Answer: (31.75%, 36.25%)
Step-by-step 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: [tex]\hat{p}=\dfrac{578}{1700}\approx0.34[/tex]
Also, z-score for 95% confidence level : 1.96
Formula to find the confidence interval for p :
[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
[tex]0.34\pm (1.96)\sqrt{\dfrac{0.34(1-0.34)}{1700}}[/tex]
[tex]=0.34\pm (1.96)\sqrt{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\%)[/tex]
Hence, a 95% confidence interval for the true percent of workers in this city who have full health insurance coverage= (31.75%, 36.25%)
