Download ECO 252 First Exam - West Chester University`s

ECO 252 First Exam
Fall 2000
Answer key
Dr. T. Andrews
West Chester University
Name________________________
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Under existing standards, most holes in grade A Swiss cheese must be between 11/16 and 13/16
of an inch in diameter. Swiss cheese with smaller holes cannot be sold as grade A which
represents a loss of between 15 and 20 cents per pound. The USDA is considering reducing the
minimum hole size to 3/8 of an inch. (Source: Philadelphia Inquirer 9/17/00)
Sam’s Sliced Swiss Inc. is gearing up for the new changes. Sam turns the hole size dial on the
Swiss machine down a notch. With the new setting, holes should average 3/8 of an inch with a
standard deviation of 1/16.
Tip: Change the fractions to decimals.
μ= 3/8 =.375 inches
σ = 1/16=.0625 inches
Assuming the machine is properly adjusted, the average size of all holes should be .375 inches
with a standard deviation of .0625 inches
Based on the new setting, how large should most of the holes be? Explain your answer and any
assumptions that you make.
Most (at least 75% based on Chebyshev’s theorem) holes should be within two standard
deviations of average. .375 ± 2*.0625 = .375 ± .125. or .25 to .5 inches. This will be true no
matter what the shape of the distribution is. Further, if the holes are normally distributed, 90%
of the holes will fall within 1.64 standard deviations of average: .375 ± 1.64*.0625 =
.375 ± .1025 = .2725 to .4775 inches
σ = .0625
90%
.2725
.375
.4775
(individual hole sizes: x)
Since a hole is simply a cross section of a bubble, the distribution will be continuous and
symmetric around the mean. This does not guarantee normality, but at least there are no obvious
discrepancies. If you read this, and you can demonstrate to me using a logical written argument
that the distribution of hole sizes is probably not normal (I am convinced that it isn’t), I will give
you a 5% bonus on your grade for this exam. In order to qualify for the bonus, you argument
must be submitted to me no later than 4:00 p.m. October 2, 2000.
How many holes should he measure to estimate the average to within 1/32 (.03125) of an inch?
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1.64 2.0625 2
1.64 2.0625 2
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 10.75 call that 11. So the formula says that
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.031252
.031252
if we sample 11 holes, we will be able to estimate the true average within 1/32 of an inch with 90
percent confidence. If holes are normally distributed, that will work. If they are not normal, We
should use at least 30 holes.
If Sam takes a few random slices and measures the diameter of 64 holes, what average should he
expect to find? Explain your answer.
Assuming the machine is properly adjusted, he should get something close to 3/8.
90% of the time the sample average should be within 1.64 standard errors of the population
average: .375 ± 1.64*.0625/√64 =
.375 ± .013 = .3672 to .3828 inches
x 
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.0625
 .0078
64
90%
.362 .375 .388
(average size of 64 holes : x )
Note that the averages are much closer to the mean than the individual observations
If Sam samples 64 holes, and discovers that the sample average is 4/8, can he be reasonably sure
that the machine needs further adjustment? What adjustments might be necessary? Explain your
answer.
Based on the results of the previous question, .5 is an unexpected sample average if the true mean
is .375 inches. Furthermore, If the sample average is .5, then the population average comes from
the confidence interval:

.0625
.5  .013 Assuming the standard deviation is unchanged, we can be 90
n
64
percent sure that the average of all holes is between .487 and .513 inches. The obvious
adjustment is to reduce the hole size. However, it is also possible that the standard deviation has
increased. If this happened, larger sample averages are possible. He should check that
measurement also.
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.5  1.64