| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2016 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Compare test procedures or parameters |
| Difficulty | Standard +0.8 This S4 question requires understanding of hypothesis test sizes, power functions, and sequential testing procedures. Part (b) involves conditional probability across two samples, and part (c) requires deriving an algebraic expression for power. While the calculations are methodical, the conceptual demand of comparing test procedures and understanding power functions places this above average difficulty for A-level, though it remains a standard S4 exercise rather than requiring novel insight. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| \(p\) | 0.1 | 0.2 | 0.3 | 0.4 |
| Power test \(A\) | 0.93 | \(r\) | 0.38 | 0.17 |
| Power test \(B\) | 0.83 | 0.55 | 0.31 | 0.15 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Size of test \(A = P(Y \leq 2) = 0.0547\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Size of test \(B = P(\text{Rejecting } H_0 \mid p=0.5)\) | M1 | For a correct expression/selection of probabilities |
| \(= P(X=0) + (1-P(X=0)) \times P(X=0)\) | A1 | For a correct expression in terms of probabilities. Allow \(0.0312 + (0.9688)(0.0312)\) |
| \(= 0.5^5 + (1-0.5^5)(0.5^5)\) | ||
| \(= 0.03125 + (0.96875)(0.03125)\) | ||
| \(= 0.0615/0.0614\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Power \(= P(0 \text{ long screws in first } 5) + P(0 \text{ long screws in second } 5 \mid >0 \text{ long screws in first } 5)\) | M1 | For a correct expression |
| \(= P(X=0\mid p) + [1-P(X=0\mid p)]P(X=0\mid p)\) | A1 | For a correct expression in terms of \(p\) |
| \(= (1-p)^5 + [1-(1-p)^5](1-p)^5\) | ||
| \(= 2(1-p)^5 - (1-p)^{10}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(r = 0.68\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Test \(A\) as it is more powerful for values of \(p < 0.4\) | M1 A1 | M1 for reason based on the power function; A1 test \(A\) |
# Question 4:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Size of test $A = P(Y \leq 2) = 0.0547$ | B1 | |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Size of test $B = P(\text{Rejecting } H_0 \mid p=0.5)$ | M1 | For a correct expression/selection of probabilities |
| $= P(X=0) + (1-P(X=0)) \times P(X=0)$ | A1 | For a correct expression in terms of probabilities. Allow $0.0312 + (0.9688)(0.0312)$ |
| $= 0.5^5 + (1-0.5^5)(0.5^5)$ | | |
| $= 0.03125 + (0.96875)(0.03125)$ | | |
| $= 0.0615/0.0614$ | A1 | |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Power $= P(0 \text{ long screws in first } 5) + P(0 \text{ long screws in second } 5 \mid >0 \text{ long screws in first } 5)$ | M1 | For a correct expression |
| $= P(X=0\mid p) + [1-P(X=0\mid p)]P(X=0\mid p)$ | A1 | For a correct expression in terms of $p$ |
| $= (1-p)^5 + [1-(1-p)^5](1-p)^5$ | | |
| $= 2(1-p)^5 - (1-p)^{10}$ | | |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $r = 0.68$ | B1 | |
## Part (e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Test $A$ as it is more powerful for values of $p < 0.4$ | M1 A1 | M1 for reason based on the power function; A1 test $A$ |
---\begin{enumerate}
\item A manufacturer produces boxes of screws containing short screws and long screws. The manufacturer claims that the probability, $p$, of a randomly selected screw being long, is 0.5
\end{enumerate}
A shopkeeper does not believe the manufacturer's claim. He designs two tests, $A$ and $B$, to test the hypotheses $\mathrm { H } _ { 0 } : p = 0.5$ and $\mathrm { H } _ { 1 } : p < 0.5$
In test $A$, a random sample of 10 screws is taken from a box of screws and $\mathrm { H } _ { 0 }$ is rejected if there are fewer than 3 long screws.
In test $B$, a random sample of 5 screws is taken from a box of screws and $\mathrm { H } _ { 0 }$ is rejected if there are no long screws, otherwise a second random sample of 5 screws is taken from a box of screws. If there are no long screws in this second sample $\mathrm { H } _ { 0 }$ is rejected, otherwise it is accepted.\\
(a) Find the size of test $A$.\\
(b) Find the size of test $B$.\\
(c) Find an expression for the power function of test $B$ in terms of $p$.
Some values, to 2 decimal places, of the power function for test $A$ and the power function for test $B$ are given in the table below.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$p$ & 0.1 & 0.2 & 0.3 & 0.4 \\
\hline
Power test $A$ & 0.93 & $r$ & 0.38 & 0.17 \\
\hline
Power test $B$ & 0.83 & 0.55 & 0.31 & 0.15 \\
\hline
\end{tabular}
\end{center}
(d) Find the value of $r$.
The shopkeeper believes that the value of $p$ is less than 0.4\\
(e) Suggest which of the tests the shopkeeper should use. Give a reason for your answer.
\hfill \mbox{\textit{Edexcel S4 2016 Q4 [9]}}