necessaryh

2021-02-21

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 $99\mathrm{%}$ confidence interval for the proportion of adult residents who are parents in a given county.
Express your answer in the form of three inequalities. Provide your responses in decimal fractions up to three places $ Express the same answer using a point estimate and a margin of error. Provide your responses as decimals, to three places.
$p=±$

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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 $p=\frac{x}{n}=\frac{192}{600}=0.32.$
From the standard normal table, the Z value at $99\mathrm{%}$ confidence interval is $Z=2.576.$
Substitute the known values in the formula $E=Z\cdot \sqrt{\frac{p\left(1-p\right)}{n}}$ to evaluate the margin of error E.
$E=Z\cdot \sqrt{\frac{p\left(1-p\right)}{n}}$
$=2.576\cdot \sqrt{\frac{0.32\left(1-0.32\right)}{600}}$
$\approx 0.049056852$
Step 2
The confidence interval is given by $\left(p-E,p+E\right).$
Here, we have
$\left(p-E,p+E\right)=\left(0.32-0.049056852,0.32+0.049056852\right)$
$=\left(0.270943147,0.369056852\right)$
$\approx \left(0.271,0.369\right)$
Hence, the confidence interval in tri-inequality form is $0.271
The confidence interval suing point estimate 0.32 and margin of error

Jeffrey Jordon

Answer is given below (on video)

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