STAT 330: 95-3
Assignment 3 Solutions
Thus use of the two sample t based confidence interval (on 33 degrees of freedom) gives the interval or -0.19 to 5.29. Using the interval which does not pool seems inadvisable here since one of the two sample sizes is small. However, the unpooled estimated standard error of the difference in means is
Thus this interval, using the normal multiplier 1.96 is -0.1 to 5.2. (Aside: there is an approximation due to Welch or Satterthwaite to the distribution of the t-statistic which leads to using the large sample statistic but with a t multiplier with degrees of freedom being
which comes to 20.16 degrees of freedom leading to the multiplier 2.085 and the interval which runs from -0.22 to 5.32.)
leading to a P-value of 0.85% which is strong evidence against equal probabilities.
or . The subscript 1 refers to foreign drivers.
Since the sample sizes are equal the pooled and unpooled versions are identical. The statistic works out to -6.4 which is overwhelming evidence that the two filling operations produce different mean weights.
yielding a one sided P-value of 13.6% from a t distribution on 29 degrees of freedom or of 13.1% from the normal curve showing that it doesn't really matter which you use. This is only very weak evidence against the null hypothesis which wold be accepted.
|First 20||Second 20||Difference|
Has the bias of the measurements changed between the first 20 and the last 20?
This is not a paired data problem; I just calculated the differences to confuse. There is no natural pairing between the first measurement and the first of the last 20. Taking the two independent samples model the question is whether the mean of the first 20 is the same as the mean of the last 20. The idea is that each mean is the true speed of light plus a bias and so the means are equal if and only if the biases are the same.
It is not obvious that the standard deviations are unchanged (and the statistical evidence is that the standard deviation has indeed changed). We thus do not pool the estimates of the standard deviations and our test statistic is
(Numerically, though, since the two sample sizes are the same the statistic has the same value as the two sample t statistic using a pooled variance estimate.) Looking in normal tables and carrying out a 2 tailed test we get a P value of 0.3% and conclude that the bias has indeed changed.
Although the data are not paired the paired comparisons t-test is still valid and the resulting statistic is 3.16 on 19 degrees of freedom yielding a P-value of 0.5% which leads to the same conclusion. The previous calculation based on the unpaired method would have had 38 degrees of freedom if the assumption of equal variances were valid and an indistinguishable P-value.
|Year||Number of||Year||Number of|
I did this in class and obtained two intervals: the first, for the parameter (which is the annual rate) is
and the second is the interval for based on solving a quadratic
This last interval translates to one for by dividing by n=20 to get
which is about the same width but centred slightly differently. The difference is negligible.
If is the per year rate and this is the same for all 20 years then both N and M have Poisson distributions with parameters . Then and the natural test statistic is
leading to a (two sided) P-value of 39% which is not significant.