The point estimate of "p" is "\\hat{p}={8 \\over30}".
If "\\hat{p}" is the proportion of successes in a random sample of size "n" and "\\hat{q}=1-\\hat{p}," then an approximate "100(1-\\alpha)\\%" confidence interval ( CI ), for the binomial parameter "\\hat{h}" is given by
where "\\hat{p}" is the sample proportion, "n" is the sample size, and "z_{\\alpha\/2}"is the appropriate value from the standard normal distribution for our desired confidence interval.
Given that "\\hat{p}=8\/30, n=30, \\alpha=0.05."
Check
Determine "z_{\\alpha\/2}"
Compute a 95% confidence interval on p
"CI=\\bigg({8 \\over30}- 1.96\\sqrt{{{8 \\over30}(1-{8 \\over30}) \\over 30}}, {8 \\over30}+ 1.96 \\sqrt{{{8 \\over30}(1-{8 \\over30}) \\over 30}}\\bigg)"
"CI=(0.108, 0.425)"
We shall calculate how many companies would need to be sampled to have a 95% confidence interval with a width "w" of only 0.05.
We must choose "n" such that
Solve for "n"
"n\\geq{(z_{\\alpha\/2})^2 \\hat{p}(1-\\hat{p})\\over w^2}"
Given that
Substitue
"n\\geq301"
We need to sample at least 301 companies to have a 95% confidence interval with a width of only 0.05.
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