## Small Sample Hypothesis Testing

The smaller a sample is, the harder it is to make decisions based on it.  In general, the larger a sample is, the less likely random variation is to make the results deceptive.  When you have to use a small sample, you can make the decision using binomial distributions.

A tree poison company claims that their new poison is much more effective than older poisons.  They support their claim with data from an experiment, in which they poisoned 10 trees with their poison.  Their poison killed 8 of the trees, while their rivals claim only a 40% kill rate on their containers. Is their poison better?

Solution

Firstly, we formulate the hypotheses:

Null hypothesis, H0

The new poison is no more effective.

Alternative hypothesis, H1

The new poison is more effective.

Since we are using binomial distributions, we must find out s and f for the old rival poison (s = chance of success, f = chance of failure):

s = 0.4, f = 0.6

We know n = 10, so we can look up the binomial table:

 X 0 1 2 3 4 5 P(X) 0.006 0.0403 0.1209 0.215 0.2508 0.2007 X 6 7 8 9 10 P(X) 0.1115 0.0425 0.0106 0.0016 0.0001

In this table, capital ‘X’ is a number of successes out of 10, and the probability below them is the chance of that number of successes happening. The P(X) means, say for the top of the first column, that the probability of there being 0 successes out of 10 trials is 0.006.

Now say we used the old poison on 10 trees.  It might kill say 4 trees – like it should according to the 40% kill claim.  Say I apply it to another 10 trees.  This time it might kill only 3, or maybe 5.  Using the table, we want to find out the probability of the old poison killing 8 or more trees out of the 10, and hence matching the new poison’s performance.

So we add up P(X = 8) + P(X = 9) + P(X = 10)

= 0.0106 + 0.0016 + 0.0001

= 0.0123

= 1.23%.

So using the old poison, there is only a 1.23 % chance that it would kill 8 or more trees out of 10 when applied.

For the claim about the new poison to be valid, this chance would usually have to be less than 5%.  In other words, there would have to be a less than 5% probability of the new result occurring randomly using the old poison.  So the experiment does prove beyond reasonable doubt that the new poison is better.  We can say there is sufficient evidence to support the claim about the new poison.

We would accept the alternate hypothesis in this case.