STAT 330: 98-1: Midterm Solutions
Midterm, 23 February 1998Instructor: Richard Lockhart
Solution: This is a one sample t type confidence interval problem. The assumption is that the measurements are a sample from a normal population whose mean is the true speed of sound in air. The interval is
Solution: The question is whether or not has changed between the two sets. You have from a population (the BEFORE) measurements and from a population (the AFTER) measurements. You want to test against a two sided alternative. The test statistic is
The critical points for a 10% level test are and and we must accept this null hypothesis. There is little evidence of a change in precision following recalibration.
Solution: This question asks for a test of with a two sided alternative. Since the recalibration happens between the first 15 measurements and the second 15 there is no sense in which the data are really paired. The pooled estimate of is
This leads to a P value (using t tables with 28 degrees of freedom) of about 0.0016. So there is very clear evidence that the bias has changed after recalibration.
Solution: This is a two sample test of against a two sided alternative. The pooled estimate of p under the null hypothesis is
which leads to a P value of 2(1-.9992)=0.0016. There is very strong evidence of a difference between men and women in support for Reform in Calgary.
Solution: This a two sample two sided test and we will see that the solution is a large sample size so that the normal distribution formulas will be just fine. On the bottom of page 351 we see that
or just 589 trips on each route. This sort of accuracy will require 4 or 5 years of experimentation!
Solution: Plan A is susceptible to a change in traffic patterns over time. If traffic levels go up over time the experiment will be biased in favour of the route I try first. Plan B is susceptible to day of the week effects - route 1 will be Monday and Wednesday. Since traffic is heaviest on Friday the design would be biased in favour of route 1. The randomization balances out (probably) any other factors which might affect the length of my trip (day of week, time of day, time of year, weather, ...).
Solution: The likelihood is
which simplifies to
Take logs to get the log likelihood:
The derivative with respect to is
Set this equal to 0 and solve to get
What integral do you have to do to show this? You need not do the integral! [1 mark]
Solution: Set or
and solve for to get