| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Calculate Type I error probability |
| Difficulty | Standard +0.3 This is a straightforward S4 hypothesis testing question requiring calculation of significance level (size) using binomial tables, algebraic manipulation to show a given power function formula, and interpretation of power function values. Part (a) is routine table lookup, part (b) requires expanding binomial probabilities but the answer is given, and parts (d-e) involve graph reading. Slightly easier than average due to structured parts and provided formula. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X < 3) = P(X \leq 2)\) when \(X \sim B(10, 0.5)\) | M1 | Finding correct probability |
| \(= 0.0547\) | A1 | Accept 0.055 |
| Answer | Marks | Guidance |
|---|---|---|
| Power \(= P(X < 3 \mid p)= P(X=0)+P(X=1)+P(X=2)\) | M1 | Attempting sum of 3 terms |
| \(= (1-p)^{10} + 10p(1-p)^9 + 45p^2(1-p)^8\) | A1 | Correct expression |
| \(= (1-p)^8[(1-p)^2 + 10p(1-p) + 45p^2]\) | M1 | Factoring out \((1-p)^8\) |
| \(= (1-p)^8[1-2p+p^2+10p-10p^2+45p^2]\) | Expanding bracket | |
| \(= (1-p)^8(1+8p+36p^2)\) | A1 | Given answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(r\): substitute \(p=0.2\): \((0.8)^8(36(0.04)+8(0.2)+1)\) | M1 | Substituting \(p=0.2\) |
| \(r = (0.8)^8(1+1.6+1.44) = 0.1678... \times 4.04 \approx 0.32\) | A1 | \(r = 0.32\) |
| \(s\): substitute \(p=0.4\): \((0.6)^8(36(0.16)+8(0.4)+1)\) | M1 | Substituting \(p=0.4\) |
| \(s=(0.6)^8(5.76+3.2+1)=0.01679...\times 9.96\approx 0.17\) | A1 | \(s = 0.17\) |
| Answer | Marks | Guidance |
|---|---|---|
| Smooth curve through all correct table points (including \(r\) and \(s\)) | B1 | Correct shape, decreasing curve |
| Passes through \((0, 1)\) or approaching 1 as \(p\to 0\); passes through \((0.5, 0.0547)\) | B1 | Correct end behaviour |
| Answer | Marks | Guidance |
|---|---|---|
| Probability of accepting \(H_0\) when biased \(= 1 - \text{Power}\) | M1 | Correct interpretation |
| Need \(1 - \text{Power} \leq 0.4\), i.e. Power \(\geq 0.6\) | M1 | Setting up inequality |
| From graph/table, Power \(\geq 0.6\) when \(p \leq 0.15\) (approx) | A1 | Reading from graph |
| \(0 < p \leq 0.15\) (or equivalent) | A1 | Correct range stated |
# Question 5:
## Part (a)
| $P(X < 3) = P(X \leq 2)$ when $X \sim B(10, 0.5)$ | M1 | Finding correct probability |
| $= 0.0547$ | A1 | Accept 0.055 |
**1 mark total** — size of test is **0.0547**
## Part (b)
| Power $= P(X < 3 \mid p)= P(X=0)+P(X=1)+P(X=2)$ | M1 | Attempting sum of 3 terms |
| $= (1-p)^{10} + 10p(1-p)^9 + 45p^2(1-p)^8$ | A1 | Correct expression |
| $= (1-p)^8[(1-p)^2 + 10p(1-p) + 45p^2]$ | M1 | Factoring out $(1-p)^8$ |
| $= (1-p)^8[1-2p+p^2+10p-10p^2+45p^2]$ | | Expanding bracket |
| $= (1-p)^8(1+8p+36p^2)$ | A1 | **Given answer** |
**3 marks total**
## Part (c)
| $r$: substitute $p=0.2$: $(0.8)^8(36(0.04)+8(0.2)+1)$ | M1 | Substituting $p=0.2$ |
| $r = (0.8)^8(1+1.6+1.44) = 0.1678... \times 4.04 \approx 0.32$ | A1 | $r = 0.32$ |
| $s$: substitute $p=0.4$: $(0.6)^8(36(0.16)+8(0.4)+1)$ | M1 | Substituting $p=0.4$ |
| $s=(0.6)^8(5.76+3.2+1)=0.01679...\times 9.96\approx 0.17$ | A1 | $s = 0.17$ |
**2 marks total**
## Part (d)
| Smooth curve through all correct table points (including $r$ and $s$) | B1 | Correct shape, decreasing curve |
| Passes through $(0, 1)$ or approaching 1 as $p\to 0$; passes through $(0.5, 0.0547)$ | B1 | Correct end behaviour |
**2 marks total**
## Part (e)
| Probability of accepting $H_0$ when biased $= 1 - \text{Power}$ | M1 | Correct interpretation |
| Need $1 - \text{Power} \leq 0.4$, i.e. Power $\geq 0.6$ | M1 | Setting up inequality |
| From graph/table, Power $\geq 0.6$ when $p \leq 0.15$ (approx) | A1 | Reading from graph |
| $0 < p \leq 0.15$ (or equivalent) | A1 | Correct range stated |
**3 marks total**
---5. A statistician believes a coin is biased and the probability, $p$, of getting a head when the coin is tossed is less than 0.5
The statistician decides to test this by tossing the coin 10 times and recording the number, $X$, of heads. He sets up the hypotheses $\mathrm { H } _ { 0 } : p = 0.5$ and $\mathrm { H } _ { 1 } : p < 0.5$ and rejects the null hypothesis if $x < 3$
\begin{enumerate}[label=(\alph*)]
\item Find the size of the test.
\item Show that the power function of this test is
$$( 1 - p ) ^ { 8 } \left( 36 p ^ { 2 } + 8 p + 1 \right)$$
Table 1 gives values, to 2 decimal places, of the power function for the statistician's test.
\begin{table}[h]
\end{table}
Table 1\\
(d) On the axes below draw the graph of the power function for the statistician's test.\\
(e) Find the range of values of $p$ for which the probability of accepting the coin as unbiased, when in fact it is biased, is less than or equal to 0.4\\
\includegraphics[max width=\textwidth, alt={}, center]{1d84c9fc-be67-45be-b439-3111c48ff1cb-09_1143_1209_945_402}
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2014 Q5 [11]}}