Problems
Filters
3 problems found
2004 Paper 1 Q13
- Three real numbers are drawn independently from the continuous rectangular distribution on \([ 0, 1 ]\,\). The random variable \(X\) is the maximum of the three numbers. Show that the probability that \(X \le 0.8\) is \(0.512\,\), and calculate the expectation of \(X\).
- \(N\) real numbers are drawn independently from a continuous rectangular distribution on \([ 0, a ]\,\). The random variable \(X\) is the maximum of the \(N\) numbers. A hypothesis test with a significance level of 5\% is carried out using the value, \(x\), of \(X \). The null hypothesis is that \(a=1\) and the alternative hypothesis is that \(a<1 \,\). The form of the test is such that \(H_0\) is rejected if \(x < c\,\), for some chosen number \(c\,\). Using the approximation \(2^{10} \approx 10^3\,\), determine the smallest integer value of \(N\) such that if \(x \le 0.8\) the null hypothesis will be rejected. With this value of \(N\), write down the probability that the null hypothesis is rejected if \(a = 0.8\,\), and find the probability that the null hypothesis is rejected if \(a = 0.9\,\).
Solution: \begin{align*} \P(X \leq 0.8) &= \P(X_1 \leq 0.8,X_2 \leq 0.8,X_3 \leq 0.8) \\ &= 0.8^3 \\ &= 0.512 \end{align*} \begin{align*} && \P(X < c) &= c^3 \\ \Rightarrow && f_X(x) &= 3x^2 \\ \Rightarrow && \E[X] &= \int_0^1 x \cdot (3x^2) \, dx \\ && &= \left [ \frac{3}{4}x^4 \right]_0^1 \\ &&&= \frac{3}{4} \end{align*} \(X\) is distributed the maximum of \(N\) numbers on \([0,a]\). \begin{align*} H_0 : & x= 1 \\ H_1 : & x < 1 \end{align*} \begin{align*} &&\P(X < c) &= c^N \\ &&&= \frac1{20} \\ \Rightarrow && N &= -\frac{\log(20)}{\log(c)} \end{align*} where \(c = 0.8\), we have \begin{align*} N &= \frac{\log(20)}{\log(5/4)} \\ &= \frac{\log(5)+\log(4)}{\log(5)-\log(4)} \\ &= \frac{ \frac{\log(5)}{\log(4)}+1}{\frac{\log(5)}{\log(4)} - 1} \end{align*} \begin{align*} && 2^{10} &\approx 10^{3} \\ && 10\log(2) &\approx 3 (\log(5) + \log(2)) \\ && 7\log(2) &\approx 3 \log(5) \\ && \frac{\log(5)}{2\log(2)} &\approx \frac{7}{6} \end{align*} \begin{align*} &= \frac{ \frac{\log(5)}{\log(4)}+1}{\frac{\log(5)}{\log(4)} - 1} &= \frac{\frac{7}{6} + 1}{\frac{7}{6} -1} \\ &= 13 \end{align*} Since \(2^{10} > 10^3\) then \(N=14\) is the value we seek. \(\P(X < 0.8 | a= 0.8) = 1\) \(\P(X < 0.8 | a= 0.9, N=14) = \frac{8^{14}}{9^{14}}\)
2004 Paper 3 Q14
In this question, \(\Phi(z)\) is the cumulative distribution function of a standard normal random variable. A random variable is known to have a Normal distribution with mean \(\mu\) and standard deviation either \(\sigma_0\) or \(\sigma_1\), where \(\sigma_0 < \sigma_1\,\). The mean, \(\overline{X}\), of a random sample of \(n\) values of \(X\) is to be used to test the hypothesis \(\mathrm{H}_0: \sigma = \sigma_0\) against the alternative \(\mathrm{H}_1: \sigma = \sigma_1\,\). Explain carefully why it is appropriate to use a two sided test of the form: accept \(\mathrm{H}_0\) if \(\mu - c < \overline{X} < \mu+c\,\), otherwise accept \(\mathrm{H}_1\). Given that the probability of accepting \(\mathrm{H}_1\) when \(\mathrm{H}_0\) is true is \(\alpha\), determine \(c\) in terms of \(n\), \(\sigma_0\) and \(z_{\alpha}\), where \(z_\alpha \) is defined by \(\displaystyle\Phi(z_{\alpha}) = 1 - \tfrac{1}{2}\alpha\). The probability of accepting \(\mathrm{H}_0\) when \(\mathrm{H}_1\) is true is denoted by \(\beta\). Show that \(\beta\) is independent of \(n\). Given that \(\Phi(1.960)\approx 0.975\) and that \(\Phi(0.063) \approx 0.525\,\), determine, approximately, the minimum value of \(\displaystyle \frac{\sigma_1}{\sigma_0}\) if \(\alpha\) and \(\beta\) are both to be less than \(0.05\,\).
