Answer:
<em>95% confidence interval for the proportion of companies likely to require higher employee contributions for health care coverage.</em>
<em>(0.5868 , 0.6532)</em>
Step-by-step explanation:
<u><em>Step(i):-</em></u>
<em>Given the survey was based on a sample of 800 companies</em>
<em>Given size 'n' = 800</em>
<em>A recent survey showed that 62% of employers are likely to require higher employee contributions for health care coverage this year relative to last year</em>
<em>sample proportion </em>
<em> p⁻ = 0.62</em>
<u><em>Step(ii):-</em></u>
The margin of error for the proportion of companies likely to require higher employee contributions for health care coverage.
![M.E= Z_{0.05} \sqrt{\frac{p^{-} (1-p^{-}) }{n} }](https://tex.z-dn.net/?f=M.E%3D%20Z_%7B0.05%7D%20%20%5Csqrt%7B%5Cfrac%7Bp%5E%7B-%7D%20%281-p%5E%7B-%7D%29%20%7D%7Bn%7D%20%7D)
![M.E= 1.96\sqrt{\frac{0.62 (1-0.62 }{800} }](https://tex.z-dn.net/?f=M.E%3D%201.96%5Csqrt%7B%5Cfrac%7B0.62%20%281-0.62%20%7D%7B800%7D%20%7D)
M.E = 0.017 X 1.96
M.E = 0.03
<u>Step(iii):- </u>
<em>95% confidence interval for the proportion of companies likely to require higher employee contributions for health care coverage.</em>
![(p^{-} - Z_{0.05} \sqrt{\frac{p^{-} (1-p^{-}) }{n} } , p^{-} +Z_{0.05} \sqrt{\frac{p^{-} (1-p^{-}) }{n} })](https://tex.z-dn.net/?f=%28p%5E%7B-%7D%20-%20Z_%7B0.05%7D%20%20%5Csqrt%7B%5Cfrac%7Bp%5E%7B-%7D%20%281-p%5E%7B-%7D%29%20%7D%7Bn%7D%20%7D%20%2C%20p%5E%7B-%7D%20%2BZ_%7B0.05%7D%20%20%5Csqrt%7B%5Cfrac%7Bp%5E%7B-%7D%20%281-p%5E%7B-%7D%29%20%7D%7Bn%7D%20%7D%29)
( 0.62 - 0.0332 , 0.62+0.0332)
<em>(0.5868 , 0.6532)</em>