Pinworm: In a random sample of 810 adults in the U.S.A., it was found that 80 of
ID: 3174781 • Letter: P
Question
Pinworm: In a random sample of 810 adults in the U.S.A., it was found that 80 of those had a pinworm infestation. You want to find the 99% confidence interval for the 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 z/2) for a 99% confidence interval? Use the value from the table or, if using software, round to 2 decimal places.
z/2 =
(c) What is the margin of error (E) for a 99% confidence interval? Round your answer to 3 decimal places.
E =
(d) Construct the 99% confidence interval for the proportion of all U.S. adults with pinworm. Round your answers to 3 decimal places.
< p <
(e) Based on your answer to part (d), are you 99% confident that more than 5% of all U.S. adults have pinworm?
No, because 0.05 is below the lower limit of the confidence interval.
Yes, because 0.05 is below the lower limit of the confidence interval.
No, because 0.05 is above the lower limit of the confidence interval.
Yes, because 0.05 is above the lower limit of the confidence interval.
(f) In Sludge County, the proportion of adults with pinworm is found to be 0.17. Based on your answer to (d), does Sludge County's pinworm infestation rate appear to be greater than the national average?
Yes, because 0.17 is below the upper limit of the confidence interval.
Yes, because 0.17 is above the upper limit of the confidence interval.
No, because 0.17 is below the upper limit of the confidence interval.
No, because 0.17 is above the upper limit of the confidence interval.
Explanation / Answer
a.-Here p=80/810
=.099
q=1-.099
=.901
Null hypothesis H0=proprtion of ringworm in adult is 1%,p0=.1
Alternate hypothesis H1(Pnot equal to .1)
test ststistic z=Observed value- expected value/Margin of error
=.099-.1/.01
= .098
since this is two tail test
zvalue at 1% significant level from table is2.33 which is more than calculated value,hence we accept null hypothesis thatringworm proportion in adult is .1
c.-Margin of Error of p= sqrt of pq/n
= sqrt.901*.099/810
=.01
d.-99% confidence interval of p=p+-2.58*Margin of error of p
= .099+-2.58*.01
=.099+-.027
.126 and .027 are confidence interval of p at99%
e.-No. because .05 is above the lower limit of confidence interval.
f.-Yes because .17 is above the upper limit of confidence interval