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    xop5's Avatar
    xop5 Posts: 1, Reputation: 1
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    #1

    Jan 17, 2006, 02:47 PM
    Statistics
    Chicken Delight claims that 90% of its orders are delivered within 10 minutes of the time the order is placed. A sample of 100 orders revealed that 82 were delivered within the promised time. At the 0.10 significance level, can we conclude that less than 90% of the orders are delivered in less than 10 minutes?
    manutd4eva's Avatar
    manutd4eva Posts: 209, Reputation: 14
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    #2

    Jan 17, 2006, 02:51 PM
    I think so but wouldn't you need a higher number of orders and at say the same time of day and the same order etc etc to make it fair
    reinsuranc's Avatar
    reinsuranc Posts: 92, Reputation: 6
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    #3

    Jan 19, 2006, 05:58 AM
    Either a delivery is on time (<= 10 minutes), or it isn't, so assume this is a binomial distribution with n=100, p=.9, q=1-p=.1

    Then, for a binomial distribution,
    mu=np=90, and s=SQRT(npq)=SQRT(100*.9*.1)=3

    The sampling distribution of XBAR is a t distribution (because we used a sample standard deviation s), but because n is large we can approximate it with a normal distribution.

    We want P[ z = (XBAR-mu)/(s/SQRT(n)) < critical value of z ] = .90

    The critical value of z is 1.282 from normal table.

    Is (XBAR-mu)/(s/SQRT(n)) < 1.282 ?

    XBAR = .82 < 1.282 * (s/SQRT(n)) + mu

    .82 < 1.282 * (3/SQRT(100)) + .82 = 1.2046

    So since .82 is less than 1.2046, we conclude that less than 90% of orders are delivered in less than 10 minutes.

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