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    Probabilty Confidence Level

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    1) Megacorp wants to determine whether the mean of all one-day travel expenses in Moscow exceeds $500. A simple random sample of 35 one-day travel expenses is obtained from Megacorp travel expense files. The average one-day Moscow expense was found to be $538 with a standard deviation of $41.

    a) Calculate the probability of observing a sample mean of at least $538 if the actual population mean is $500.

    b) Based on you answer to a, do you think the mean of all one-day travel expenses for Megacorp employees in Moscow exceeds $500? Explain.

    2) Suppose that, for the sample size of n=100 measurements, we find a sample mean of 50 and a standard deviation of 2. Calculate confidence intervals for the population mean with an 82% confidence level.

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    1.
    <br>a.)
    <br>To calculate:
    <br>p[ >= $538] = ?
    <br>z = 538 = 500 + t(a/2) * SD/sqrt(n)
    <br>where 1-a = p[ 500-t(a/2)*SD/sqrt(n) < <x> < 500+t(a/2)*SD/sqrt(n)]
    <br>n =35 (close to 30 : small sample)
    <br>SD = $41
    <br>therefore
    <br>t(a/2) = (538-500)*sqrt(35)/41 = 38*sqrt(35)/41 = 5.483
    <br>corresponding to t(a/2) = 5.483 => a/2 ~ 0.00001 (even less: from ...

    Solution Summary

    1) Megacorp wants to determine whether the mean of all one-day travel expenses in Moscow exceeds $500. A simple random sample of 35 one-day travel expenses is obtained from Megacorp travel expense files. The average one-day Moscow expense was found to be $538 with a standard deviation of $41.

    a) Calculate the probability of observing a sample mean of at least $538 if the actual population mean is $500.

    b) Based on you answer to a, do you think the mean of all one-day travel expenses for Megacorp employees in Moscow exceeds $500? Explain.

    2) Suppose that, for the sample size of n=100 measurements, we find a sample mean of 50 and a standard deviation of 2. Calculate confidence intervals for the population mean with an 82% confidence level.

    $2.19

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