Given a level of confidence of 99% and a population standard deviation of 13, answer the following:
(A) What other information is necessary to find the sample size (n)?
(B) Find the Maximum Error of Estimate (E) if n = 61. Show all work.
(A) In order to find the sample size (n) with a level of confidence of 99% and a population standard deviation of 13, you would also need to know the desired margin of error. The margin of error represents the maximum allowed difference between the sample mean and the population mean.
(B) To find the maximum error of estimate (E) with a sample size (n) of 61, you can use the formula:
E = Z * (σ / sqrt(n))
where:
- E is the maximum error of estimate (margin of error)
- Z is the z-score corresponding to the desired level of confidence
- σ is the population standard deviation
- n is the sample size
Since the level of confidence is 99%, the corresponding z-score can be found using a standard normal distribution table or a statistical software. For a 99% level of confidence, the z-score is approximately 2.576.
Plugging in the values:
E = 2.576 * (13 / sqrt(61))
Calculating the square root of 61:
sqrt(61) ≈ 7.810
E = 2.576 * (13 / 7.810)
E ≈ 4.279
Therefore, the maximum error of estimate (E) is approximately 4.279 when the sample size (n) is 61.