MATH 311, SPRING 2006, LAB 3
Part I: the amazing confidence interval
In this part of the lab, we’re going to have Minitab compute many, many 95% confidence intervals and see
what percentage of these confidence intervals capture the true mean of the distribution (hopefully you can
guess).
The population we will be pulling our data from will be uniformly distributed between 9 and 11.
Thus the mean of the population will be μ = 10 (obvious) and the standard deviation will be σ = 0.57735
(not so obvious… yet!).
Step 1: Start Minitab
Step 2: Have Minitab compute 10,000 rows of 36 columns (c1-c36) of uniformly distributed data with a
lower endpoint of 9.0 and an upper endpoint of 11.0. Recall menu series is Calc>Random Data>Uniform…
This gives us 10,000 samples of size n = 36
If 10,000 take too long, try 5,000.
Step 3: Have Minitab compute the mean of each row and store that mean in column c38. Title this column
“Sample Means.”
To do this, select Calc>Row statistics
Click Mean
In the Input variables: box enter C1-C36
In the Store result in: box enter C38
Select OK
The mean of each of the 10,000 rows should appear in C38. We now have 10,000 means for 10,000 samples
of size 36.
Step 4: Now we’re going to have Minitab compute a confidence interval for each of these 10,000 means.
First we compute the lower end of each interval and store it in c40:
Do this as follows: select Calc>Calculator and then type c40 in the “Store result in variable” box.
Then type c38 – 1.96*0.09622504 in the “Expression:” box.
Note: 0.09622504 = 0.57735/ 36   / n
Title column c40 “Lower bound.”
Next compute upper bound for each of our 10,000 confidence intervals and store this in column c41
Do this as follows: select Calc>Calculator and then type c41 in the “Store result in variable” box.
Then type c38 + 1.96*0.09622504 in the “Expression:” box.
Title column c41 “Upper bound.”
Step 5: Now we need to figure out which, if any, of the confidence intervals we just computed captured the
true mean (recall μ = 10) of the population. So, select Calc>Calculator and then type c43 in the “Store result
in variable” box. Then type 10 >= c40 And (it’s a button) 10 <= c41 in the “Expression:” box.
This command will return a 1 in column c43 if μ = 10 is in the confidence interval. It will return a 0 if μ = 10
is not in the confidence interval.
Step 6: Now let’s see how many times μ = 10 was captured by selecting Calc>Column Statistics… then
select Sum and an input variable of c43. Leave the “Store result in:” box empty. This command will return
the number of times μ = 10 was captured.
QUESTION 1: What percentage of the 10,000 confidence intervals you computed captured μ = 10? How does
this compare to the 95% confidence you were trying to establish?
Part II: insurance gumshoes – from the files of “real life.”
Download the dataset Firefraud.mtw from the class webpage (http://www.cwu.edu/~englundt/). Here is the
(true-life!) story behind the data:
A wholesale furniture retailer stores in-stock items at a large warehouse located in Florida. Several years ago,
a fire destroyed the warehouse and all the furniture in it. After determining the fire was an accident, the
retailer sought to recover costs by submitting a claim to its insurance company.
As is typical in a fire insurance policy of this type, the furniture retailer must provide the insurance company
with an estimate of “lost” profit for the destroyed items. Retailers calculate profit margin in percentage form
using the Gross Profit Factor (GPF). By definition, the GPF for a single sold item is the ratio of the profit to
the item’s selling price measured as a percentage. That is,
Item GPF = (Profit / Sales Price) * 100%.
Of interest to both the retailer and the insurance company is the average GPF for all of the items in the
warehouse. Since these furniture pieces were all destroyed, their eventual selling prices and profit values are
obviously unknown. Consequently, the average GPF for all the warehouse items is unknown.
One way to estimate the mean GPF of the destroyed items is to use the mean GPF of similar, recently sold
items. The retailer had sold 3,005 furniture items in the year prior to the fire and kept paper invoices on these
sales. Rather than calculate the mean GPF for all 3,005 items (the data were not computerized), the retailer
sampled a total of 253 of the invoices and computed the mean GPF for these items as 50.8%. The retailer
applied this average GPF to the costs of the furniture items destroyed in the fire to obtain an estimate of the
“lost” profit.
According to experienced claims adjusters at the insurance company, the GPF for sale items of the type
destroyed in the fire rarely exceeds 48%. Consequently, the estimate of 50.8% appeared to be unusually high.
(A 1% increase in GPF for items of this type equates to, approximately, an additional $16,000 in profit.)
Consequently, a dispute arose between the furniture retailer and the insurance company, and a lawsuit was
filed. In one portion of the suit, the insurance company accused the retailer of fraudulently representing its
sampling methodology. Rather than selecting a sample randomly, the retailer was accused of selecting an
unusual number of “high profit” items from the population in order to increase the average GPF of the
sample.
To support its claim of fraud, the insurance company hired a CPA firm to independently assess the retailer’s
true GPF. Through the discovery process, the CPA firm legally obtained the paper invoiced for the entire
population of 3,005 items sold the year before the fire and input the information in to a computer. The selling
price, the profit, profit margin, and month sold for these 3,005 furniture items are available in the file
Firefraud.
In the following, you’ll re-create the CPA’s statistical analysis:
Suppose we want to know how likely it is to obtain a GPF value that exceeds the estimated mean GPF of
50.8%. Since the data for all 3,005 items are available in the file, we can find the actual mean and standard
deviation for the entire 3,005 gross profit margins (you should do so now – Stat>Basic Statistics>Display
Descriptive Statistics). Assume that these data are normally distributed and represent the ENTIRE
population (i.e. you’re finding μ and σ, not just x and s).
QUESTION 2: Find the probability that a single randomly selected item will have a GPF that exceeds 50.8%.
QUESTION 3: Now find the probability that a sample of size 253 would have a mean GPF that exceeds 50.8%
QUESTION 4: If you were the statistician retained by the CPA firm, what would you recommend?