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).

Step-by-step explanation:

Since we have given that

n = 97

x = 63

So, we get that

$\hat{p}=\dfrac{x}{n}=\dfrac{63}{97}=0.65$

At 95% confidence interval , z = 1.96

so, Margin of error would be

$z\times \sqrt{\dfrac{p(1-p)}{n}}\\\\=1.96\times \sqrt{\dfrac{0.65\times 0.35}{97}}\\\\=0.095$

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).