Asked by chalulu
A researcher at a major hospital wishes to estimate the proportion of the adult population of the U.S. that has high blood pressure. How large a sample is needed in order to be 99% confident that the sample proportion will not differ from the tgrue proportion by more than 4%?
Answers
Answered by
MathGuru
Formula:
n = [(z-value^2 * p * q)]/E^2
...where n = sample size, z-value will be 2.58 using a z-table to represent the 99% confidence interval, p and q represent proportions, E = .04, ^2 means squared, and * means to multiply.
Note: p and q will both be .5 if no value is stated.
Plug the values into the formula and finish the calculation. Round your answer to the next highest whole number.
n = [(z-value^2 * p * q)]/E^2
...where n = sample size, z-value will be 2.58 using a z-table to represent the 99% confidence interval, p and q represent proportions, E = .04, ^2 means squared, and * means to multiply.
Note: p and q will both be .5 if no value is stated.
Plug the values into the formula and finish the calculation. Round your answer to the next highest whole number.
Answered by
Anonymous
3e
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