How do we calculate the probability of committing Type II error if the “alternate value for μ” is 201,000 and α = 0.025? (Assume sample mean lies between 197759,202241)
Hi Shakti.... I got the concept and can find type II error on excel... means using NORMDIST fx's and all.
However, I want to try solving through mathematical formula... wherein first we find Xcritical value and then find Power which is nothing but type II error.
Below is what I did... plz correct me where I go wrong.....
Xbar cric = mean-SD/sqrt of n
200000-(1.96*8000)/SQRT(64)
198040
P[Z>(Xbar cric - µ1)/(SD/sqrt of n)
(198040-201000)/(8000/SQRT(64))
-2.96
0.5-0.4985
0.0015
Ramd yes its seems correct but not exactly correct . But Type II error=1-Power of test.
and how you got the critical value, i got it this way
z=(Xbar cric - µ1)/(SD/sqrt of n) => (Xbar cric - µ1)=z*(SD/sqrt of n) => Xbar cric=µ1+z*(SD/sqrt of n) now critical values of X are at z(α = 2.5%) and z(α = 100-2.5%=97.5%) beyond these values at 95% CL we can say that we reject the null hypothesis that mean is 200000. Now if we assume mean to be 201000 then what is the probability of commiting type II error.The values of X beyond critical values will notgiveType II error as we correctely reject the samples but between these critical values there are chances that we accept a false sample to be correct. THe power of test is sum of areas beyond the critical values of X. And thus the type II error is 1-power of test. Xbar cric(α = 97.5%)=200000+1.96*(8000/8)=200000+1.96*(1000)=200000+1960=201960 Xbar cric(α = 2.5%)=200000-1.96*(8000/8)=200000-1.96*(1000)=200000-1960=198040 Now value of X should be such that 201960<X<198040 to commit type II error. if µ=201000 than, z1=201960-201000/1000=960/1000=.96 z2=198040-201000/1000=960/1000=-2.96 area of tail for z>.96=16.85% area of tail for z<-2.96=.15% power of test, P=16.85%+.15%=17% Type II error=1-P=1-.17=.83 or 83%. So there is 83% chances that we commit type II error when we assume mean to be 201000 at 95% CL.
Hi Shakti - your illustration is very interesting, I admit i have not seen the calculation for the power of the test. So, it appears that your key assumption is "if we assume mean to be 201000," such that if you assume a different true (population) mean, the power will vary (hence the assertion that it is difficult to calculate the power)? Thanks,
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