# 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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##### 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.

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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 ...

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