| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2016 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Expected value and most likely value |
| Difficulty | Standard +0.3 This is a straightforward S2 binomial distribution question with standard parts: finding n from given standard deviation (routine formula manipulation), calculating probabilities using binomial distribution, conditional probability, and a normal approximation hypothesis test. All techniques are textbook exercises requiring recall and careful calculation but no novel insight or complex problem-solving. |
| Spec | 2.04d Normal approximation to binomial5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.02d Binomial: mean np and variance np(1-p)5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\sqrt{n(0.04)(1-0.04)} = 1.44\) or \(n(0.04)(1-0.04) = 1.44^2\) | M1 | Use of s.d. \(= \sqrt{np(1-p)} = 1.44\) with a value of \(p\) in \((0,1)\) or equation with variance |
| \(0.0384n = 2.07(36)\), \(n = 54\) | dM1, A1 | Dep on 1st M1 for solving equation as far as \(an = 2.07(36)\) or \(n = \frac{1.44^2}{a}\) (Ans only 3/3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| mean \(= 54 \times 0.04 = 2.16\) or \(\frac{54}{25}\) | B1cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \([(0.96)^{20}] = 0.44200\ldots\) awrt 0.442 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(X \sim B(20, 0.04)\) | B1 | For writing or using \(B(20, 0.04)\). May be implied by e.g. \(P(X=3) = 0.036449\) |
| \([P(X=3 \mid X \geqslant 1)] = \frac{P(X=3)}{P(X \geqslant 1)}\) | M1 | 1st M1 for a correct ratio expression for conditional prob. \(P(X \geqslant 1)\) may be \(1 - P(X=0)\) |
| \(= \frac{20C3(0.04)^3(0.96)^{17}}{1-(0.96)^{20}}\) or \(\frac{20C3(0.04)^3(0.96)^{17}}{1 - \text{their (b)}} = 0.065322\ldots\) awrt 0.0653 | dM1 A1 | 2nd M1 dep on first M1 for correct attempt at either \(P(X=3)\) or \(P(X \geqslant 1)\) (may ft (b)) (Ans only 4/4) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(H_0: p = 0.04\), \(H_1: p > 0.04\) | B1 | For both hypotheses correct (\(p\) or \(\pi\)). Allow \(H_0: \lambda=5\), \(H_1: \lambda > 5\) \([\geqslant\) is B0] |
| \([X \sim B(125, 0.04)]\) approximated by \(Po(5)\) | B1 | For using \(Po(5)\). May need to check tables e.g. \(P(X \leqslant 10) = 0.9863\) would imply B1 |
| \(P(X \geqslant 10) = 1 - P(X \leqslant 9)\) or \(P(X \leqslant 9) = 0.9682\) | M1 | For writing or using \(1 - P(X \leqslant 9)\) or giving \(P(X \leqslant 9) = 0.9682\) or \(P(X \geqslant 10) = 0.0318\) |
| \(= 1 - 0.9682 = 0.0318\), \(P(X \geqslant 10) = 0.0318\), CR \(X \geqslant 10\) | A1 | For 0.0318 or CR \(X \geqslant 10\) |
| Reject \(H_0\) or Significant/10 lies in the Critical region | dM1 | Dep on 1st M1. Do not allow contradictory statements e.g. "significant, accept \(H_0\)" |
| Evidence that proportion/number/rate/%/probability of cars failing the test is more … or the car mechanic's claim is supported | A1cso | 2nd A1cso for correct contextual conclusion and no errors seen |
# Question 7:
## Part (a)(i):
| Working | Mark | Guidance |
|---------|------|----------|
| $\sqrt{n(0.04)(1-0.04)} = 1.44$ or $n(0.04)(1-0.04) = 1.44^2$ | M1 | Use of s.d. $= \sqrt{np(1-p)} = 1.44$ with a value of $p$ in $(0,1)$ or equation with variance |
