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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 check out times is 1 minute. with a .95 probability, the sample mean will provide a margin of error

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Answer:

with a .95 probability, the sample mean will provide a margin of error 0.196

Step-by-step explanation:

margin of error (ME) from the mean can be calculated using the formula

ME=[tex]\frac{z*s}{\sqrt{N} }[/tex] where

  • z is the corresponding statistic in the .95 confidence level (1.96)
  • s is the population standard deviation (1 min.)
  • N is the sample size (100)

ME=[tex]\frac{1.96*1}{\sqrt{100} }[/tex]=0.196

95% confidence interval for check-out time would be 3 ±0.196 min

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