There is a certain method of conducting hypothesis tests which makes it very suitable for automation. If we could know the set of values of a statistic – typically a single occurrence or mean – which results in the null hypothesis being rejected, these could be written down on the side of a machine and used by someone who knows nothing about hypothesis testing or statistics. If the null hypothesis is rejected, the machine could be stopped and checked for alignment for example..

Suppose we are conducting a 5% one tailed test. The null hypothesis is that the mean is 9. We are conducting a two tailed test based on a sample size of 1, so we split the 5% into two parts of 2.5% each. We now have to find the set of observations corresponding to these two 2.5% per cents. Actually we find:

a valuesuch that

a valuesuch thator

From the cumulative tables for the Poisson distribution we find, for the lower critical value that the closest probability to 0.025 and also less than 0.025 at the lower end is 0.021 corresponding toand for the upper critical valuethat the closest probability to 0.975 that is also greater than 0.975 is 0.978, and this issince the Poisson tables are cumulative, so we takeas the critical region andas the critical value.

The significance level is the sum of the areas of the upper and lower critical regions: 0.021+0.022=0.043. The significance is always less than or equal to the stated required value at the start of the hypothesis test.

Hence for a Poisson distribution, we reject the null hypothesis thatif we have a single observationor

Example. Bulbs are packed in boxes of 20. Over a long period of time it has been observed that too of the bulbs are faulty – lesson being, don’t buy from China. The factory is revamped and the management wants to assume that such a disgraceful state of affairs never occurs again. They say that the probability of a bulb being faulty must be no higher than 0.10. Find the critical region and comment.

This will be a one tailed test since we wish to protect against the possibility of the proportion of faulty bulbs rising.

From the binomial distribution tables, assuming p=0.1 we findThe critical value is

The test is not really useful since we would reject the null hypothesis always. To improve the test, we need to increaseso that the test is based on a sample of more than 20.

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