Question

In: Statistics and Probability

A recent survey showed that 65% of U.S. employers were likely to require higher employee contributions...

A recent survey showed that 65% of U.S. employers were likely to require higher employee contributions for health care coverage. Suppose the survey was based on a sample of 600 companies. Find the 98% confidence interval for the proportion of all companies likely to require higher employee contributions for health care coverage.

Find left endpoint and right endpoint.

Expert Solution

Solution :

Given that,

n = 600

Point estimate = sample proportion = = 0.65

1 - = 1 - 0.65 = 0.35

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z_/2 = Z _0.01 = 2.326

Margin of error = E = Z_{/ 2} * $\sqrt$ (( * (1 - )) / n)

= 1.96 * ( $\sqrt$ ((0.65 * 0.35) / 600)

= 0.045

A 98% confidence interval for population proportion p is ,

- E < p < + E

0.65 - 0045 < p < 0.65 + 0.045

0.605 < p < 0.695

Left endpoint = 0.605

right endpoint = 0.695

orchestra answered 2 years ago

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