| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2015 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Multi-stage or conditional testing |
| Difficulty | Challenging +1.8 This S4 question requires understanding of conditional sampling, power functions, and multi-stage hypothesis testing—topics beyond standard A-level. Students must trace through a complex two-stage procedure, derive the power function algebraically (involving binomial probabilities across two samples), and interpret Type II errors. While systematic, it demands careful probability reasoning and is more sophisticated than typical S1/S2 hypothesis tests. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{reject } H_0) = P(X \geq 2) + P(X=1) \times P(Y \geq 1)\) where \(X \sim B(6,p)\), \(Y \sim B(6,p)\) | M1 | Need to identify two routes to rejection |
| \(= 1 - P(X=0) - P(X=1) + P(X=1)\times[1-P(Y=0)]\) | M1 | Combining terms correctly |
| \(= 1-(1-p)^6 - 6p(1-p)^5 + 6p(1-p)^5[1-(1-p)^6]\) | ||
| \(= 1-(1-p)^6 - 6p(1-p)^5 \cdot (1-p)^6\) | ||
| \(= 1-(1-p)^6 - 6p(1-p)^{11}\) | A1 | Correct completion |
| Answer | Marks |
|---|---|
| Substitute \(p = 0.05\): \(1-(0.95)^6 - 6(0.05)(0.95)^{11}\) | M1 |
| \(= 1 - 0.73509... - 6(0.05)(0.56869...) = 0.0386\) (awrt) | A1 |
| Answer | Marks |
|---|---|
| \(E(\text{eggs}) = 6 \times P(\text{only 1 box sampled}) + 12 \times P(\text{2 boxes sampled})\) | M1 |
| \(P(\text{only box 1}) = P(X=0) + P(X \geq 2) = (0.9)^6 + [1-6(0.1)(0.9)^5-(0.9)^6]\) | M1 |
| \(= 1 - 6(0.1)(0.9)^5 = 1-0.354294 = 0.645706...\) | |
| \(P(\text{2 boxes}) = 6(0.1)(0.9)^5 = 0.354294...\) | |
| \(E = 6(0.645706) + 12(0.354294) = 3.87424 + 4.25152 = 8.13\) (3 s.f.) | A1 |
| Answer | Marks |
|---|---|
| Type II error: \(H_0\) accepted when \(p=0.1\) | M1 |
| \(P(\text{Type II}) = 1 - \text{Power at } p=0.1 = 1-[1-(0.9)^6-6(0.1)(0.9)^{11}]\) | |
| \(= (0.9)^6 + 6(0.1)(0.9)^{11} = 0.53144 + 0.19371 = 0.725\) (awrt) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| The probability of a Type II error is high (0.725), so the test is poor/not good at detecting when \(p=0.1\) | B1 | Must relate to the high probability/poor test in context |
# Question 4:
## Part (a):
| $P(\text{reject } H_0) = P(X \geq 2) + P(X=1) \times P(Y \geq 1)$ where $X \sim B(6,p)$, $Y \sim B(6,p)$ | M1 | Need to identify two routes to rejection |
|---|---|---|
| $= 1 - P(X=0) - P(X=1) + P(X=1)\times[1-P(Y=0)]$ | M1 | Combining terms correctly |
| $= 1-(1-p)^6 - 6p(1-p)^5 + 6p(1-p)^5[1-(1-p)^6]$ | | |
| $= 1-(1-p)^6 - 6p(1-p)^5 \cdot (1-p)^6$ | | |
| $= 1-(1-p)^6 - 6p(1-p)^{11}$ | A1 | Correct completion |
## Part (b):
| Substitute $p = 0.05$: $1-(0.95)^6 - 6(0.05)(0.95)^{11}$ | M1 | |
|---|---|---|
| $= 1 - 0.73509... - 6(0.05)(0.56869...) = 0.0386$ (awrt) | A1 | |
## Part (c): (given $p=0.1$)
| $E(\text{eggs}) = 6 \times P(\text{only 1 box sampled}) + 12 \times P(\text{2 boxes sampled})$ | M1 | |
|---|---|---|
| $P(\text{only box 1}) = P(X=0) + P(X \geq 2) = (0.9)^6 + [1-6(0.1)(0.9)^5-(0.9)^6]$ | M1 | |
| $= 1 - 6(0.1)(0.9)^5 = 1-0.354294 = 0.645706...$ | | |
| $P(\text{2 boxes}) = 6(0.1)(0.9)^5 = 0.354294...$ | | |
| $E = 6(0.645706) + 12(0.354294) = 3.87424 + 4.25152 = 8.13$ (3 s.f.) | A1 | |
## Part (d):
| Type II error: $H_0$ accepted when $p=0.1$ | M1 | |
|---|---|---|
| $P(\text{Type II}) = 1 - \text{Power at } p=0.1 = 1-[1-(0.9)^6-6(0.1)(0.9)^{11}]$ | | |
| $= (0.9)^6 + 6(0.1)(0.9)^{11} = 0.53144 + 0.19371 = 0.725$ (awrt) | A1 | |
## Part (e):
| The probability of a Type II error is high (0.725), so the test is poor/not good at detecting when $p=0.1$ | B1 | Must relate to the high probability/poor test in context |
|---|---|---|
---4. A poultry farm produces eggs which are sold in boxes of 6 . The farmer believes that the proportion, $p$, of eggs that are cracked when they are packed in the boxes is approximately 5\%. She decides to test the hypotheses
$$\mathrm { H } _ { 0 } : p = 0.05 \text { against } \mathrm { H } _ { 1 } : p > 0.05$$
To test these hypotheses she randomly selects a box of eggs and rejects $\mathrm { H } _ { 0 }$ if the box contains 2 or more eggs that are cracked. If the box contains 1 egg that is cracked, she randomly selects a second box of eggs and rejects $\mathrm { H } _ { 0 }$ if it contains at least 1 egg that is cracked. If the first or the second box contains no cracked eggs, $\mathrm { H } _ { 0 }$ is immediately accepted and no further boxes are sampled.
\begin{enumerate}[label=(\alph*)]
\item Show that the power function of this test is
$$1 - ( 1 - p ) ^ { 6 } - 6 p ( 1 - p ) ^ { 11 }$$
\item Calculate the size of this test.
Given that $p = 0.1$
\item find the expected number of eggs inspected each time this test is carried out, giving your answer correct to 3 significant figures,
\item calculate the probability of a Type II error.
Given that $p = 0.1$ is an unacceptably high value for the farmer,
\item use your answer from part (d) to comment on the farmer's test.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2015 Q4 [11]}}