1. Eagle Credit Union (ECU) has experienced a 10% default rate with its commercial
loan customers (that is, 90% of commercial loan customers pay back their loans). ECU
has developed a statistical test to assist in predicting which commercial loan customers
will default. The test assigns either a rating of “Approve” or “Reject” to each loan
applicant. When applied to recent commercial loan customers who paid their loans, the
test gave an “Approve” rating in 80% of the cases examined. When applied to recent
commercial loan customers who defaulted, it gave a “Reject” rating in 70% of the cases
examined.
a. Use this data to construct a joint probability table.
The probabilities are:
Default and approved = (0.10)(0.30) = 0.03
Default and rejected = (0.1)(0.7) = 0.07
Paid and approved = (0.90)(0.80) = 0.72
Paid and rejected = (0.90)(0.20) = 0.18
The joint probability table is:
Default
Paid
Total
Approved
0.03
0.72
0.75
Rejected
0.07
0.18
0.25
Total
0.10
0.90
1.00
b. What is the conditional probability of a “Reject” rating given that the customer
defaulted?
P ( Reject | Default ) =
0.07
= 0.70
0.10
c. What is the conditional probability of an “Approve” rating given that the customer
defaulted?
P ( Approved | Default ) =
0.03
= 0.30
0.10
d. Suppose a new customer receives a “Reject” rating. If they are given the loan anyway,
what is the probability that they will default?
P ( Default | Rejected ) =
P ( Default and Rejected ) 0.07
=
= 0.28
P ( Rejected )
0.25
3. A soft drink machine can be regulated (discharge level m) so that it dispenses an
average of m ounces per cup. If the ounces of fill are normally distributed with mean m
and standard deviation equal to 0.3 ounces. Find the setting of the discharge level m so
that eight ounce cups will overflow only one percent of the time.
Overflowing only one percent of the time corresponds to a right-tail area of 0.01.
The z-value for this tail area is 2.326.
The corresponding fill volume is:
x = mean + z * standard deviation
8 = m + 2.326(0.3)
8 = m + 0.6978
m = 8 – 0.6978
m = 7.30
The discharge level should be set to m = 7.3 ounces.
6. The United States Golf Association requires that the weight of a golf ball must not
exceed 1.62 oz. The association periodically checks golf balls sold in the United States by
sampling specific brands stocked by pro shops. Suppose that a manufacturer claims that
no more than 1 percent of its brand of golf balls exceeds 1.62 oz. in weight. Suppose that
24 of this manufacturer’s golf balls are randomly selected, and let X denote the number
of the 24 randomly selected golf balls that exceed 1.62 oz.
a. Find the probability that none of the randomly selected golf balls exceed 1.62 oz.
If the manufacturer’s claim is true, then the probability of any single golf ball
exceeding the 1.62 oz weight is p = 0.01.
In 24 trials, the probability that 0 golf balls will exceed that weight is:
P ( 0 ) = C ( 24,0 ) * ( 0.01) (1- 0.01) = 0.7857
0
24
b. Find the probability that at least one of the randomly selected golf balls exceeds 1.62
oz.
The probability that at least one of the selected golf balls exceeds 1.62 oz. is the
complement of the probability that none of the balls exceeds that weight.
P ( x ³ 1) = 1- P ( x = 0 ) = 1- 0.7857 = 0.2143
c. Suppose that two of the randomly selected golf balls are found to exceed 1.62 oz. Do
you believe the claim that no more than 1 percent of this brand of golf balls exceed 1.62
oz. in weight?
If the manufacturer’s claim is true, the probability that two balls will exceed the
weight limit is:
P ( x = 2 ) = C ( 24,2 ) * ( 0.01) * (1- 0.01) = 0.0221
2
22
Since there is only a 2.21% chance of finding two balls that exceed the weight
limit, out of 24 randomly selected balls, it is likely that the manufacturer’s claim
is not true.
7. Owing to several major ocean oil spills by tank vessels, Congress passed the 1990 Oil
Pollution Act, which requires tankers to be designed with thicker hulls. Further
improvements in the structural design of a tank vessel have been implemented since then,
each with the objective of reducing the likelihood of an oil spill and decreasing the
amount of outflow in the event of hull puncture. To aid in this development, J.C. Daidola
reported on the spillage amount and cause of puncture for 50 recent major oil spills from
tankers and carriers. The file OilSpill.sgd contains the data for the 50 spills reported.
a. Is any one cause more likely to occur than any other? Justify the answer using
hypothesis tests.
There are four identified causes of oil spills in the data set: collision, fire, hull
failure, and grounding. There are two cases that are identified as having an
“unknown” cause. These two cases will be ignored, and the sample will be
considered to consist of the 48 spills that have one of the four identified causes.
If there is no difference in the likelihood of each of the four causes, then each
would be expected to occur 25% of the time. The expected number of spills for
each cause would then be 48 * 0.25 = 12.
The following hypothesis will test whether there is a significant difference
between the expected distribution of causes and the observed distribution.
Hypotheses:
Null:
H 0 :Causes of oil spills are equally likely.
Alternative:
H 0 :Causes of oil spills are not equally likely.
Critical value:
This will be a X2 test, with 4 – 1 = 3 degrees of freedom.
