Question

We wish to estimate what percent of adult residents in a certain
county are parents. Out of 100 adult residents sampled, 40 had
kids. Based on this, construct a 95% 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

point estimate of Sample proportion, = 40/100 = 0.4

Standard error of sample proportion, SE = = = 0.049

Z value for 95% confidence interval is 1.96

Margin of error = SE * Z = 0.049 * 1.96 = 0.096

95% confidence interval for the proportion p of adult residents who are parents in this county is,

0.4 - 0.096 < p < 0.4 + 0.096

**0.304 < p < 0.496**

**p = 0.4
0.096**

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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.
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certain county are parents. Out of 400 adult residents sampled, 296
had kids. Based on this, construct a 95% 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.
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county are parents. Out of 600 adult residents sampled, 270 had
kids. Based on this, construct a 90% confidence interval for the
proportion p p of adult residents who are parents in this
county.

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county are parents. Out of 500 adult residents sampled, 400 had
kids. Based on this , construct a 90% confidence interval for the
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Enter your answer as an open-interval
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