Signs test

[barbados]

Hi all,

I've got a question regarding the signs test described in Section 7.3 of Chapter 12.

In the first example of this section there are 20 groups, 6 positive and 14 negative. We look for 2P(P<=6).

In Question 12.11 (a) there are 13 positive and 27 negative. We look for P(>= 27 negatives), see Solution 12.11.

In Question 12.11 (b) there are 5 positive and 3 negative. We look for P(>= 5 positives), see Solution 12.11.

Can somebody explain why it is what we are looking for. For example, why is it we look for P(<=6 positives) in one case, and P(>= 5 positives) in the other. And why P(>= 27 negatives) in Q12.11(a) and P(>= 5 positives) in Q12.11(b).

Hope you can help me out. Thanks.

 

[Mark Mitchell]

Hi there!

Firstly, whether we calculate the probability based on positive deviations or negative deviations won't affect our result - we can do either. The distribution of the number of positive deviations is Bin(m,0.5). The distribution of the number of negative deviations is also Bim(m,0.5)

So, if we had 10 deviations, and were interested in the probability that they were all positive:

P(10 positive) = 0.5^10

But if the deviation is positive it must not be negative, giving:

P(10 positive) = P(0 negative) = 0.5^10.

So, either way, considering positives or negatives, gives the same result. (This is because the probability of observing each is one half.)

As for your other question - why do we sometimes calculate the probabilty of there being less than a certain number of deviations and sometimes more than a certain number of deviations - this is because we are looking for the probability of seeing a result as extreme as the one that we have observed.

So in cases where we have a low observed value of positive deviations we are interested in the likelihood of seeing a more extreme result i.e. that number or lower (eg. Example on page 38). In cases where we have a high observed value of positive deviations we are interested in the likelihood of seeing a more extreme result i.e. that number or higher (Q12.11(b)).

Hope this helps,

Mark.

 

[kika258]

I had the same query as barbados below and I found your reply very helpful. However, I have a minor doubt still. You explained that we always look for extreme result, but how do you identify the extreme situation? For instance, in the example in the notes where there are 20 groups, 6 positive deviations and 14 negative deviations, why is it obvious that the 6 positive devations are the extreme? I mean to have no bias we should have 10 positive and 10 negative, so obviously what is lost from one side is gained in another... we have 4 fewer positives and 4 more negatives... how can you conclude that the 6 positives are the most extreme?

Would the result be good aswell if we tested the Probability of having more than 14 negatives?

 

[didster]

Would the result be good aswell if we tested the Probability of having more than 14 negatives?

Yes it's exactly the same.

 

[kika258]

perfect thanks!