A random sample of 1900 workers in a particular city found 399 workers who had full health insurance coverage.

A random sample of 1900 workers in a particular city found 399 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. Express your results to the nearest hundredth of a percent.

Answer: to %

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Expert Solution

sample size = n = 1900

x = 399

p = x/n = 0.21

zα = 1.96 at 95% confidence

standard error = s.e. = √[p(1-p)/n] = 0.00934429181

margin of error = E = s.e.× zα = 0.00934429181 × 1.96 = 0.0183148119

95% confidence interval for this proportion ( p – E; p + E ) = ( 0.1916851880 ,0.2283148119 )

= (0.19; 0.23) [Approx.]

orchestra answered 1 year ago

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