To test the hypothesis that a coin is fair(i.e. p = probability of heads = 0.5) by a number of tosses of the coin, we wish to impose the following conditions:
(I) the probabilityof rejecting the hypothesis when it is actually correct must be less than 0.05 ,
(II) the probability of accepting the hypothesis when actually p differs from 0.5 by 0.1 or more (i.e. p>0.6 or p<0.4) must be less than 0.05.
Determine the minimum sample size which is necessary and state the resulting decision rule.
Statistical inference 2
2009-07-22 8:02 pm
回答 (1)
2009-07-24 12:10 am
✔ 最佳答案
According to the requirment, we want the type I error is α=0.05 and also Type II error is β=0.05. The situation is illustrated at the following figure.http://in.geocities.com/myisland8132/yahooknowledge/7009072200679.bmp
Let n denote the required sample size and x the number of heads tossed which we rejected the hypothesis p=0.5
Then
(1) Area under normal curve p=0.5 to right of (x-np)/√(npq) =(x-0.5n)/0.5√n is 0.025
(2) Area under normal curve p=0.6 to left of (x-np)/√(npq) =(x-0.6n)/0.49√n is 0.05
Note: Actually, we need to caculate the area between
(n-x-0.6n)/0.49√n and (x-0.6n)/0.49√n and set it equal to 0.05. But (2) is a good approximation.
Notice that by making the type II error in the worst case, we have p=0.6. That means for other values of p which is >0.6 or p<0.4, the values of β will be smaller.
Now from (1) x=0.5n+0.98√n, from (2) x=0.6n-0.806√n
Solve it, we have n=318.98. It follows that the least sample size is 319. Putting n=319 into (1) x=177.
When p=0.5, x-np=177-159.5=17.5. So the decision rule is
(1) Accept p=0.5, of the number of heads in 319 tosses is within the range 159.5+/-17.5, i.e. between 142 and 177
(2) Reject the hypothesis otherwise.
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