The manager of a grocery store has taken a random sample of 144 customers. The average length of time it took these 144 customers to check out was 3 minutes. It is known that the standard deviation of the population of checkout times is 1 minute. The standard error of the mean is .083. Compute a 90% confidence interval. At 90% confidence, the size of the margin of error is? The 90% confidence interval estimate of the population mean check out time is?
Added by Elizabeth H.
Step 1
For a 90% confidence interval, the z-score is 1.645 (you can find this value in a standard z-table or use a calculator that provides it). The formula for the margin of error is: Margin of Error = Z-score * Standard Error So, we plug in the values we Show more…
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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 3.0 minutes. It is known that the standard deviation of the population of checkout times is one minute. The 95% confidence interval for the true average checkout time (in minutes) is 1.00 to 5.00 2.804 to 3.196 1.36 to 4.64 3.00 to 5.00
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The owner of a local chain of grocery stores is always trying to minimize the time it takes her customers to check out. In the past, she has conducted many studies of the checkout times, and they have displayed a normal distribution with a mean time of 12 minutes and a standard deviation of 2.3 minutes. She has implemented a new schedule for cashiers in hopes of reducing the mean checkout time. A random sample of 28 customers visiting her store this week resulted in a mean of 10.9 minutes. Does she have sufficient evidence to claim the mean checkout time this week was less than 12 minutes? Use $\alpha=0.02$
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