Lecture 9 Chapter 5. Non-Normal Populations. 5.1 Introduction Throughout the course ( in Chapters 2, 3 and 4) we have focussed on data which we can assume comes from the Normal distribution. However, some experiments give results that cannot sensibly be modelled by a Normal distribution.
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We first consider the situation where our data are continuous but may not be Normally distributed, and in fact we do not know what distribution might be appropriate.
In these cases, the methods that we have studied so far in this course, t-tests, ANOVA etc. are not appropriate, and must be replaced by tests which do not assume the Normal distribution, or indeed any other distribution.
Methods which do not assume the data come from any distribution are called distribution-free, or non-parametric.
The speech of two groups of speech-impaired children is assessed following two different programmes of treatment:
Group A: Active Speech Therapy
Group B: Conversation Sessions
The following data are scores on a scale in which higher values represent greater difficulty in speaking.
1.7 2.8 1.5 2.2 2.7 1.7 1.8 2.2
1.8 3.2 1.7 2.0 2.2 2.1
3.4 2.2 3.7 3.1 2.0 2.6 2.8 2.1
Let’s look at histograms…
If we could assume these data to be normally distributed, we could use a two-sample t-test. However, this assumption is difficult to justify here.
So, we use the appropriate non-parametric test for comparing two independent samples, which is called the Mann-Whitney test.
The details of how to do the calculations for this are not necessary here. We omit them and go straight to the implementation in Minitab using the command:
For our data, we get p = 0.0307. This is significant at the 5% level, so we have evidence for a difference in the two treatments. Active speech therapy appears to be more effective than conversation sessions.
We now consider a different kind of data altogether. Instead of numbers measured on a continuous scale, we consider situations where our data are counts of different kinds.
In this section we consider what happens when we count the successes from a number of trials.
The Binomial distribution is used to model the number of successes in a series of nindependent trials, where each trial results in either a ‘success’ or a ‘failure’.
Let’s first see how this works. we could use a two-sample t-test. However, this assumption is difficult to justify here.
A drug is known to be 80% effective, i.e. the probability that each person with the disease will be cured is 0.8.
Suppose four people with the disease are given the drug. What is the probability distribution for the number of people cured?
Notation we could use a two-sample t-test. However, this assumption is difficult to justify here.
Let X = number of people cured.
Let s denote a success.
Let f denote a failure.
Consider a typical outcome of the experiment for the four people, e.g. that the first two are cured, and the second two are not. We would write this outcome: s s f f.
Since each person is cured (or not cured) independently, we can calculate the probability of this outcome as
Pr (s s f f) = Pr(s) x Pr(s) x Pr(f) x Pr(f)
= 0.8 x 0.8 x 0.2 x 0.2
We could do similar calculations for we could use a two-sample t-test. However, this assumption is difficult to justify here.all of the possible outcomes:
1. ssss 0.8 x 0.8 x 0.8 x 0.8 = 0.4096
2. sssf etc.
4. ssff 0.8 x 0.8 x 0.2 x 0.2 = 0.0256
5. sfss etc.
6. sfsf 0.8 x 0.2 x 0.8 x 0.2 = 0.0256
7. sffs 0.8 x 0.2 x 0.2 x 0.8 = 0.0256
8. sfff etc.
10. fssf 0.2 x 0.8 x 0.8 x 0.2 = 0.0256
11. fsfs 0.2 x 0.8 x 0.2 x 0.8 = 0.0256
12. fsff etc.
13. ffss 0.2 x 0.2 x 0.8 x0.8 = 0.0256
14. ffsf etc.
16. ffff 0.2 x 0.2 x 0.2 x 0.2 = 0.0016
Now suppose we want to know the probability that exactly two of the four patients are cured (not necessarily the first two), i.e. Pr (X=2).
We can obtain this probability by adding up the probabilities for all of the outcomes in the table for which X=2. There are six of these, i.e. outcomes 4, 6, 7, 10, 11 and 13. Each of these outcomes has probability 0.0256.
Pr (X = 2) = 6 x 0.0256 = 0.1536.
We can do similar calculations to obtain: of the four patients are cured (not necessarily the first two), i.e. Pr (X=2).
Pr (X = 4) = 0.4096
Pr (X = 3) = 0.4096
Pr (X = 1) = 0.0256
Pr (X = 0) = 0.0016
In practice we get Minitab to do the calculations.