JustAnswer
a)If a population has a standard deviation of 20, what is the minimum number of samples that need to be averaged in order to be 95% confident that the average of the means is within 3 of the true mean?
|Confidence Level % = 95 |
|z- score = 1.96 |
|Population SD, σ = 20 |
|Error, E = 3 |
|Sample Size, N = (z * σ / E)^2 = (1.96 * 20/3)^2 = 171 |
b)A sample of 35 different payroll departments found that employees worked an average of 240.6 days a year. If the population standard deviation is 18.8 days, find the 90% confidence interval for the average number of days worked by all employees who are paid through payroll departments.
|Data: |
|n =35 |
|x-bar = 240.6 |
|σ ’ 18.8 |
|% = 90 |
|Standard Error, SE = σ/√n = 18.8/√35 = 3.1778 |
|z- score = 1.645 |
|Width of the confidence interval = z * SE = 1.645 * 3.1778 = 5.227 |
|Lower Limit of the confidence interval = x-bar - width = 235.37 |
|Upper Limit of the confidence interval = x-bar + width = 245.83 |
|The confidence interval is [235.37 days, 245.83 days]. |
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