Question

A random sample of 1700 workers in a particular city found 578 workers who had full health insurance coverage. Find a 95% confidence interval for the true percent of workers in this city who have full health insurance coverage.

Answer 1

Answer: (31.75%, 36.25%)

Explanation:

Let p be the proportion of workers in this city who have full health insurance coverage.

As per given,

Sample size : n= 170

Number of workers who had full health insurance coverage.=578

i.e. sample proportion:
`\hat{p}=(578)/(1700)\approx0.34`

Also, z-score for 95% confidence level : 1.96

Formula to find the confidence interval for p :

`\hat{p}\pm z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}`

`0.34\pm (1.96)\sqrt{(0.34(1-0.34))/(1700)}`

`=0.34\pm (1.96)√(0.000132)\\\\=0.34\pm (1.96)(0.011489125)\\\\\approx 0.34\pm0.0225\\\\=(0.34-0.0225,\ 0.34+0.0225)\\\\=(0.3175,\ 0.3625) =(31.75\%,\ 36.25\%)`

Hence, a 95% confidence interval for the true percent of workers in this city who have full health insurance coverage= (31.75%, 36.25%)