Writing null hypothesis and deciding on rejection criteria The number of faults in one metre of a thread is Poisson distributed.It is claimed that the average number of faults is 0.02 per metre.A random sample of 100 one metre lengths of the thread reveals a total of 6 faults.Does this information support the claim?
What I want to know whether my H0 should be, H0:lambda=0.02 or should it be lamda=2
And my next question is in deciding the rejection criteria, Is it enough to check to reject H0,P[X>=6]<0.025 (Testing at 5% significance) or should I have looked for P[X<=6]<0.025 as well and checked whether either one of these is satisfied.
Or calculate 2*P[X>=6] and check if it is less than 0.05
 A: For question one, I assume you want to test if the claim in the question is true or not. The claim is that there are 0.02 faults per meter. In other words, the expected value is 0.02. However, the observed value is 6 faults in 100 meters, or 0.06 faults/meter. So,


*

*$H_0 : \lambda = 0.02 $ 

*$H_1 : \lambda \gt 0.02 $


You can also write the alternative as $\lambda \ne 0.02$. 
In the second question, the rejection criteria depends on what level of $ \alpha $ you choose. It also depends on which of the two alternative hypotheses you chose. Since the observed value is greater, it is natural to chose greater than, not equal to. Assume a 5% type-1 error. Then, the sum of iid poissons variables is a new poisson random variable. You want to test 
$ P\{ X_1 + X_2 + \dots + X_{100} \ge 6 | \lambda' = 2 \} 
    = 1 - P\{X_1+\dots \le 6 | \lambda'=2  \}$
You are still testing if $ \lambda = 0.02 $, but indirectly by calculating if the new poisson variable $ \lambda' = 100\lambda $. Since this will be a one-tailed distribution, you still want P[X>=5] if it is less than 5%, not 2*P, which corresponds to the two-tailed test
Hope this helps.
