Answer:
95% confidence interval for the proportion of companies likely to require higher employee contributions for health care coverage.
(0.5868 , 0.6532)
Step-by-step explanation:
Step(i):-
Given the survey was based on a sample of 800 companies
Given size 'n' = 800
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
sample proportion
p⁻ = 0.62
Step(ii):-
The margin of error for the proportion of companies likely to require higher employee contributions for health care coverage.
[tex]M.E= Z_{0.05} \sqrt{\frac{p^{-} (1-p^{-}) }{n} }[/tex]
[tex]M.E= 1.96\sqrt{\frac{0.62 (1-0.62 }{800} }[/tex]
M.E = 0.017 X 1.96
M.E = 0.03
Step(iii):-
95% confidence interval for the proportion of companies likely to require higher employee contributions for health care coverage.
[tex](p^{-} - Z_{0.05} \sqrt{\frac{p^{-} (1-p^{-}) }{n} } , p^{-} +Z_{0.05} \sqrt{\frac{p^{-} (1-p^{-}) }{n} })[/tex]
[tex](0.62 - 1.96\sqrt{\frac{0.62 (1-0.62 }{800} } ,0.62+1.96\sqrt{\frac{0.62 (1-0.62 }{800} }[/tex]
( 0.62 - 0.0332 , 0.62+0.0332)
(0.5868 , 0.6532)