Answered

IDNLearn.com makes it easy to find accurate answers to your specific questions. Join our community to access reliable and comprehensive responses to your questions from experienced professionals.

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

Sagot :

Abidemiokinidn

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

We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Accurate answers are just a click away at IDNLearn.com. Thanks for stopping by, and come back for more reliable solutions.