Question 825249: Executives of a supermarket chain are interested in the amount of time that customers spend in the stores during shopping trips. The executives hire a statistical consultant and ask her to determine the mean shopping time(mu) of customers at the supermarkets. The consultant will collect a random sample of shopping times at the supermarkets and use the mean of these shopping times to estimate (mu). Assuming that the standard deviation of the population of shopping times at the supermarkets is 26 minutes, what is the minimum sample size she must collect in order for her to be 95% confident that her estimate is within 5 minutes of (mu)?
Here's what I worked out:
n>=
(1.96)^2(26)^2
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25
= (3.8416)(676)
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25
=27.04
I'm hoping the math is correct. However, I'm not sure if the minimum would be 28 (since you can't have .04 of a person) or 27. A little help, please.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Executives of a supermarket chain are interested in the amount of time that customers spend in the stores during shopping trips. The executives hire a statistical consultant and ask her to determine the mean shopping time(mu) of customers at the supermarkets. The consultant will collect a random sample of shopping times at the supermarkets and use the mean of these shopping times to estimate (mu). Assuming that the standard deviation of the population of shopping times at the supermarkets is 26 minutes, what is the minimum sample size she must collect in order for her to be 95% confident that her estimate is within 5 minutes of (mu)?
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n = [z*s/E]^2
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n = [1.96*26/5]^2 = 104 when rounded up
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Cheers,
Stan H.
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