BusinessWeek magazine surveyed MBA alumni 10 years after gra

BusinessWeek magazine surveyed MBA alumni 10 years after graduation. One finding was that alumni spend an average of $115.50 per week eating out socially. You have been asked to conduct a follow-up study by taking a sample of 40 of these MBA alumni. Assume the population standard deviation is $35. First write the probability statement using the random variable, then find the probability.

1) Show the sampling distribution of x, which is the sample mean weekly expenditure for the 40 MBA alumni. In other words, find the mean and standard deviation for x.

2) What is the probability the sample mean will be within $10 of the population mean?

3) Suppose you find a sample mean of $100. What is the probability of finding a sample mean of $100 or less? Would you consider this sample to be an unusually low spending group of alumni? Why or why not?

Solution

1)
Mean of the sampling distribution of means is the population mean = $115.50
Standard deviation of the sampling distribution of means = ? / ?n = 35 / ?40 = 5.534

2)
P( 105.50 < sample mean < 125.50)
Mean ? = 115.5
Standard deviation ? = 35
Standard error ? / ? n = 35 / ? 40 = 5.5339859
standardize xbar to z = (xbar - ?) / (? / ? n )
P( 105.5 < xbar < 125.5) = P[( 105.5 - 115.5) / 5.533986 < z < ( 125.5 - 115.5) / 5.533986]
P( -1.807 < z < 1.807) = 0.9649 - 0.0351 = 0.9298
(from normal probability table)

3)
Mean ? = 115.5
Standard deviation ? = 35
Standard error ? / ? n = 35 / ? 40 = 5.5339859
standardize xbar to z = (xbar - ?) / (? / ? n )
P(xbar < 100) = P( z < (100-115.5) / 5.533986)
= P(z < -2.8009) = 0.0026
(from normal probability table)