Given a level of confidence of 99% & a population standard deviation of 10, 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=77. show all work.
To find the sample size (n) given a level of confidence and a population standard deviation, you would also need to know the desired margin of error.
(A) The margin of error is the maximum distance between the sample estimate and the true population value. It represents the uncertainty in the estimate. So, to determine the sample size (n), you would need to specify the desired margin of error.
(B) To find the maximum error of estimate (E), you can use the formula:
E = Critical value * (population standard deviation / √n)
where the critical value is determined based on the level of confidence.
Assuming a level of confidence of 99%, this corresponds to a significance level of α = 0.01. For a two-tailed test, the critical value can be found in the z-table or calculated using a statistical software.
Let's assume the critical value for a 99% confidence level is approximately 2.576.
Next, substitute the values into the formula:
E = 2.576 * (10 / √77)
Calculating the square root of 77, we get:
E = 2.576 * (10 / 8.77496439)
Evaluating the division, we have:
E = 2.576 * 1.138958625
Calculating the product, we find:
E ≈ 2.9291
Therefore, the maximum error of estimate (E) is approximately 2.9291, assuming a sample size (n) of 77, a population standard deviation of 10, and a 99% confidence level.