Question

In Sludge County, a sample of 50 randomly selected citizens were tested for pinworm. Of these, 11 tested positive. The CDC reports that the U.S. average pinworm infection rate is 12%. Test the claim that Sludge County has a pinworm infection rate that is greater than the national average. Use a 0.05 significance level. (a) What is the sample proportion of Sludge County residents with pinworm

Answer 1

Answer: (a) 0.22

Explanation:

Given : In Sludge County, a sample of 50 randomly selected citizens were tested for pinworm. Of these, 11 tested positive.

Then, the sample proportion of Sludge County residents with pinworm will be :-

`\hat{p}=(11)/(50)=0.22`

Hence, the sample proportion of Sludge County residents with pinworm = 0.22

Hypothesis for test :

`H_0: p=0.12\\\\ H_a: p>0.12`, since alternative hypothesis is right tailed so the test is a right-tailed test.

Test statistic :
`z=\frac{0.22-0.12}{\sqrt{(0.12(1-0.12))/(50)}}\approx2.18`

p-value : P(z>2.18)=0.0146287

Since p-value (0.0146287) is less than the significance level (0.05) , so we reject the null hypothesis.

Conclusion : Sludge County has a pinworm infection rate that is greater than the national average.