) Implied by sight of 95 or better
- (d)(ii) A1 for 95 (accept awrt 94.9)
- (d)(ii) **NB:** May see 0.814 × 80 + 0.186 × 160 = awrt 94.9 or 95, which is M1A1
- (e) B1 for a choice backed up by a suitable reason:
- If concluding B, they need to mention similar power and a smaller sample size
- If concluding A, they need to mention smaller size and greater power
- (e) **NB:** We do not fit the candidate's incorrect values for size or power
- **SC: Use of Poisson approximation**
- (b) Allow M1 for Po(3) and A1 for answer of 0.1847 or better
- (c) Allow M1 and 1st A1 for awrt 0.21 but 2nd A0 (Po(2.4) gives 0.2099…)
- (d)(i) Allow B1 for 0.7149 or better from Po(6)
- (d)(ii) Allow M1 for expression (answer should be 94.56…) A0 for answer.
---\begin{enumerate}
\item Some of the components produced by a factory are defective. The management requires that no more than $3 \%$ of the components produced are defective.\\
Niluki monitors the production process and takes a random sample of $n$ components.\\
(a) Write down the hypotheses Niluki should use in a test to assess whether or not the proportion of defective components is greater than 0.03
\end{enumerate}
Niluki defines the random variable $D _ { n }$ to represent the number of defective components in a sample of size $n$. She considers two tests $\mathbf { A }$ and $\mathbf { B }$
In test $\mathbf { A }$, Niluki uses $n = 100$ and if $D _ { 100 } \geqslant 5$ she rejects $H _ { 0 }$\\
(b) Find the size of test $\mathbf { A }$
In test B, Niluki uses $n = 80$ and
\begin{itemize}
\item if $D _ { 80 } \geqslant 5$ she rejects $\mathrm { H } _ { 0 }$
\item if $D _ { 80 } \leqslant 3$ she does not reject $\mathrm { H } _ { 0 }$
\item if $D _ { 80 } = 4$ she takes a second random sample of size 80 and if $D _ { 80 } \geqslant 1$ in this second sample then she rejects $\mathrm { H } _ { 0 }$ otherwise she does not reject $\mathrm { H } _ { 0 }$\\
(c) Find the size of test $\mathbf { B }$
\end{itemize}
Given that the actual proportion of defective components is 0.06\\
(d) (i) find the power of test $\mathbf { A }$\\
(ii) find the expected number of components sampled using test $\mathbf { B }$
Given also that, when the actual proportion of defective components is 0.06 , the power of test $\mathbf { B }$ is 0.713\\
(e) suggest, giving your reasons, which test Niluki should use.
\hfill \mbox{\textit{Edexcel FS1 2024 Q5 [10]}}