| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2003 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Explain Type I or II error |
| Difficulty | Moderate -0.8 This question tests standard definitions of Type I and II errors plus straightforward binomial probability calculations. Part (i) is pure recall, and part (ii) involves direct application of binomial probability formula with given parameters—no problem-solving insight required, just careful arithmetic with P(X=0) under two different probability values. |
| Spec | 5.05a Sample mean distribution: central limit theorem |
| Answer | Marks | Guidance |
|---|---|---|
| (i) (a) Rejecting \(H_0\) when it is true; (b) Accepting \(H_0\) when it is false | B1, B1 | Or equivalent |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{Type I error}) = 1 - 0.7339 = 0.266\) | M1*, A1, M1*, A1ft dep* | For evaluating P(NNNNN) under \(H_0\); For correct answer (could be implied); For identifying the Type I error outcome; For correct final answer. SR If M0M0 allow B1 for Bin(5,0.94) used |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{Type II error}) = 0.168\) | M1, M1, A1 | For Bin(5,0.7) used; For P(NNNNN) under \(H_1\); For correct final answer |
| [3] |
**(i)** (a) Rejecting $H_0$ when it is true; (b) Accepting $H_0$ when it is false | B1, B1 | Or equivalent
| [2]
**(ii)** (a) $P(NNNNN)$ under $H_0 = (0.94)^5 = 0.7339$
$P(\text{Type I error}) = 1 - 0.7339 = 0.266$ | M1*, A1, M1*, A1ft dep* | For evaluating P(NNNNN) under $H_0$; For correct answer (could be implied); For identifying the Type I error outcome; For correct final answer. SR If M0M0 allow B1 for Bin(5,0.94) used
| [4]
(b) $P(NNNNN)$ under $H_1 = (0.7)^5 = 0.168$
$P(\text{Type II error}) = 0.168$ | M1, M1, A1 | For Bin(5,0.7) used; For P(NNNNN) under $H_1$; For correct final answer
| [3]6 (i) Explain what is meant by
\begin{enumerate}[label=(\alph*)]
\item a Type I error,
\item a Type II error.\\
(ii) Roger thinks that a box contains 6 screws and 94 nails. Felix thinks that the box contains 30 screws and 70 nails. In order to test these assumptions they decide to take 5 items at random from the box and inspect them, replacing each item after it has been inspected, and accept Roger's hypothesis (the null hypothesis) if all 5 items are nails.\\
(a) Calculate the probability of a Type I error.\\
(b) If Felix's hypothesis (the alternative hypothesis) is true, calculate the probability of a Type II error.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2003 Q6 [9]}}