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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.

##### 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.

##### Solution Preview

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 ...

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• MSc , Pune University, India
• PhD (IP), Pune University, India
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