Pinwomm in a random sample of 810 adults in the US A.\' it was found that 68 of
ID: 3318038 • Letter: P
Question
Pinwomm in a random sample of 810 adults in the US A.' it was found that 68 of those had a pinworm infestation You want to find the 90 % confidence interval ir th proportion of all U.S. adults with pinworm. a) What is the point estimate for the proportion of all U.S. adults with pinworm? Round your answer to 3 decimal places b) what is the critical value of z denoted zn for a 90% confidence interval? use the value from the table or, if using software u nd to 2 deci nal plac (c) what is the margin of error (E) for a 90% confidence interval? Round your answer to 3 decimal places. d) Construct the g0% confidence interval for th proportion of all us adults with pinworm. Round your answers to 3 deci nal places. (e) Based on your answer to part (d), are you 90% confident that more than 5% of all US adults have pirwom? O No, because 0.05 is below the lower limit of the confidence interval O Yes, because 0.05 is above the lower limit of the confidence interval. O Yes, because 0.05 is below the lower limit of the confidence interval O No, because 0.05 is above the lower limit of the confidence interval () In Sludge County, the proportion of adults with pinworm is found to be 0.14. Based on your answer to (d), does Sludge County's pinworm infestation rate appe to be greater than the national average? O Yes, because 0.14 is below the upper limit of the confidence interval. O No, because 0.14 is above the upper limit of the confidence interval. O No, because 0.14 is below the upper imit of the confidence interval. O Yes, because 0.14 is above the upper limit of the confidence interval.Explanation / Answer
Ans:
a)Point estimate=68/810=0.084
b)critical z value=1.64
c)Margin of error=1.645*sqrt(0.084*(1-0.084)/810)=0.016
d)90% CI for p
=0.084+/-0.016
=(0.068,0.100)
e)Yes,because 0.05 is below the lower limit of the confidence interval.
f)Yes,because 0.14 is above the upper lmit of the confidence interval.