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: (a) 0.22Step-by-step 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 :-[tex]\hat{p}=\dfrac{11}{50}=0.22[/tex]Hence, the sample proportion of Sludge County residents with pinworm = 0.22Hypothesis for test :[tex]H_0: p=0.12\\\\ H_a: p>0.12[/tex], since alternative hypothesis is right tailed so the test is a right-tailed test.Test statistic : [tex]z=\dfrac{0.22-0.12}{\sqrt{\dfrac{0.12(1-0.12)}{50}}}\approx2.18[/tex]p-value : P(z>2.18)=0.0146287Since 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.