MATH 376 -- Probability and Statistics II
Type II Error Probabilities for a large-sample lower-tail Z-test
March 22, 2006
We consider testing for value of a proportion. Say the null
hypothesis is 
 :  
p = 
.3  and the alternative hypothesis is
 
 :  
p < .3.  
We will assume  
n  
is always large enough
so that our general discussions of the large-sample tests applies.
The test statistic is   z =
  and the rejection region for
an 
-level test is  RR = { 
z < z_.05 
} = { z < -1.645 } 
or  {
Y/n < 
.3 - 1.645
 }.
Set-up:
| > | sigma[1]:=n->sqrt((.3)*(.7)/n); | 
The normal pdf under the assumption that  
  is true:
| > | f[1]:=n->1/sqrt(2*Pi*sigma[1](n)^2)*exp(-(y-.3)^2/(2*sigma[1](n)^2)); | 
But now suppose that the actual value of p is .24. We ask: How likely are
we to make a Type II error with the 
 = .05-level rejection region given above?
Here's the normal pdf that approximates the distribution with p = .24:
| > | sigma[2]:=n->sqrt((.24)*(.76)/n); | 
| > | f[2]:=n->1/sqrt(2*Pi*sigma[2](n)^2)*exp(-(y-.24)^2/(2*sigma[2](n)^2)); | 
| > | RR:=n->.3-1.645*sigma[1](n); | 
The following plot shows the normal densities for the sampling distribution
of Y/ 100:
    blue plot (shorter normal curve) = if 
  is 
true
    red plot (taller normal curve) = if 
  is 
false and actual  p = .24
The vertical black line is plotted at the value .2246 (the upper edge of the RR taking
n = 100. )
| > | RR(100); | 
| > | Dists:=plot([f[1](100),f[2](100)],y=0..0.5,color=[blue,red]): | 
| > | RRPlot:=plot([RR(100),t,t=0..10],color=black): | 
| > | with(plots): | 
| > | display(Dists,RRPlot); | 
The area under the shorter normal curve to the left of the black line is the probability
of a Type I error:
| > | evalf(Int(f[1](100),y=-infinity..RR(100))); | 
This checks with the desired value  
 = .05 The area under the taller normal curve to the 
right of the black line is the probability of a Type II error (always assuming p = .24):
| > | evalf(Int(f[2](100),y=RR(100)..infinity)); | 
This is certainly unacceptably large! What can we do to decrease it?
The answer is: Increase the sample size(!) Increasing n decreases the variances of both
distributions (so the pdf's get taller and skinnier) , and moves the cut-off for the RR for the
.05-level test farther to the right. For instance the picture for n = 500 is:
| > | RR(500); | 
| > | Dists:=plot([f[1](500),f[2](500)],y=0..0.5,color=[blue,red]): | 
| > | RRPlot:=plot([RR(500),t,t=0..15],color=black): | 
| > | display(Dists,RRPlot); | 
| > | evalf(Int(f[2](500),y=RR(500)..infinity)); | 
| > | 
So the Type II error probability  
  has been reduced to .084 by increasing  
n  
to 500,
while 
 = .05 still.  In class we showed that 
  can be reduced to any given value
by taking n sufficiently large.