Statistics 2MA3 - Test #2 Solutions

2002-03-07 [Q2b solution revised 2002-03-14]

The marking scheme is indicated in red. Full Marks = 40.

Q1 [10]

  1. Give five interesting facts about the life and work of Siméon Denis Poisson (1781-1840). (Click here for pictures.)
  2. A parameter is a scalar or vector that indexes a family of probability distributions. A sample is a subset of observations selected from the population. A statistic is any function of the observations in a sample; it may not include any unknown parameters. The distribution of a statistic is called a sampling distribution; it describes how the statistic will vary from one sample to another.

Q2 [10]

  1. The distribution of X is discrete uniform, with each possible value from 1 to 6 having the same probabilty 1/6. Hence
    E[X] = 1*(1/6) + 2*(1/6) + ... + 6*(1/6) = 3.5
    E[X2] = 1*(1/6) + 4*(1/6) + ... + 36*(1/6) = 91/6
    Var[X] = E[X2] - E[X]2 = 35/12 = 2.916667
  2. Since Y is the sum of 10 independent realizations of X, E[Y] = 10*E[X] = 35 and Var[Y] = 10*Var[X] = 350/12 = 29.16667. By the Central Limit Theorem the distribution of Y will be approximately normal so, approximately, P[Y > 45] = 1 - F((45.5 - 35)/sqrt(350/12)) = 0.02593. Note that I have used a continuity correction: because the actual score must be an integer, I have approximated the probability of getting 45 or less by taking the area under the normal curve up to 45.5. You will get a slightly different answer if you omit the continuity correction. Finally, the maximum score on any one die is 6 so we can say exactly that P[Y > 60] = 0.

Q3 [10]

> diedat
 [1] 3 6 3 6 4 4 2 5 3 6 2 3 6 5 1 5 3 3 5 3
> mean(diedat)
[1] 3.9
> var(diedat)
[1] 2.305263
> var(diedat)/mean(diedat)
[1] 0.5910931
> hist(diedat,breaks=seq(-.5,7.5,by=1),prob=T,col="green")
> lines(0:7,dpois(0:7,mean(diedat)),lwd=3,type="h",col="red")

The data appear to be underdispersed, since the variance is much less than the mean. Compared to the Poisson distribution, the data are more concentrated in the range 1 to 6. [In fact, the data were generated to simulate 20 rolls of a 6-sided die.]

Q4 [10]

> diff <- (Lv2hrtrt-Bashrtrt)[Trtgrp=="P"]
> diff
 [1]  4 NA  5 NA  5 -8 NA -2  8  0 12 13  0  0 NA NA
> n <- sum(!is.na(diff))
> xbar <- mean(diff,na.rm=T)
> s <- sqrt(var(diff,na.rm=T))
> c(n=n, xbar=xbar, s=s)
        n      xbar         s 
11.000000  3.363636  6.217278 
> xbar + c(-1,1)*qt(.975,n-1)*s/sqrt(n)
[1] -0.8131878  7.5404605
> xbar + c(-1,1)*qnorm(.975)*s/sqrt(n)
[1] -0.3104726  7.0377453
  1. Whether computed as (-0.81, 7.54) with the t-distribution or, approximately, as (-0.31, 7.04) with the normal distribution, the 95% confidence interval for the mean difference in heart rate includes 0 so at the 95% level of confidence there is no evidence that the true mean difference between Level 2 heart rate and Baseline heart rate is not zero. This analysis assumes that the subjects are independent and the differences are normally distributed.
  2. The zero differences carry no information one way or the other; it would bias the result to treat them as positive or negative, so it is best to omit them and say that we have 6 positive differences out of 8 non-zero differences. Under the hypothesis that the median difference is zero, the number of positive differences will be distributed Bin(8, 0.5). The probability of getting 6 or more positive differences out of 8 would then be 28/256 + 8/256 + 1/256 = 0.1445 which is quite large so there is no reason to claim that the median difference between Level 2 heart rate and Baseline heart rate is not zero.
Statistics 2MA3