Given a level of confidence of 99% and a population standard deviation of 12, 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 = 78.

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To find the sample size (n) with a given level of confidence of 99% and a population standard deviation of 12, you will need to know the desired margin of error (E) or the population size (N).

(A) What other information is necessary to find the sample size (n)?
- To find the sample size (n), you will need to know either the desired margin of error (E) or the population size (N). If you have the desired margin of error, you can use the formula:

n = (Z * σ / E)^2

where:
n is the sample size,
Z is the Z-score corresponding to the desired level of confidence,
σ is the population standard deviation, and
E is the desired margin of error.

If you have the population size (N), you can use the formula:

n = (Z^2 * σ^2 * N) / [(Z^2 * σ^2) + (E^2 * (N-1))]

where:
n is the sample size,
Z is the Z-score corresponding to the desired level of confidence,
σ is the population standard deviation,
E is the desired margin of error, and
N is the population size.

(B) Find the Maximum Error of Estimate (E) if n = 78:
To find the Maximum Error of Estimate (E) with a given sample size (n), you can rearrange the sample size formula to solve for E:

E = Z * σ / sqrt(n)

where:
E is the Maximum Error of Estimate,
Z is the Z-score corresponding to the desired level of confidence,
σ is the population standard deviation, and
n is the sample size.

Using the given values:
Z-score for a 99% confidence level is approximately 2.576 (look up in the Z-table or use a calculator),
σ = 12,
n = 78,

E = 2.576 * 12 / sqrt(78)
E ≈ 3.698

Therefore, the Maximum Error of Estimate (E) is approximately 3.698 when the sample size (n) is 78.