Asked by Heather
A large corp wants to estimate the proportion of independent contractors who use a certain device. They want to be within 4% of the true proportion when using a confidence interval of 95%.
How many contractors must be sampled if no preliminary estimate is available.
I came up with 110. Just looking for some confirmation please.
How many contractors must be sampled if no preliminary estimate is available.
I came up with 110. Just looking for some confirmation please.
Answers
Answered by
MathGuru
Try this formula:
n = [(z-value)^2 * p * q]/E^2
= [(1.96)^2 * .5 * .5]/.04^2
= .9604/.0016
= 601 (rounding to the next highest whole number)
Note: n = sample size needed; .5 for p and .5 for q are used if no value is stated in the problem. E = maximum error, which is .04 (4%) in the problem. Z-value is found using a z-table (for 95%, the value is 1.96).
n = [(z-value)^2 * p * q]/E^2
= [(1.96)^2 * .5 * .5]/.04^2
= .9604/.0016
= 601 (rounding to the next highest whole number)
Note: n = sample size needed; .5 for p and .5 for q are used if no value is stated in the problem. E = maximum error, which is .04 (4%) in the problem. Z-value is found using a z-table (for 95%, the value is 1.96).
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