| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Find sample size for test |
| Difficulty | Standard +0.3 This is a standard S2 hypothesis testing question with routine binomial probability calculations in part (a), a straightforward one-tailed test in part (b), and a slightly more challenging inequality to solve in part (c). All techniques are textbook exercises requiring no novel insight, making it slightly easier than average overall. |
| 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 |
| \(X \sim B(10, 0.1)\); \(P(X \geq 4) = 1 - P(X \leq 3) = 1 - 0.9872\) | M1 | For writing or using \(P(X \geq 4) = 1 - P(X \leq 3)\) |
| \(= 0.0128\) | A1 | awrt 0.0128 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(1 < X < 5) = P(X \leq 4) - P(X \leq 1) = 0.9984 - 0.7361\) or \(P(X=2)+P(X=3)+P(X=4) = 0.1937 + 0.0574 + 0.0112\) | M1 | For writing or using \(P(X \leq 4) - P(X \leq 1)\) or \(P(X=2)+P(X=3)+P(X=4)\) |
| \(= 0.2623\) | A1 | awrt 0.262 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: p = 0.1\), \(H_1: p < 0.1\) | B1 | Both hypotheses correct. Must be in terms of \(p\) or \(\pi\). Must be attached to \(H_0\) and \(H_1\) |
| \(X \sim B(50, 0.1)\); \(P(X \leq 2) = 0.1117\) or CR \(X \leq 1\) | B1 | awrt 0.112 or CR \(\leq 1\) |
| Do not reject \(H_0\) / Not in the critical region | M1 | Any correct ft statement consistent with their \(p\)-value and 0.05 or their CR and 2 |
| There is insufficient evidence to suggest that this result supports the managing director's claim / not enough evidence to suggest a reduction in the probability of a tennis ball failing the bounce test | A1 | Correct conclusion in context which must be not rejecting \(H_0\). Must use underlined words (oe). No hypotheses then A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(X \sim B(n, 0.1)\) and we reject \(H_0\) if \(P(X=0) < 0.01\); \(P(X=0) = [^nC_0 \times 0.1^0] \times 0.9^n [< 0.01]\) | M1 | For recognising \(P(X=0) = 0.9^n\) |
| \(0.9^{44} = 0.00969...[< 0.01]\) or \(n > \frac{\ln 0.01}{\ln 0.9} \Rightarrow n > 43.7\) | M1 | For \(0.9^{44}\ (= 0.00969...)\) or \(0.9^{43}\ (= 0.01077...)\) or rearranging to \(n > \frac{\ln 0.01}{\ln 0.9}\) (Allow \(=\)) |
| \(n = 44\) | A1 | Cao. SC: Use of tables only, \(n=40, p=0.0148\) and \(n=50, p=0.0052\) scores M1M0A0 |
# Question 3:
## Part (a)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $X \sim B(10, 0.1)$; $P(X \geq 4) = 1 - P(X \leq 3) = 1 - 0.9872$ | M1 | For writing or using $P(X \geq 4) = 1 - P(X \leq 3)$ |
| $= 0.0128$ | A1 | awrt 0.0128 |
## Part (a)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(1 < X < 5) = P(X \leq 4) - P(X \leq 1) = 0.9984 - 0.7361$ or $P(X=2)+P(X=3)+P(X=4) = 0.1937 + 0.0574 + 0.0112$ | M1 | For writing or using $P(X \leq 4) - P(X \leq 1)$ or $P(X=2)+P(X=3)+P(X=4)$ |
| $= 0.2623$ | A1 | awrt 0.262 |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: p = 0.1$, $H_1: p < 0.1$ | B1 | Both hypotheses correct. Must be in terms of $p$ or $\pi$. Must be attached to $H_0$ and $H_1$ |
| $X \sim B(50, 0.1)$; $P(X \leq 2) = 0.1117$ or CR $X \leq 1$ | B1 | awrt 0.112 or CR $\leq 1$ |
| Do not reject $H_0$ / Not in the critical region | M1 | Any correct ft statement consistent with their $p$-value and 0.05 or their CR and 2 |
| There is insufficient evidence to suggest that this result supports the managing director's claim / not enough evidence to suggest a reduction in the probability of a tennis ball failing the bounce test | A1 | Correct conclusion in context which must be **not rejecting** $H_0$. Must use underlined words (oe). No hypotheses then A0 |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $X \sim B(n, 0.1)$ and we reject $H_0$ if $P(X=0) < 0.01$; $P(X=0) = [^nC_0 \times 0.1^0] \times 0.9^n [< 0.01]$ | M1 | For recognising $P(X=0) = 0.9^n$ |
| $0.9^{44} = 0.00969...[< 0.01]$ or $n > \frac{\ln 0.01}{\ln 0.9} \Rightarrow n > 43.7$ | M1 | For $0.9^{44}\ (= 0.00969...)$ or $0.9^{43}\ (= 0.01077...)$ or rearranging to $n > \frac{\ln 0.01}{\ln 0.9}$ (Allow $=$) |
| $n = 44$ | A1 | Cao. SC: Use of tables only, $n=40, p=0.0148$ **and** $n=50, p=0.0052$ scores M1M0A0 |
---\begin{enumerate}
\item Superbounce is a manufacturer of tennis balls.
\end{enumerate}
It knows from past records that 10\% of its tennis balls fail a bounce test.\\
(a) Find the probability that from a random sample of 10 of these tennis balls\\
(i) at least 4 fail the bounce test\\
(ii) more than 1 but fewer than 5 fail the bounce test.
The managing director makes changes to the production process and claims that these changes will reduce the probability of its tennis balls failing the bounce test.
After the changes were made a random sample of 50 of the tennis balls were tested and it was found that 2 failed the bounce test.\\
(b) Test, at the $5 \%$ significance level, whether or not this result supports the managing director's claim.
In a second random sample of $n$ tennis balls it was found that none failed the bounce test. As a result of this sample, the managing director's claim is supported at the 1\% significance level.\\
(c) Find the smallest possible value of $n$
\hfill \mbox{\textit{Edexcel S2 2023 Q3 [11]}}