Assuming a = 0.05, the critical value is 7.815.
The null hypothesis will be rejected if the test statistic is greater than
7.815.
Test statistic:
Observed
10
12
12
14
Expected
12
12
12
12
(O – E)2/E
0.3333
0.0000
0.0000
0.3333
0.6667
Decision:
Since the test statistic is less than the critical value, the decision is to fail
to reject the null hypothesis.
Summary:
There is insufficient evidence at the 0.05 level of significance to support a
claim that one cause is more likely than another.
b. Construct a 90 percent confidence interval for the difference between the mean spillage
amount of accidents caused by collision and the mean spillage amount of accidents
caused by fire/explosion. Interpret the result.
Data:
Collision:
Mean:
x1 = 76.6
StdDev:
s1 = 70.3629
Sample size:
n1 = 10
Fire/Explosion:
Mean:
x2 = 70.9167
StdDev:
s2 = 60.6757
Sample size:
n2 = 12
Since the standard deviations are approximately equal, the number of degrees of
freedom is n1 + n2 – 2 = 10 + 12 – 2 = 20.
The critical t-value for a 90% confidence interval with 20 degrees of freedom is
tcrit = ±1.7247.
The limits of the confidence interval are then calculated from:
( x1 - x2 ) ± ( ta /2 )
( n1 - 1) s12 + ( n2 - 1) s2 2
n1 + n2 - 2
( 76.6 - 70.9167 ) + (1.7247 )
1 1
+
n1 n2
(10 - 1) ( 70.3629 )2 + (12 - 1) ( 60.6757 )2
10 + 12 - 2
1 1
+
10 12
5.6833 ± (1.7247 ) ( 65.2133 ) ( 0.4282 )
5.6833 ± 48.1582
( -42.4749, 53.8145 )
The 90% confidence interval is (-42.47, 52.81).
This confidence interval is interpreted to mean that we can be 90% confident that
the true difference between the two means lies within the limits of this interval.
c. Can we say that the mean spillage amount of accidents caused by grounding is the
same as the corresponding mean of accidents caused by hull failure?
Data:
Grounding:
Mean:
x1 = 47.7857
StdDev:
s1 = 28.4664
Sample size:
n1 = 14
Mean:
x2 = 54.4167
StdDev:
s2 = 56.3874
Sample size:
n2 = 12
Hull Failure:
Hypotheses:
Null:
H 0 : µ1 = µ2
Alternative:
H1 : µ1 ¹ µ2
Critical value:
Treating the variances as equal, the number of degrees of freedom is 14 +
12 – 2 = 24.
Assuming that a = 0.05, the critical value is tcrit = ±2.064.
The null hypothesis will be rejected if the test statistic is less than -2.064,
or greater than 2.064.
Test statistic:
ttest =
x1 - x2
( n1 - 1) s12 + ( n2 - 1) s2 2
n1 + n2 - 2
ttest =
1 1
+
n1 n2
47.7857 - 54.4167
(14 - 1) ( 28.4664 )2 + (12 - 1) ( 56.3874 )2
14 + 12 - 2
1 1
+
14 12
ttest = -0.3871
Decision:
Since the test statistic is between the two critical values, the decision is to
fail to reject the null hypothesis.
Summary:
There is insufficient evidence to reject a claim that the mean spillage
amount of accidents caused by grounding is the same as the corresponding
mean of accidents caused by hull failure
d. State any assumptions required for the inferences derived from the analyses to be valid.
Are these assumptions reasonably satisfied?
The X2 “goodness of fit” test performed in part a requires that the data be
obtained from a random sample, and that the expected frequency of each category
be at least 5. As there is no information presented in the problem statement
regarding how the sample was obtained it is not possible to make a determine
about the validity of the assumption. The requirement concerning the expected
frequencies is met however, as the expected frequency for each category was 12.
The t-tests in part c and d require that the samples be independent, and that they
be drawn from a normally distributed population. The inferences assume that
these requirements are met. Again, there is insufficient information about the
sample to determine whether the requirement of independence is valid. It is
possible that some of the reported spills involved the same ships, thereby
hampering the independence of the samples.
e. Is the variation in spillage amounts for accidents caused by collision the same as the
variation in spillage amounts for accidents caused by grounding?
Data:
Collision:
Variance:
s12 = 4590.9333
Sample size:
n1 = 10
Variance:
s2 2 = 810.3352
Sample size:
n2 = 14
Grounding:
Hypotheses:
Null:
H 0 : s 12 = s 2 2
Alternative:
H1 : s 12 ¹ s 2 2
Critical values:
A test of variances uses the F statistic.
For this test, the number of degrees of freedom in the numerator is
10 – 1 = 9. The number of degrees of freedom in the denominator is
14 – 1 = 13.
Assuming a = 0.05, the critical value is Fcrit = 3.3120
Test statistic:
s12 4590.9333
Ftest = 2 =
= 264.76
s2
810.3352
Decision:
Since the test value is greater than the critical value, the decision is the
reject the null hypothesis.
Summary:
There is sufficient evidence at the 0.05 level of significance to reject a
claim that the variation in spillage amounts for accidents caused by
collision is the same as the variation in spillage amounts for accidents
caused by grounding.