Modelling and Hypothesis Testing

Showing 1-3 of 3 problems
2004 Paper 1 Q12
D: 1500.0 B: 1529.3

In a certain factory, microchips are made by two machines. Machine A makes a proportion \(\lambda\) of the chips, where \(0 < \lambda < 1\), and machine B makes the rest. A proportion \(p\) of the chips made by machine A are perfect, and a proportion \(q\) of those made by machine B are perfect, where \(0 < p < 1\) and \(0 < q < 1\). The chips are sorted into two groups: group 1 contains those that are perfect and group 2 contains those that are imperfect. In a large random sample taken from group 1, it is found that \(\frac 2 5\) were made by machine A. Show that \(\lambda\) can estimated as \[ {2q \over 3p + 2q}\;. \] Subsequently, it is discovered that the sorting process is faulty: there is a probability of \(\frac 14\) that a perfect chip is assigned to group 2 and a probability of \(\frac 14\) that an imperfect chip is assigned to group 1. Taking into account this additional information, obtain a new estimate of \(\lambda\,\).

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\begin{align*} && \frac25 &= \frac{\lambda p}{\lambda p + (1-\lambda) q} \\ \Rightarrow && 2(1-\lambda)q &= 3\lambda p \\ \Rightarrow && \lambda(3p+2q) &= 2q \\ \Rightarrow && \lambda &= \frac{2q}{3p+2q} \end{align*} \begin{align*} && \frac25 &= \frac{\lambda (p + \frac14(1-p))}{\lambda (p + \frac14(1-p))+(1-\lambda) (q + \frac14(1-q))} \\ &&&= \frac{\lambda(\frac34p + \frac14)}{\lambda(\frac34p + \frac14)+(1-\lambda)(\frac34q + \frac14)} \\ \Rightarrow && \lambda &= \frac{2(\frac34q+\frac14)}{3(\frac34p + \frac14)+2(\frac34q+\frac14)} \\ &&&= \frac{\frac32q + \frac12}{\frac94p + \frac32q + \frac54} \\ &&&= \frac{6q+2}{9p+6q+5} \end{align*}
2003 Paper 2 Q14
D: 1600.0 B: 1484.8

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 \({\alpha}\,\).

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\(X \sim Geo(p)\). \(\alpha = \mathbb{P}(X > M | p = B) = (1-B)^{M}\) \(\beta = \mathbb{P}(X \leq M | p = A) = 1 - \mathbb{P}(X > M | p = A) = 1 - (1-A)^{M}\) \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 \(0 < K < 1\)
TikZ diagram
1988 Paper 1 Q15
D: 1500.0 B: 1484.0

In Fridge football, each team scores two points for a goal and one point for a foul committed by the opposing team. In each game, for each team, the probability that the team scores \(n\) goals is \(\left(3-\left|2-n\right|\right)/9\) for \(0\leqslant n\leqslant4\) and zero otherwise, while the number of fouls committed against it will with equal probability be one of the numbers from \(0\) to \(9\) inclusive. The numbers of goals and fouls of each team are mutually independent. What is the probability that in some game a particular team gains more than half its points from fouls? In response to criticisms that the game is boring and violent, the ruling body increases the number of penalty points awarded for a foul, in the hope that this will cause large numbers of fouls to be less probable. During the season following the rule change, 150 games are played and on 12 occasions (out of 300) a team committed 9 fouls. Is this good evidence of a change in the probability distribution of the number of fouls? Justify your answer.

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\begin{array}{c|c|c|c} k & \P(k \text{ goals}) & \P(\geq 2k+1 \text{ fouls}) & \P(k \text{ goals and } \geq 2k+1 \text{ fouls}) \\ \hline 0 & \frac{3-|2|}{9} = \frac19 & \frac{9}{10} & \frac{9}{90}\\ 1 & \frac{3-|2-1|}{9} = \frac29 & \frac{7}{10} & \frac{14}{90} \\ 2 & \frac{3-|2-2|}{9} = \frac39 & \frac{5}{10} & \frac{15}{90} \\ 3 & \frac{3-|2-3|}{9} = \frac29 & \frac{3}{10} & \frac{6}{90} \\ 4 & \frac{3-|2-4|}{9} = \frac19 & \frac{1}{10} & \frac{1}{90} \\ \hline &&& \frac{9+14+15+6+1}{90} = \frac12 \end{array} The probability a team scores more than half its points from fouls is \(\frac12\). Letting \(X\) be the number of times a team committed \(9\) fouls, then \(X \sim B(300, p)\). Consider two hypotheses: \(H_0: p = \frac1{10}\) \(H_1: p < \frac1{10}\) Under \(H_0\), we are interested in \(\P(X \leq 9)\). Since \(300 \frac{1}{10} > 5\) it is appropriate to use a normal approximation, \(N(30, 27)\). Therefore, \begin{align*} && \P(X \leq 9) &\approx \P(3\sqrt{3}Z + 30 \leq 9.5) \\ &&&= \P( Z \leq \frac{9.5-30}{3\sqrt{3}}) \\ &&&= \P(Z \leq \frac{-20.5}{3\sqrt{3}}) \\ &&&< \P(Z \leq -\frac{7}{2}) \end{align*} Which is very small. Therefore there is good evidence to believe there has been a change in the number of fouls.