Question

Of 97 adults selected randomly from one city, 63 have health insurance. Create a 95% confidence interval for the true proportion of all adults in the town who have health insurance.

Answer 1

Answer: 95% confidence interval would be (0.555,0.745).

Explanation:

Since we have given that

n = 97

x = 63

So, we get that

`\hat{p}=(x)/(n)=(63)/(97)=0.65`

At 95% confidence interval , z = 1.96

so, Margin of error would be

`z* \sqrt{(p(1-p))/(n)}\\\\=1.96* \sqrt{(0.65* 0.35)/(97)}\\\\=0.095`

So, interval would be

`p\pm 0.095\\\\=0.65\pm 0.095\\\\=(0.65-0.095,0.65+0.095)\\\\=(0.555,0.745)`

Hence, 95% confidence interval would be (0.555,0.745).