Answer :
The 95% confidence interval for the proportion of adults residents who are parents is; 0.381 < p < 0.519
The confidence interval expressed using the point estimate and margin of error is; 0.45 - 6.895 × 10⁻² < p < 0.45 + 6.895 × 10⁻²
What is a confidence interval?
A confidence interval is a specification of the interval within which a parameter of the population can be found for a specified proportion time duration.
The number of respondent in the sample, n = 200
The number of respondent that had kids = 80
The confidence level = 95%
The confidence interval of a population proportion can be found as follows;
[tex]\hat{p} \pm z ^* \times \sqrt{\dfrac{\hat {p}\cdot \left(1 - \hat{p}\right)}{n} }[/tex]
Where;
[tex]\hat{p} = \dfrac{90}{200} = 0.45[/tex]
z = The z-score at 95% confidence level ≈ 1.96
Therefore;
[tex]\hat{p} \pm z ^* \times \sqrt{\dfrac{\hat {p}\cdot \left(1 - \hat{p}\right)}{n} } =0.45\pm 1.96 \times \sqrt{\dfrac{0.45\cdot \left(1 - 0.45\right)}{200} }\\[/tex]
Therefore, the 95% confidence interval is 0.381 < p < 0.519
The margin of error = [tex]1.96 \times \sqrt{\dfrac{0.45\cdot \left(1 - 0.45\right)}{200} }\\[/tex] ≈ 6.895 × 10⁻²
The point estimate is 90/100 = 0.45
Therefore;
0.45 - 6.895 × 10⁻² < p < 0.45 + 6.895 × 10⁻²
Learn more on the margin of error here: https://brainly.com/question/2595840
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