| $0.0384n = 2.07(36)$, $n = 54$ | dM1, A1 | Dep on 1st M1 for solving equation as far as $an = 2.07(36)$ or $n = \frac{1.44^2}{a}$ (Ans only 3/3) |
## Part (a)(ii):
| Working | Mark | Guidance |
|---------|------|----------|
| mean $= 54 \times 0.04 = 2.16$ or $\frac{54}{25}$ | B1cao | |
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $[(0.96)^{20}] = 0.44200\ldots$ awrt **0.442** | B1 | |
## Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| $X \sim B(20, 0.04)$ | B1 | For writing or using $B(20, 0.04)$. May be implied by e.g. $P(X=3) = 0.036449$ |
| $[P(X=3 \mid X \geqslant 1)] = \frac{P(X=3)}{P(X \geqslant 1)}$ | M1 | 1st M1 for a correct ratio expression for conditional prob. $P(X \geqslant 1)$ may be $1 - P(X=0)$ |
| $= \frac{20C3(0.04)^3(0.96)^{17}}{1-(0.96)^{20}}$ or $\frac{20C3(0.04)^3(0.96)^{17}}{1 - \text{their (b)}} = 0.065322\ldots$ awrt **0.0653** | dM1 A1 | 2nd M1 dep on first M1 for correct attempt at either $P(X=3)$ or $P(X \geqslant 1)$ (may ft (b)) (Ans only 4/4) |
## Part (d):
| Working | Mark | Guidance |
|---------|------|----------|
| $H_0: p = 0.04$, $H_1: p > 0.04$ | B1 | For both hypotheses correct ($p$ or $\pi$). Allow $H_0: \lambda=5$, $H_1: \lambda > 5$ $[\geqslant$ is B0] |
| $[X \sim B(125, 0.04)]$ approximated by $Po(5)$ | B1 | For using $Po(5)$. May need to check tables e.g. $P(X \leqslant 10) = 0.9863$ would imply B1 |
| $P(X \geqslant 10) = 1 - P(X \leqslant 9)$ or $P(X \leqslant 9) = 0.9682$ | M1 | For writing or using $1 - P(X \leqslant 9)$ or giving $P(X \leqslant 9) = 0.9682$ or $P(X \geqslant 10) = 0.0318$ |
| $= 1 - 0.9682 = 0.0318$, $P(X \geqslant 10) = 0.0318$, CR $X \geqslant 10$ | A1 | For 0.0318 or CR $X \geqslant 10$ |
| Reject $H_0$ or Significant/10 lies in the Critical region | dM1 | Dep on 1st M1. Do not allow contradictory statements e.g. "significant, accept $H_0$" |
| Evidence that proportion/number/rate/%/probability of cars failing the test is more … **or** the car mechanic's claim is supported | A1cso | 2nd A1cso for correct contextual conclusion and no errors seen |
**Normal approximation:** (can score B1B0M1A0M1A0) 1st M1 for $P(X \geqslant 10)$ and standardising with 9.5 or 10.5
**Two-Tail:** Allow max of B0B1M1A1 (2 for CR of $X \geqslant 11$ otherwise 0) dM1 (for accepting $H_0$) A0cso7. Last year $4 \%$ of cars tested in a large chain of garages failed an emissions test.
A random sample of $n$ of these cars is taken. The number of cars that fail the test is represented by $X$
Given that the standard deviation of $X$ is 1.44
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item find the value of $n$
\item find $\mathrm { E } ( X )$
A random sample of 20 of the cars tested is taken.
\end{enumerate}\item Find the probability that all of these cars passed the emissions test.
Given that at least 1 of these cars failed the emissions test,
\item find the probability that exactly 3 of these cars failed the emissions test.
A car mechanic claims that more than $4 \%$ of the cars tested at the garage chain this year are failing the emissions test. A random sample of 125 of these cars is taken and 10 of these cars fail the emissions test.
\item Using a suitable approximation, test whether or not there is evidence to support the mechanic's claim. Use a $5 \%$ level of significance and state your hypotheses clearly.\\
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2016 Q7 [15]}}