Question

Binomial Hypothesis Test.

While this method is limited in the type of data it can handle (binary - "success/failure" outcomes), it is powerful in providing figures of authority with scientific information off of which to base important decisions.

There is a limitation to using this method and, statistically, this notion refers to the "power" of a test.

First, let's suppose that you are a disease outbreak coordinator for the Center for Disease Control and Prevention. A recent flu outbreak has led to a successful recovery without long-term problems with probability 93%. Your team runs a test in a small town and determines that, in 20 people, 19 successfully recovered. If you hypothesize that this small town has a higher recover rate, will you be able to reject the null hypothesis with alpha = 0.05? What if 20 successfully recovered? Provide your alpha-observed in both cases.

What your figures in the above two questions reveal is a lack of statistical power. That is, with such a huge probability of success and small sample size, it is hard to fall below the set level of alpha. One option proposed is to increase the sample size.

What is the smallest sample size needed so that it would be possible to reject the null hypothesis? Out of this sample size, how many "successes" would it take to fall in the cut-off region (i.e. alpha-observed <= alpha)?

Answer 1

i.
Given that,
possible chances (x)=19
sample size(n)=20
success rate ( p )= x/n = 0.95
success probability,( po )=0.93
failure probability,( qo) = 0.07
null, Ho:p=0.93
alternate, H1: p!=0.93
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.95-0.93/(sqrt(0.0651)/20)
zo =0.351
| zo | =0.351
critical value
the value of |z α| at los 0.05% is 1.96
we got |zo| =0.351 & | z α | =1.96
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 0.35055 ) = 0.72592
hence value of p0.05 < 0.7259,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.93
alternate, H1: p!=0.93
test statistic: 0.351
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.72592
we do not have enough evidence to support the claim that recent flu outbreak has led to a successful recovery without
long-term problems with probability 93%.
type 2 error is possible for the above the test because fail to reject the null hypothesis.

ii.
power of the test
Alpha = p(Z>=((po-p)-Zalpha*(sqrt(po*(1-po)/n))/(p*(1-p)/n)
beta = p(Z>=((0.93-0.95)-Z0.05*(sqrt(0.93*(1-0.93)/20)/(0.95*(1-0.95)/20))
beta= p(Z>=((0.93-0.95)-1.64*((sqrt(0.93*(1-0.93)/20)/(0.95*(1-0.95)/20)))
beta =p(Z>= -1.939)
beta = 0.0525
power of the test = 1- type 2 error
power of the test = 1- 0.0525
power of the test =0.9475

iii.
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.05 is = 1.96
Sample Proportion = 0.5
ME = 0.95
n = ( 1.96 / 0.95 )^2 * 0.5*0.5
= 1.064 ~ 2