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

Question

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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%.

Mathematics

1 answer:

tester [12.3K] · Answer 1 · 2026-04-11T06:13:30+03:00

Response with clarification:

Let p denote the proportion of adults in the town who have encountered this flu strain.

According to the provided information

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

∵ $H_a$ this is a two-tailed test.

Test statistic:

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

, where p= denotes the population proportion

${\hat{p}$ = signifies the sample proportion

n= represents the sample size

Setting 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 the 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) [from the z-table]

=0.6782

Decision: Because the p-value(0.6782) exceeds the significance level of 0.01, we do not reject the null hypothesis.

This leads us to conclude that there is insufficient evidence to back the assertion that the percentage of all adults in this town exposed to this flu strain deviates from the national average of 8%.