(2 pts) A random sample of 2000 workers in a particular city found 1000 workers who had full health insurance?

(2 pts) A random sample of 2000 workers in a particular city found 1000 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.
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ANSWER: 95% Resulting Confidence Interval for 'true mean': = [0.478, 0.522] Why??? POPULATION PROPORTION, CONFIDENCE INTERVAL, NORMAL DISTRIBUTION n = Number of samples 2000 Number of Successes 1000 p = POPULATION PROPORTION 0.5 significant digits 3 Confidence Level 95 "Look-up" Table 'z-critical value' 1.960 from Excel function: NORMSINV(probability) Returns the inverse of the standard normal cumulative distribution. The distribution has a mean of zero, a standard deviation of one and is symmetrical. 95% Resulting Confidence Interval for 'true mean': p +/- (z critical value) * SQRT[p * (1 - p)/n] = 0.5 +/- 1.96 * SQRT[0.5 * (1 - 0.5)/2000] = [0.478, 0.522]
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