How a large sample should be surveyed to estimate the true proportion of college students who do a laundry once a week within 3% with 95% confidence?
"ME=z_{0.025}\\sqrt{\\frac{p(1-p)}{n}}"
"n=(\\frac{z_{0.025}}{ME})^2p(1-p)=(\\frac{1.96}{0.03})^20.5(1-0.5)=1068."
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