Question
A manager at a drug manufacturing wants to estimate what proportion of the adult population of India has high blood pressure. He wants to be 99% sure that the error of his estimate will not exceed 0.02. Census reports indicate that about 0.20 of all adults have high blood pressure. What sample size shall he take?
Answer :
Word Count : 584
To determine the sample size for estimating a population proportion with a specified margin of error and confidence level, we use the standard formula from statistics for proportions: $$ n = \frac{Z^2 \cdot p \cdot (1-p)}{E^2} $$ Where: * $n$ = required sample size * $Z$ = Z-score corresponding to the desired confidence level * $p$ = estimated population proportion * $E$ = maximum allowable error (margin of error) From the question: * Desired confidence level = 99% → Z-score ($Z$) = 2.576 (from standard Z-tables) * Estimated population proportion ($p$) = 0.20 * Margin of error ($E$) = 0.02 Step 1: Plug values into the formula $$ n = \frac{(2.576)^2 \cdot 0.20 \cdot (1-0.20)}{(0.02)^2} $$ ___ ___ __________ _______ __________.
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To determine the sample size for estimating a population proportion with a specified margin of error and confidence level, we use the standard formula from statistics for proportions: $$ n = \frac{Z^2 \cdot p \cdot (1-p)}{E^2} $$ Where: * $n$ = required sample size * $Z$ = Z-score corresponding to the desired confidence level * $p$ = estimated population proportion * $E$ = maximum allowable error (margin of error) From the question: * Desired confidence level = 99% → Z-score ($Z$) = 2.576 (from standard Z-tables) * Estimated population proportion ($p$) = 0.20 * Margin of error ($E$) = 0.02 Step 1: Plug values into the formula $$ n = \frac{(2.576)^2 \cdot 0.20 \cdot (1-0.20)}{(0.02)^2} $$ ___ ___ __________ _______ __________.
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