Question

Pinworm: In a random sample of 790 adults in the U.S.A., it
was found that 76 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 zα/2) for a 99%
confidence interval? Use the value from the table or, if using
software, round to 2 decimal places.

zα/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.

Yes, because 0.05 is above the lower limit of the confidence
interval.

No, 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 0.16. Based on your answer to (d), does Sludge County's
pinworm infestation rate appear to be greater than the national
average?

No, because 0.16 is below the upper limit of the confidence
interval.

No, because 0.16 is above the upper limit of the confidence
interval.

Yes, because 0.16 is below the upper limit of the confidence
interval.

Yes, because 0.16 is above the upper limit of the confidence
interval.

Answer #1

Pinworm: In a random sample of 790 adults in
the U.S.A., it was found that 80 of those had a pinworm
infestation. You want to find the 95% 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
zα/2) for a 95% confidence interval?
Use the value from...

Pinworm: In a random sample of 810 adults in
the U.S.A., it was found that 84 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) Construct the 99% confidence interval for the proportion of all
U.S. adults with pinworm. Round your answers to...

Pinworm: In a random sample of 830 adults in
the U.S.A., it was found that 70 of those had a pinworm
infestation. You want to find the 90% 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?
(b) What is the critical value of z (denoted
zα/2) for a 90% confidence
interval?
zα/2 =
(c) What is the margin of error (E) for...

Pinworm: In a random sample of 830 adults in the U.S.A., it was
found that 74 of those had a pinworm infestation. You want to find
the 90% 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) Construct the 90% confidence interval for the proportion of
all U.S. adults with pinworm. Round your answers to...

In a random sample of 830 adults in the U.S.A., it was found
that 84 of those had a pinworm infestation. You want to find the
95% 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
zα/2) for a 95% confidence interval?
Use the value from the...

In a random sample of 830 adults in the U.S.A., it was found
that 84 of those had a pinworm infestation. You want to find the
95% 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) Construct the 95% confidence interval for the proportion of all
U.S. adults with pinworm. Round your answers to 3...

Corn: In a random sample of 84 ears of corn, farmer Carl finds
that 7 of them have worms. He wants to find the 90% confidence
interval for the proportion of all his corn that has worms.
(a) What is the point estimate for the proportion of all of
Carl's corn that has worms? Round your answer to 3 decimal places.
____
(b) What is the critical value of z (denoted zα/2) for a 90%
confidence interval? Use the value...

Corn: In a random sample of 88 ears of corn,
farmer Carl finds that 15 of them have worms. He wants to find the
99%confidence interval for the proportion of all his corn that has
worms.
(a) What is the point estimate for the proportion of all of
Carl's corn that has worms? Round your answer to 3 decimal
places.
(b) What is the critical value of z (denoted
zα/2) for a 99% confidence interval?
Use the value from the...

Pinworm: In Sludge County, a sample of 60
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? Round your answer to 3 decimal
places.
p̂ =
(b) What is the...

Corn: In a random sample of 92 ears of corn, farmer Carl finds
that 14 of them have worms. He wants to find the 95% confidence
interval for the proportion of all his corn that has worms.
(a) What is the point estimate for the proportion of all of
Carl's corn that has worms? Round your answer to 3 decimal
places.
(b) What is the critical value of z (denoted zα/2) for a 95%
confidence interval? Use the value from...

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