Statistice help??? Confidence interval & sampling error.

Question

A company surveyed 600 people for a new retirement plan. 235 employees prefered the new plan. What is the employee population proportion that would agree to the new plan? Find a confidence interval for 99% confidence. What is the sampling error? What is the sample size required to reduce the sampling error to 55? Use a 99% confidence.

Answer 1

A.      First, you need to label all the relevant data in the question:

·       Sample size (n) = 600 people

·       Population (sample) proportion (p̂) = 235/600 = 0.3917 (39.17%)

·       Z-score for 99% confidence interval (*This is a constant*) = 2.576

B.      Next, let's calculate the confidence interval using this equation:

CI = p̂ ± Z (√(p̂ (1 – p̂)/n))

** The ± means we will use the same equation twice. Once adding and then subtracting**

CI = 0.3917 + 2.579(√(0.3917(1-0.3917)/600)

CI = 0.3917 + 2.579(√(0.3917(0.6083)/600)

CI = 0.3917 + 2.579(√(0.2383)/600)

CI = 0.3917 + 2.579(√(0.0003971)

CI = 0.3917 + 2.579(0.01993)

CI = 0.3917 + 0.0513; CI = 0.3917 - 0.0513

CI = (0.34, 0.44) at 99% confidence; 34% to 44% of employees prefer the new plan.

C.    On to sampling error: The formula we will use for this part is:

SE = Z(√(p̂(1-p̂)/n)

SE = 2.579(√(0.3917(1-0.3917)/600)

SE = 2.579(√(0.3917(0.6083)/600)

SE = 2.579(√(0.2383)/600)

SE = 2.579(√(0.0003971)

SE = 2.579(0.01993)

SE = 0.0514 or 5.14%

D.    Finally, let's find the sample size needed to reduce the sampling error to 5.5 (I’m assuming 55 was a typo). We will reverse engineer the above equation like this to find n:

SE = Z(√(p̂(1-p̂)/n)  -> n = (Z²/SE²) * p̂(1-p̂)

N = (2.579²/0.055²) * 0.3917(1-0.3917)

N = (6.651/0.003025) * 0.3917(0.6083)

N = 2198.68 * 0.238

N = 523.3 or 523 people surveyed are needed to reduce the sampling error to 5.5.