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 o

Question

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

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Kinsley 3 months 2021-09-03T19:08:29+00:00 1 Answer 0 views 0

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  1. Ava
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    2021-09-03T19:10:15+00:00

    Answer:

    Margin of error = 0.196

    Step-by-step explanation:

    We are given that 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.

    That means; Sample size, n = 100

                        Population standard deviation = 1 minute

                        Sample mean = 3 minutes

    Now, the Margin of error means how much deviation is there from the sample mean to calculate the confidence interval for true population mean.

    Margin of error formula is given by = Z_\frac{\alpha}{2} \times \frac{\sigma}{\sqrt{n} }

    where, \sigma = population standard deviation

                n = sample size of customers

                \alpha = significance level = 5%

    Now, the z probability area at 5% significance level is given to be 1.96 from z table.

    So, Margin of error = 1.96 \times \frac{1}{\sqrt{100} } = \frac{1.96}{10} = 0.196 .

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

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