Need help calculating the margin of error for the question below. I show n=100, sigma = 1, and 95% prob equls a z score of 1.96 for z_alpha/2. Just having issues calcuating the Margin of Error value. The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took these 100 customers to check out was 3 minutes. It is known that the standard deviation of the population of checkout times is 1 minute. With a 0.95 probability, the sample mean will provide a margin of error of 0.196. If we wanted to cut this margin of error in half, how many customers would we need to include in our sample?
Need help calculating the margin of error for the question below. I show n=100, sigma = 1, and 95% prob equls a z score of 1.96 for z_alpha/2. Just having issues calcuating the Margin of Error value. The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took these 100 customers to check out was 3 minutes. It is known that the standard deviation of the population of checkout times is 1 minute. With a 0.95 probability, the sample mean will provide a margin of error of 0.196. If we wanted to cut this margin of error in half, how many customers would we need to include in our sample?
Practical Management Science
6th Edition
ISBN:9781337406659
Author:WINSTON, Wayne L.
Publisher:WINSTON, Wayne L.
Chapter2: Introduction To Spreadsheet Modeling
Section: Chapter Questions
Problem 20P: Julie James is opening a lemonade stand. She believes the fixed cost per week of running the stand...
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Need help calculating the margin of error for the question below.
I show n=100, sigma = 1, and 95% prob equls a z score of 1.96 for z_alpha/2. Just having issues calcuating the Margin of Error value.
The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took these 100 customers to check out was 3 minutes. It is known that the standard deviation of the population of checkout times is 1 minute. With a 0.95 probability, the sample mean will provide a margin of error of 0.196. If we wanted to cut this margin of error in half, how many customers would we need to include in our sample?
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