| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2005 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Explain Type I or II error |
| Difficulty | Standard +0.3 This question tests standard definitions and routine calculations in hypothesis testing. Parts (a)-(c) are pure recall/definition, part (d) requires straightforward binomial probability calculation (though labeled 'show that'), and parts (e)-(g) involve direct substitution and standard conceptual knowledge. While it covers multiple concepts, each step is procedural with no novel problem-solving required. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.05a Hypothesis testing language: null, alternative, p-value, significance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (a) A Type I error occurs when \(H_0\) is rejected when in fact it is true. | B1 (i) | |
| (b) The size of a test is the probability of a Type I error. | B1 (i) | |
| (c) \(X \sim B(5, 0.25)\) | B1 | Can be implied |
| Size \(= P(X > 6) = 1 - P(X \leq 6 \mid n=8, p=0.25)\) | M1 | |
| \(= 1 - 0.9994 = 0.0006\) | A1 (3) | |
| (d) Power \(= P(X > 6 \mid t = \frac{1}{2}, n = 8)\) | ||
| \(= P(X = 7) + P(X = 8)\) | M1 | |
| \(= \binom{8}{7}(0.5)^7 + \binom{8}{8}(0.5)^8\) | A1 | |
| \(= 8 \cdot 0.5^7 - 0.5^7 + 0.5^8\) | A1 (3) | |
| = \(\frac{0.0313}{...}\) | ||
| (e) Power when \(p = 0.3\) \(= \binom{8}{0.3}^8 - 1 \times 0.3^8\) | B1 (i) | |
| \(= 0.0013\) | AWRT 0.0013 | |
| (f) P(Type I error) \(= 1 -\) Power\((0.3)\) | M1 | |
| \(= 0.9987...\) | A1 (2) | |
| (g) Increase the number of trials | B1 | |
| Increase the critical region | B1 (2) |
| Answer/Working | Marks | Guidance |
|---|---|---|
| **(a) A Type I error occurs when $H_0$ is rejected when in fact it is true.** | B1 (i) | |
| **(b) The size of a test is the probability of a Type I error.** | B1 (i) | |
| **(c)** $X \sim B(5, 0.25)$ | B1 | Can be implied |
| Size $= P(X > 6) = 1 - P(X \leq 6 \mid n=8, p=0.25)$ | M1 | |
| $= 1 - 0.9994 = 0.0006$ | A1 (3) | |
| **(d) Power** $= P(X > 6 \mid t = \frac{1}{2}, n = 8)$ | | |
| $= P(X = 7) + P(X = 8)$ | M1 | |
| $= \binom{8}{7}(0.5)^7 + \binom{8}{8}(0.5)^8$ | A1 | |
| $= 8 \cdot 0.5^7 - 0.5^7 + 0.5^8$ | A1 (3) | |
| **= $\frac{0.0313}{...}$** | | |
| **(e) Power when $p = 0.3$** $= \binom{8}{0.3}^8 - 1 \times 0.3^8$ | B1 (i) | |
| $= 0.0013$ | | AWRT 0.0013 |
| **(f) P(Type I error)** $= 1 -$ Power$(0.3)$ | M1 | |
| $= 0.9987...$ | A1 (2) | |
| **(g)** Increase the number of trials | B1 | |
| Increase the critical region | B1 (2) | |
---5. Define
\begin{enumerate}[label=(\alph*)]
\item a Type I error,
\item the size of a test.
Jane claims that she can read Alan's mind. To test this claim Alan randomly chooses a card with one of 4 symbols on it. He then concentrates on the symbol. Jane then attempts to read Alan's mind by stating what symbol she thinks is on the card. The experiment is carried out 8 times and the number of times $X$ that Jane is correct is recorded.
The probability of Jane stating the correct symbol is denoted by $p$.\\
To test the hypothesis $\mathrm { H } _ { 0 } : p = 0.25$ against $\mathrm { H } _ { 1 } : p > 0.25$, a critical region of $X > 6$ is used.
\item Find the size of this test.
\item Show that the power function of this test is $8 p ^ { 7 } - 7 p ^ { 8 }$.
Given that $p = 0.3$, calculate
\item the power of this test,
\item the probability of a Type II error.
\item Suggest two ways in which you might reduce the probability of a Type II error.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2005 Q5 [12]}}