Question

**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
*z*_{α/2}) for a 95% confidence interval?
**Use the value from the table or, if using software, round
to 2 decimal places.**

*z*_{α/2} =

(c) What is the margin of error (*E*) for a 95% confidence
interval? **Round your answer to 3 decimal
places.**

*E* =

(d) Construct the 95% 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 95% confident that
more than 6% of all U.S. adults have pinworm?

No, because 0.06 is below the lower limit of the confidence interval. Yes, because 0.06 is above the lower limit of the confidence interval. Yes, because 0.06 is below the lower limit of the confidence interval. No, because 0.06 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?

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

Answer #1

a)

point estimate for the proportion=0.101

b)

critical value of *z =1.96*

*c)*

margin of error (*E*) =0.021

d)

95% confidence interval for the proportion =0.080 < p <0.122

e)

Yes, because 0.06 is below the lower limit of the confidence interval.

f)

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

Pinworm: In a random sample of 780 adults in the U.S.A., it was
found that 76 of those had a pinworm infestation. You want to find
the 95% confidence interval for the proportion of all U.S. adults
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(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 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...

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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...

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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 =
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Pinworm: In a random sample of 830 adults in the U.S.A., it was
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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 780 adults in the U.S.A., it was found
that 76 of those had a pinworm infestation. You want to find the
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