Question

In a sample of 88 adults selected randomly from one town, it is found that 6 of them have been exposed to a particular strain of the flu. At the 0.01 significance level, test the claim that the proportion of all adults in the town that have been exposed to this strain of the flue differs from the nationwide percentage of 8%.

Answer 1

Answer with explanation:

Let p be the proportion of of all adults in the town that have been exposed to this strain of the flue .

As per given , we have

$H_0:p=0.08\\\\ H_a: p\neq0.08$

∵ $H_a$ is two-tailed , so the test is a two-tailed test.

Test statistic :

$z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$

, where p= population proportion

${\hat{p}$ = sample proportion

n= sample size

Put n= 6 and ${\hat{p}=\dfrac{6}{88}\approx0.068$  and p=0.08

$z=\dfrac{0.068-0.08}{\sqrt{\dfrac{0.08(1-0.08)}{88}}}$

$z=\dfrac{-0.012}{0.0289199522192}\approx-0.415$

P-value for two-tailed test : 2P(Z>|z|)

=2P(Z>|-0.415|)

=2P(Z>0.415)

=2[1-P(Z≤0.415)]   [∵ P(Z>z)=1-P(Z≤z)]

=2(1-0.6609)   [ by z-table]

=0.6782

Decision : Since the p-value(0.6782) is > significance level(0.01) , so we fail to reject the null hypothesis.

We conclude that that we do not have sufficient evidence to support the claim that the proportion of all adults in the town that have been exposed to this strain of the flue differs from the nationwide percentage of 8%.