One area of concern is that the Bulk Mailing Entry Units (BMEUs)/Business Mailin

Question

One area of concern is that the Bulk Mailing Entry Units (BMEUs)/Business Mailing Units (BMUs) are under collecting the proper amounts of postage for bulk mailings. They want to audit a sample of BMEUs to estimate the additional amount of revenue that possibly could be collected across the entire system. I have placed a file of 2,693 records in Minitab on my webpage. Each record is from a make believe U.S. Postal Service population of BMEUs with total revenue in excess of $100,000. Unrealistically, each record contains the additional postage revenue that the unit could have collected if they had done everything right. It is known from a small pilot study that the standard deviation of the additional revenue that could be collected is $2,157,594.90 Your report should include: • A description of the population of interest using the population data provided • A simple random sample of sufficient size to construct a 90% confidence interval with a length of no more than $300,000 dollars for the average amount of additional revenue that a BMEU could collect. You must specify how the sample was drawn and list the record numbers in your sample. (Please put the record numbers in several columns on a page and NOT just one column per page. This would be considered an attachment.). You must compute and give the confidence interval and any other statistics of your choosing using your simple random sample. • Using your point estimate of the average amount of additional revenue that could be collected, give a point estimate of the total amount of additional revenue that possibly could be collected by the Postal Service. • Test the hypothesis that the mean amount of additional revenue that could be collected by a BMEU is $1,000,000 against the alternative that the mean is greater than $1,000,000. Give the assumptions you made to use the test you selected. • In paragraph form, draw conclusions about BMEUs that are supported by your analyses. The sample data is $5,778.43 $3,624.02 $957.46 $5,029.40 $893,765.83 $2,722.69 $3,585,848.92 $7,276.81 $1,850.02 $3,331.40 $5,015.40 $4,362.81 $4,335.35 $13,671,122.96 $5,841.57 $1,568.90 $0.00 $3,327,321.24 $7,498.81 $2,064.51 $60,638.82 $130,204.25 $7,669.51 $9,078.52 $6,909,682.17 $6,639.97 $1,818.70 $0.00 $4,192.44 $3,422.81 $9,046,688.61

Explanation / Answer

Solution:

Here, we have to find the point estimate for the population mean and 90% confidence interval for the population mean.

The point estimate for the population mean is nothing but the sample mean. From the given data, we have

Estimate for µ = Point estimate Xbar = sample mean = 1216753.301

Now, we have to find the 90% confidence interval for population mean for given data. We are given a value for population standard deviation so we have to use the Z critical value or normal distribution. The formula for confidence interval is given as below:

Confidence interval = Xbar -/+ Z*/sqrt(n)

We are given

Confidence level = 90%

So, critical Z value = 1.6449

(By using Z-table or excel)

Sample mean = Xbar = 1216753.301

Sample size = n = 31

Population standard deviation = = 2157594.9

Confidence interval = 1216753.301 -/+ 1.6449* 2157594.9/sqrt(31)

Confidence interval = 1216753.301 -/+ 1.6449*387515.4837

Confidence interval = 1216753.301 -/+ 637406.2488

Lower limit = 1216753.301 - 637406.2488

Lower limit = 579347.05

Upper limit = 1216753.301 + 637406.2488

Upper limit = 1854159.55

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