| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| 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 requiring routine application of binomial distribution tables and normal approximation. Part (a) involves systematic trial of values to find critical region boundary, and part (b) is a textbook two-tailed test with continuity correction. Both parts follow well-practiced procedures with no novel insight required, making it slightly easier than average. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \([P(Y = 0) < 0.05]\) | M1 | |
| \((1 - 0.07)^n < 0.05\) | M1 | |
| \(n \log(0.93) < \log(0.05)\) | M1 | |
| \(n > 41.28..., n = 42\) | A1 | 1st M1 For 0.93^n or 0.93^42 or 0.93^41. 2nd M1 for \(n \log(0.93) < \log(0.05)\) or log₀.₉₃ 0.05, n Allow = or ..., condone ≥ or ... or 0.93^42 = 0.0474...or 0.0475 (min 4 dp) Implied by 41.28... or awrt 41.3. A1 42 cao NB An answer of 42 gains 3/3. |
| (b) \(H_0: p = 0.08\), \(H_1: p ≠ 0.08\) | B1 | both hypotheses correct (may use p or π but do not allow p(x)) Allow 8% connected to H₀ and H₁ correctly |
| \(X \sim B(75, 0.08) → Po(6)\) | M1 | |
| \(P(X ...11) = 1 - P(X_{,n} 10)\) | M1 | |
| \(= 1 - 0.9574 = 0.0426\) [> 0.025] | A1 | |
| Do not Reject \(H_0\) or not significant or 11 does not lie in the CR | dM1 | |
| There is not significant evidence to suggest that the proportion of pears weighing more than 180g has changed | A1 | 3rd dM1 Independent of their hypotheses dependent on 2nd M1 but A correct contextual statement. Need proportion oe and changed oe Allow the farmers' belief (oe) is not supported (bold words). Do not accept contradicting statements. No hypotheses is A0 |
| Answer | Marks |
|---|---|
| dM1A0 |
**(a)** $[P(Y = 0) < 0.05]$ | M1 |
$(1 - 0.07)^n < 0.05$ | M1 |
$n \log(0.93) < \log(0.05)$ | M1 |
$n > 41.28..., n = 42$ | A1 | 1st M1 For 0.93^n or 0.93^42 or 0.93^41. 2nd M1 for $n \log(0.93) < \log(0.05)$ or log₀.₉₃ 0.05, n Allow = or ..., condone ≥ or ... or 0.93^42 = 0.0474...or 0.0475 (min 4 dp) Implied by 41.28... or awrt 41.3. A1 42 cao NB An answer of 42 gains 3/3.
**(b)** $H_0: p = 0.08$, $H_1: p ≠ 0.08$ | B1 | both hypotheses correct (may use p or π but do not allow p(x)) Allow 8% connected to H₀ and H₁ correctly
$X \sim B(75, 0.08) → Po(6)$ | M1 |
$P(X ...11) = 1 - P(X_{,n} 10)$ | M1 |
$= 1 - 0.9574 = 0.0426$ [> 0.025] | A1 |
Do not Reject $H_0$ or not significant or 11 does not lie in the CR | dM1 |
There is not significant evidence to suggest that the **proportion** of pears weighing more than 180g has **changed** | A1 | 3rd dM1 Independent of their hypotheses dependent on 2nd M1 but A correct contextual statement. Need proportion oe and changed oe Allow the farmers' belief (oe) is not supported (bold words). Do not accept contradicting statements. No hypotheses is A0
NB Award d M1A1 for a correct contextual statement on its own.
SC1: Use of one-tailed test may score B0M1M1A1M1A0 for rejecting H₀.
SC2: Use of Binomial throughout max (3/6) B1M0M1A0dM1A0
SC3: normal approximation prob = 0.0277 (maximum 3 out of 6) B1 M0 M1 for writing or using $1 - P(X_{,n} 10.5)$ allow < implied by awrt 0.027/0.028 A0
dM1A0 |
**[9 marks]**
---\begin{enumerate}
\item Past evidence shows that $7 \%$ of pears grown by a farmer are unfit for sale.
\end{enumerate}
This season it is believed that the proportion of pears that are unfit for sale has decreased. To test this belief a random sample of $n$ pears is taken. The random variable $Y$ represents the number of pears in the sample that are unfit for sale.\\
(a) Find the smallest value of $n$ such that $Y = 0$ lies in the critical region for this test at a $5 \%$ level of significance.
In the past, $8 \%$ of the pears grown by the farmer weigh more than 180 g . This season the farmer believes the proportion of pears weighing more than 180 g has changed. She takes a random sample of 75 pears and finds that 11 of them weigh more than 180 g .\\
(b) Test, using a suitable approximation, whether there is evidence of a change in the proportion of pears weighing more than 180 g .\\
You should use a $5 \%$ level of significance and state your hypotheses clearly.
\hfill \mbox{\textit{Edexcel S2 2022 Q4 [9]}}