## Hypothesis Testing, Critical Values and Critical Regions

There is a certain method of conducting hypothesistests which makes it very suitable for automation. If we could knowthe set of values of a statistic – typically a single occurrence ormean – which results in the null hypothesis being rejected, thesecould be written down on the side of a machine and used by someonewho knows nothing about hypothesis testing or statistics. If the nullhypothesis is rejected, the machine could be stopped and checked foralignment for example..

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

a value suchthat a value suchthat or 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 thelower end is 0.021 corresponding to andfor the upper critical value thatthe closest probability to 0.975 that is also greater than 0.975 is0.978, and this is sincethe Poisson tables are cumulative, so we take asthe critical region and asthe critical value.

The significance level is the sum of the areas of the upper andlower critical regions: 0.021+0.022=0.043. The significance is alwaysless than or equal to the stated required value at the start of thehypothesis test.

Hence for a Poisson distribution, we reject the null hypothesisthat ifwe have a single observation or Example. Bulbs are packed in boxes of 20. Over a long period oftime it has been observed that too of the bulbs are faulty – lessonbeing, don't buy from China. The factory is revamped and themanagement wants to assume that such a disgraceful state of affairsnever occurs again. They say that the probability of a bulb beingfaulty must be no higher than 0.10. Find the critical region andcomment.

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

From the binomial distribution tables, assuming p=0.1 we find Thecritical value is The test is not really useful since we would reject the nullhypothesis always. To improve the test, we need to increase sothat the test is based on a sample of more than 20. 