Answer: (a) 0.22
Step-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 = 0.22
Hypothesis 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)=0.0146287
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.
