Question

We wish to estimate what percent of adult residents in a certain
county are parents. Out of 400 adult residents sampled, 128 had
kids. Based on this, construct a 99% confidence interval for the
proportion p of adult residents who are parents in this
county.

Express your answer in tri-inequality form. Give your answers as
decimals, to three places.

____ < p < ____ Express the same answer using the point
estimate and margin of error. Give your answers as decimals, to
three places.

p = ____ ± ____

Answer #1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 128 / 400 = 0.320

Z_{/2}
= 2.576

Margin of error = E = Z_{
/ 2} *
((
* (1 -
)) / n)

= 2.576 * (((0.320 * 0.680) / 400)

= 0.060

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.320 - 0.060 < p < 0.320 + 0.060

**0.260 < p < 0.380**

**p = 0.320 0.060**

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Express your answer in tri-inequality form. Give your answers as
decimals, to three places.
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Enter your answer as an open-interval
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QUESTION PART A: Assume that a sample is used to estimate a
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M.E. = %
Answer should be obtained without any preliminary rounding.
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Round final answer to one decimal place
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