Asked by Monique
The estimate of the population proportion is to be within plus or minus .10, with a 99 percent level of confidence. The best estimate of the population proportion is .45. How large a sample is required?
Answers
Answered by
MathGuru
Here is one formula you might use for this problem:
n = [(z-value)^2 * p * q]/E^2
With your data:
n = [(2.575)^2 * .45 * .55]/.10^2
I'll let you finish the calculation (round to the next highest whole number).
Note: n = sample size needed; .45 for best estimate of the population proportion; .55 for q, which is (1 - p). E = maximum error, which is .10 from the problem. Z-value is found using a z-table (for 99%, the value is 2.575).
I hope this helps.
n = [(z-value)^2 * p * q]/E^2
With your data:
n = [(2.575)^2 * .45 * .55]/.10^2
I'll let you finish the calculation (round to the next highest whole number).
Note: n = sample size needed; .45 for best estimate of the population proportion; .55 for q, which is (1 - p). E = maximum error, which is .10 from the problem. Z-value is found using a z-table (for 99%, the value is 2.575).
I hope this helps.
Answered by
Monique
THANK YOU
Answered by
LUCIA
165
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