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12 October, 19:28

A grocery store has an average sales of $8000 per day. The store introduced several advertising campaigns in order to increase sales. To determine whether or not the advertising campaigns have been effective in increasing sales, a sample of 100 days of sales was selected. It was found that the average was $8300 per day. From past information, it is known that the standard deviation of the population is $1200. The p-value is

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  1. 12 October, 19:55
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    0.0062

    Explanation:

    p = 1 - P (Z < Zcal)

    Zcal = (8300-8000) / (1200/√100)

    Zcal = 2.5

    p = 1 - P (Z < Zcal)

    = 1 - P (Z < 2.5)

    = 1 - 0.9938

    = 0.0062
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