Solution: If \(\sigma\) is smaller we should expect our sample to have a mean closer to the true mean. Therefore we should use a two sided test which accepts \(\mathrm{H}_0\) if the mean is very close to the true mean. Suppose \(\textrm{H}_0\) is true, ie \(\sigma = \sigma_0\), then note that \(X \sim N(\mu, \frac{\sigma_0^2}{n})\) \begin{align*} && 1-\alpha &= \mathbb{P}(\mu - c < X < \mu + c) \\ &&&= \mathbb{P}(\mu - c < \frac{\sigma_0}{\sqrt{n}} Z + \mu < \mu + c) \\ &&&= \mathbb{P}(- \frac{c\sqrt{n}}{\sigma_0} < Z<\frac{\sqrt{n}c}{\sigma_0}) \\ &&&= \mathbb{P}(Z<\frac{\sqrt{n}c}{\sigma_0}) -\mathbb{P}( Z<-\frac{\sqrt{n}c}{\sigma_0}) \\ &&&= \mathbb{P}(Z<\frac{\sqrt{n}c}{\sigma_0}) -(1-\mathbb{P}( Z<\frac{\sqrt{n}c}{\sigma_0})) \\ &&&= 2\mathbb{P}(Z<\frac{\sqrt{n}c}{\sigma_0})-1 \\ \Rightarrow && \Phi(\frac{\sqrt{n}c}{\sigma_0})&=1 - \tfrac12 \alpha \\ \Rightarrow && \frac{\sqrt{n}c}{\sigma_0} &= z_{\alpha} \\ && c &= \frac{\sigma_0 z_{\alpha}}{\sqrt{n}} \end{align*} Under \(\mathrm{H}_1\), \(\sigma = \sigma_1\) so \begin{align*} && \beta &= \mathbb{P}(\mu - c < X < \mu + c) \\ &&&= \mathbb{P}(-\frac{c\sqrt{n}}{\sigma_1} < Z < \frac{\sqrt{n}c}{\sigma_1}) \\ &&&= \mathbb{P}(-\frac{\sigma_0}{\sigma_1} z_{\alpha}< Z < \frac{\sigma_0}{\sigma_1} z_{\alpha}) \\ &&&= 2\Phi(\frac{\sigma_0}{\sigma_1} z_{\alpha})-1 \end{align*} which does not depend on \(n\). Suppose both \(\alpha<0.05\) and \(\beta<0.05\), then \(z_{\alpha} > 1.96\) and \(\Phi(\frac{\sigma_0}{\sigma_1}1.96)<0.525 \Rightarrow \frac{\sigma_0}{\sigma_1}1.96 < 0.063 \Rightarrow \frac{\sigma_1}{\sigma_0} > \frac{1.96}{0.063} = 31.1 \) so the ratio of variances needs to be larger than \(31.1\).
2003 Paper 2 Q14
The probability of throwing a 6 with a biased die is \(p\,\). It is known that \(p\) is equal to one or other of the numbers \(A\) and \(B\) where \(0 < A < B < 1 \,\). Accordingly the following statistical test of the hypothesis \(H_0: \,p=B\) against the alternative hypothesis \(H_1: \,p=A\) is performed. The die is thrown repeatedly until a 6 is obtained. Then if \(X\) is the total number of throws, \(H_0\) is accepted if \(X \le M\,\), where \(M\) is a given positive integer; otherwise \(H_1\) is accepted. Let \({\alpha}\) be the probability that \(H_1\) is accepted if \(H_0\) is true, and let \({\beta}\) be the probability that \(H_0\) is accepted if \(H_1\) is true. Show that \({\beta} = 1- {\alpha}^K,\) where \(K\) is independent of \(M\) and is to be determined in terms of \(A\) and \(B\,\). Sketch the graph of \({\beta}\) against \({\alpha}\,\).
Solution: \(X \sim Geo(p)\). \(\alpha = \mathbb{P}(X > M | p = B) = (1-B)^{M}\) \(\beta = \mathbb{P}(X \leq M | p = A) = 1 - \mathbb{P}(X > M | p = A) = 1 - (1-A)^{M}\) \begin{align*} \ln \alpha &= M \ln(1-B) \\ \ln (1-\beta) &= M \ln(1-A) \\ \frac{\ln \alpha}{\ln (1-\beta)} &= \frac{\ln(1-B)}{\ln(1-A)} \\ \ln(1-\beta) &= \ln \alpha \frac{\ln (1-A)}{\ln(1-B)} \\ \beta &= 1- \alpha^{ \frac{\ln (1-A)}{\ln(1-B)} } \end{align*} and \(K = \frac{\ln (1-A)}{\ln(1-B)} \) Since \(0 < A < B < 1\) we must have that \(0 < 1 - B < 1-A < 1\) and \(\ln(1-B) < \ln(1-A) < 0\) so \(0 < K < 1\)
