We wish to estimate what percent of adult residents in a certain county are parents. Out of 600 adult residents sampled, 192 had kids. Based on this,

necessaryh 2021-02-21 Answered

We wish to estimate what percent of adult residents in a certain county are parents. Out of 600 adult residents sampled, 192 had kids. Based on this, plot a 99% confidence interval for the proportion of adult residents who are parents in a given county.
Express your answer in the form of three inequalities. Give your answers in decimal fractions up to three places <p< Express the same answer using a point estimate and a margin of error. Give your answers as decimals, to three places.
p=±

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Expert Answer

ottcomn
Answered 2021-02-22 Author has 97 answers

Step 1
Given that the sample size n=600, the favourable size x=192.
Let p denotes the proportion of adult residents who are parents in the country.
Thus, the value of p is given by p=xn=192600=0.32.
From the standard normal table, the Z value at 99% confidence interval is Z=2.576.
Substitute the known values in the formula E=Zp(1p)n to evaluate the margin of error E.
E=Zp(1p)n
=2.5760.32(10.32)600
0.049056852
Step 2
The confidence interval is given by (pE,p+E).
Here, we have
(pE,p+E)=(0.320.049056852,0.32+0.049056852)
=(0.270943147,0.369056852)
(0.271,0.369)
Hence, the confidence interval in tri-inequality form is 0.271<p<0.369.
The confidence interval suing point estimate 0.32 and margin of error E0.049 is p=0.32±0.049.

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Jeffrey Jordon
Answered 2021-10-27 Author has 2027 answers

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