Asked by Eric
Find the minimum sample size that should be chosen to assure that the proportion estimate p will be within the required margin of error, .06. Use a 95% confidence interval and a population proportion of .7. The critical value for a 95% confidence level is 1.96
Answers
Answered by
MathGuru
Try this formula:
n = [(z-value)^2 * p * q]/E^2
= [(1.96)^2 * .7 * .3]/.06^2
= ? (round to the next highest whole number)
I'll let you finish the calculation.
Note: n = sample size needed. Use .7 for p and .3 for q (q = 1 - p). E = maximum error, which is .06 and z-value is 1.96.
I hope this helps.
n = [(z-value)^2 * p * q]/E^2
= [(1.96)^2 * .7 * .3]/.06^2
= ? (round to the next highest whole number)
I'll let you finish the calculation.
Note: n = sample size needed. Use .7 for p and .3 for q (q = 1 - p). E = maximum error, which is .06 and z-value is 1.96.
I hope this helps.
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