Statistics Paper 4 November 2004

1. A$F(x)=0$ for $x\le-\pi$; $F(x)=\dfrac{x}{2\pi}$ for $-\pi<x<\pi$; $F(x)=1$ for $x\ge\pi$
2. B$F(x)=0$ for $x\le-\pi$; $F(x)=\dfrac{x+2\pi}{2\pi}$ for $-\pi<x<\pi$; $F(x)=1$ for $x\ge\pi$
3. C$F(x)=0$ for $x\le-\pi$; $F(x)=\dfrac{x+\pi}{2\pi}$ for $-\pi<x<\pi$; $F(x)=1$ for $x\ge\pi$
4. D$F(x)=0$ for $x\le-\pi$; $F(x)=\dfrac{1}{2\pi}$ for $-\pi<x<\pi$; $F(x)=1$ for $x\ge\pi$

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1. A$f(x)\ge0$ on $0\le x\le1$, and $\displaystyle\int_{0}^{1}\frac{3}{2}x(2-x)\,dx=1$
2. B$f(x)\ge0$ on $0\le x\le1$, and $\displaystyle\int_{0}^{2}\frac{3}{2}x(2-x)\,dx=1$
3. C$f(0)=0$ and $f(1)=\dfrac{3}{2}$, so $f$ increases across the interval
4. D$\displaystyle\int_{0}^{1}\frac{3}{2}x(2-x)\,dx=\frac{2}{3}$, and $f$ is continuous

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1. A$N(20,\ 13.3)$
2. B$N(20,\ 3.65)$
3. C$N(30,\ 15)$
4. D$N(20,\ 40)$

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1. AReject $H_0$, but only because the last class has a small expected frequency.
2. BReject $H_0$: a Poisson distribution with $\lambda=1$ is not a suitable model.
3. CAccept $H_0$: a Poisson distribution with $\lambda=1$ fits the data well.
4. DThe test is inconclusive, because $\chi^{2}_{\text{cal}}$ is more than twice the critical value.

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1. A$y=0.618x+6.92$
2. B$y=0.511x+8.05$
3. C$y=0.511x-8.05$
4. D$y=8.05x+0.511$

1. AThere is a strong positive relationship: the rainfall of a season almost completely determines the yield of the crop.
2. BThere is no relationship, because the coefficient is closer to 0 than to 1.
3. CHigher rainfall causes a proportionally higher yield, in the ratio 0.562 to 1.
4. DThere is a moderate positive relationship: more rainfall tends to go with a higher yield, but the link is loose.

1. AThe two lines coincide exactly, because both of them pass through the point $(\bar{x},\ \bar{y})$.
2. BThe two lines are parallel, and 0.562 is the distance between them.
3. CThe two lines are perpendicular, because 0.562 is roughly halfway between 0 and 1.
4. DThey are well apart: they cross at $(\bar{x},\ \bar{y})$ and diverge on either side of it.

1. AYes: a sample of 35 is large enough for a one-tailed test to be valid.
2. BNo: a one-tailed test is never appropriate whenever the population variance has to be estimated from the sample data.
3. CNo: 'have changed' names no direction, so the alternative hypothesis is two-sided and the test is two-tailed.
4. DYes: the sample mean came out below 5 870, so the test should be one-tailed in the lower tail.

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1. AReject $H_0$, because $z_{\text{cal}}$ is negative and the critical value quoted is positive.
2. BDo not reject $H_0$: at the 5% level there is no evidence that the mean has changed.
3. CReject $H_0$: the sample gives evidence at the 5% level that the mean has changed.
4. DDo not reject $H_0$, which proves that the population mean is exactly 5 870.

1. A(5 707, 5 873)
2. B(5 748, 5 832)
3. C(5 540, 6 040)
4. D(5 681, 5 899)

1. A$W\sim N(80,\ 35.41)$
2. B$W\sim N(100,\ 7.9)$
3. C$W\sim N(100,\ 35.41)$
4. D$W\sim N(100,\ 62.41)$

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1. A$\bar{X}\sim N(38,\ 0.05)$
2. B$\bar{X}\sim N(3.8,\ 0.25)$
3. C$\bar{X}\sim N(38,\ 0.25)$
4. D$\bar{X}\sim N(38,\ 25)$

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1. AIt is biased: a mean that high arises in only about 1.4% of random samples, so the delegates are not representative.
2. BIt is a random sample: the 100 questionnaires were issued out randomly, so the result must be representative of the population.
3. CIt is biased, because the sample size of 100 is too small for the central limit theorem to apply.
4. DIt is representative, because 39.1 is within one standard deviation of 38.

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1. A(52.3, 53.1)
2. B(51.5, 53.9)
3. C(50.9, 54.5)
4. D(51.9, 53.5)

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