Mathematics: 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.

The variable must be raised to the second power...

A random sample of 1700 workers in a particular

Ответ:

aaronolivera200161

31.01.2022

Mathematics

(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: $\hat{p}=\dfrac{578}{1700}\approx0.34$

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

Formula to find the confidence interval for p :

$\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$0.34\pm (1.96)\sqrt{\dfrac{0.34(1-0.34)}{1700}}$

$=0.34\pm (1.96)\sqrt{0.000132}\\\\=0.34\pm (1.96)(0.011489125)\\\\\approx 0.34\pm0.0225\\\\=(0.34-0.0225,\ 0.34+0.0225)\\\\=(0.3175,\ 0.3625) =(31.75\%,\ 36.25\%)$

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

0,0(0 оценок)

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.

Ответ:

Ответ:

Ask your question to the AI advisor

More tips

Answers on questions: Mathematics

Ask an AI advisor a question