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Of 92 adults selected randomly from one town, 61 have health insurance. Find a 90% confidence interval for the true proportion of all adults in the town who have health insurance. Group of answer choices 0.582 < p < 0.744 0.548 < p < 0.778 0.536 < p < 0.790 0.566 < p < 0.760

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abidemiokin

A 90% confidence interval for the true proportion of all adults in the town who have health insurance is 0.582 < p < 0.744

The formula for calculating the confidence interval is expressed as;

[tex]CI=p \pm z \cdot\sqrt{\frac{P(1-p)}{n} }[/tex]

  • p is the proportion = 61/92 = 0.66
  • n is the sample size = 92
  • z is the z-score at 90% = 1.645

Substitute the given parameters into the formula to have:

[tex]CI=0.66 \pm 1.645 \cdot\sqrt{\frac{0.66(1-0.66)}{92} }\\CI=0.66 \pm 1.645 \cdot\sqrt{\frac{0.66(0.34)}{92} }\\CI =0.66\pm 1.645(0.0495)\\CI=0.66 \pm 0.0814\\CI = (0.582, 0.744)[/tex]

Hence a 90% confidence interval for the true proportion of all adults in the town who have health insurance is 0.582 < p < 0.744

Learn more on confidence interval here: https://brainly.com/question/15712887

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