Question

**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 3 decimal
places.**

< *p* <

(c) Based on your answer to part (b), are you 99% confident that
more than 6% of all U.S. adults have pinworm?

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

No, because 0.06 is below 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.

(d) In Sludge County, the proportion of adults with pinworm is
found to be 0.16. Based on your answer to (b), does Sludge County's
pinworm infestation rate appear to be above the national
average?

No, 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.

Yes, 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.

Answer #1

a) For i from 1 to 810, le Xi = 1 if the ith person has pinworm and 0 otherwise. These are Bernoulli(p) independent samples.

The point estimate of mean is the sample mean= = 84/810 = 0.1037

b) The estimatedstandard deviation of the mean is

The 99% confidence interval is bounded by points where is the point over which in N(0,1)

the probability would be 0.005. This point is 2.575829. Thus the confidence interval is (0.0761071, 0.1312929) on substituting these values. Rounding off,

0.076 < p < 0.131

c) Yes, because 0.06 is below the lower limit.

d) Yes, because 0.16 is above the upper limit.

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In a random sample of 830 adults in the U.S.A., it was found
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(b) What is the critical value of z (denoted
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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.
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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
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adults with pinworm? Round your answer to 3 decimal places.
(b) Construct the 90% confidence interval for the proportion of
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