construct a confidence interval of the population proportion at the given level

Question

construct a confidence interval of the population proportion at the given level of confidence x=860 n=1100 95%confidence
The lower bound of the confidence interval is? (Round to the nearest thousandth as needed)

The upper bound of the confidence interval is? (Round to the nearest thousandth as needed)


construct a confidence interval of the population proportion at the given level of confidence x=860 n=1100 95%confidence
The lower bound of the confidence interval is? (Round to the nearest thousandth as needed)

The upper bound of the confidence interval is? (Round to the nearest thousandth as needed)


construct a confidence interval of the population proportion at the given level of confidence x=860 n=1100 95%confidence
The lower bound of the confidence interval is? (Round to the nearest thousandth as needed)

The upper bound of the confidence interval is? (Round to the nearest thousandth as needed)

Explanation / Answer

p = x/n = 860/1100 = .7818

The standard error of p is SE = sqrt(p*(1-p)/n) = sqrt((0.7818*(0.2181/1100)) = sqrt(0.000155)

The confidence interval has the form p +- (z*) * SE

z* is the critical z-value from a Z table for which PR is the upper (1 - 0.95)/2 tail of the distribution.

P(Z< 1.96) = 0.975 so z* = 1.96

So the 95% confidence interval is

0.7818 +- 1.96*sqrt(0.000155) = 0.7818 +- 0.0244 = (0.757,0.806)

The lower bound of the confidence interval is = .757

The upper bound of the confidence interval is = .806

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