Université d’Ottawa/University of Ottawa
Faculté des sciences sociales /Faculty of Social Sciences
Science Économique/Economics
ECO 3150B
Final exam
Lecture 2:
In a time study, conducted at a manufacturing plant, the length of time to complete a specified
operation is measured for each on n = 40 workers. The mean and standard deviation are found
to be 15.2 and 1.40, respectively.
1. Describe the sample data using the Empirical Rule.
ANSWER:
To describe the data, calculate these intervals:
( x  s ) = 15.2  1.40, or 13.8 to 16.6
( x  2s ) = 15.2  2.80, or 12.4 to 18.0
( x  3s ) = 15.2  4.20, or 11.0 to 19.4
If the distribution of measurements is mound-shaped, you can apply the Empirical Rule
and expect approximately 68% of the measurements to fall into the interval from 13.8
to 16.6, approximately 95% to fall into the interval from 12.4 to 18.0, and all or almost
all to fall into the interval from 11.0 to 19.4.
Lecture 3:
A review of the personnel records of a small corporation has revealed the following information
about the number of sick days taken per year and the corresponding probabilities.
Number of Sick Days
Probability
0
1
2
3
4
5
0.05
0.22
0.31
0.27
0.13
0.02
Let A be the event that an employee takes more than 2 sick days. Let B be the event that an
employee takes less than five sick days.
2. Are events A and B collectively exhaustive and/or mutually exclusive? Why?
ANSWER:
Yes, since A B = the sample space S
No, since P( A B )  0 = P(3) + P(4) = 0.27 + 0.13 = 0.40
Lecture 4:
3. An auditor reviewing the invoices of a small company finds that there are errors in 1.5% of
them. If the auditor looks at 500 invoices, what is the probability that he finds more than 3
invoices with errors? Use the Poisson approximation to the binomial distribution.
ANSWER:
 = nP = 7.5, then P(X>3) = 1 – P(X  3) = 1 – (0.0006+0.0041+0.0156+0.0389) = 0.94
Lecture 5:
4. Suppose x is normally distributed with a mean of 75 and a standard deviation of 4.
a. Find the 90th percentile.
b. Find the 5th percentile.
ANSWER:
a. Find x0 such that P(X  x0 ) = 0.90 = P(Z  z0 ).
Since the area to the left of z0 is greater than 0.5, z0 must be larger than zero such that
P(0  Z  z0 ) = 0.40, so z0 = 1.28. Thus, x0 = 75 + (1.28)(4) = 80.12
b. Find
x0
such that P(X 
x0
) = 0.05 = P(Z 
z0
).
Since the area to the left of z0 is smaller than 0.5, z0 must be smaller than zero such that
P( z0  Z  0) = 0.45, so z0 = -1.645. Thus, x0 = 75 + (-1.645)(4) = 68.42
5. A normal random variable x has an unknown mean  and standard deviation   2.5 If the
probability that x exceeds 7.5 is 0.8289, find  .
ANSWER:
It is given that x is normally distributed with  = 2.5 but with unknown mean  , and
that P (x > 7.5) = 0.8289. In terms of the standard normal random variable z, we can
write
P( X  7.5)  P[ Z  (7.5   ) / 2.5]  0.8289
Since the area to the right of (7.5   ) / 2.5 is greater than 0.5, then (7.5   ) / 2.5 must
be negative, and that P[ (7.5   ) / 2.5 < z < 0] = 0.3289. Hence,
(7.5   ) / 2.5 = -.095. This implies  = 9.875.
Lecture 6:
A random sample of 10 stock market mutual funds was taken. Suppose that rates of returns on
the population of stock market mutual funds follow a normal distribution.
6. The probability is 0.10 that sample variance is bigger than what percentage of the
population variance?
ANSWER:
P( 92  14.684)  0.10  14.684 = 9 (Difference)  Difference = 1.6316 or 163.16%
Lecture 7:
From a random sample of 500 registered voters in Los Angeles, 400 indicated that they would
vote in favor of a proposed policy in an upcoming election.
7. Calculate the width of the 90% confidence interval estimate for the population proportion
in favor of this policy.
ANSWER:
In this case, z / 2  z0.05  1.645,
Margin or error ME = z / 2 pˆ (1  pˆ ) / n  1.645 (0.80)(0.20) / 500 = 0.0294
Width of the 90% confidence interval estimate = 2  ME = (2)(0.0294) = 0.0588
Lecture 8:
8. An economist is interested in determining the proportion of employees who lost their job in
2008 due to the economic crises the country is dealing with for the last two years. How large
of a sample should the economist take in order to estimate the proportion of all employees
who have lost their job to within 2.5%? Assume that the economist wants to have 97.8%
reliability in his results.
ANSWER:
2
 0.5  2.29  
 0.5 z / 2 
2
n

   45.8  2097.64  2098

 ME 
 0.025 
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Lecture 9: THE NEXT TWO QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
An insurance company employs agents on a commission basis. It claims that, in their first year,
agents will earn a mean commission of at least $42,000 and that the population standard
deviation is no more than $6,800. A random sample of nine agents found, for commissions in
9
the first year,

xi  351,000 and
i 1
9
 x  x 
i
2
 338,000,000 The population distribution can be
i 1
assumed to be normal.
9. What are the appropriate null and alternative hypotheses?
ANSWER:
H 0 :   42,000 and H1 :   42,000
10. Test at the 5% level the null hypothesis in the first question.
ANSWER: first we calculate the sample mean and sample standard deviation.
9
x   xi / 9  351,000 / 9  39,000 and s 
i 1
9
 x  x 
i 1
i
2
/ 8  338,000,000 / 8  6,500
Reject H 0 if t  t8,0.05  1.860
t
x  0
s/ n

39,000  42,000
 1.385 , therefore, fail to reject H 0 . We conclude that
6,500 / 9
insurance company agents, in their first year, earn a mean commission of at least
$42,000.
BONUS QUESTION:
From lecture 3
A student has access to professor evaluations. Overall, he has enjoyed 70% of all classes he has
taken. He finds that of the courses he has enjoyed, 13% were taught by professors with poor
evaluations. 84% of the courses he has taken were taught by professors with good evaluations.
Suppose a student signed up for three courses next semester, all of which are taught by
professors with good evaluation. What is the probability he enjoys all three?
ANSWER: find first what is the probability that the class was taught by a professor with
poor evaluation and that the student enjoyed the class?
Event A: Student enjoyed class
Event B: Class taught by professors with good evaluation
Since P(A) = 0.70 and P( B | A) = 0.13
Then, P( B A) = P( B |A)  P(A) = (0.13)(0.70) = 0.091
Then: What is the probability that the class was taught by a professor with good
evaluation and that the student enjoyed the class?
Since P(A) = P(A
B )  0.70 = P(A B)+0.091, then, P(A B) = 0.609.
Then: What is the probability that the student enjoyed the class given that it was taught
by a professor with good evaluation?
P(B) = 0.84
P(A | B) = P(A B) / P(B) = 0.609/0.84 = 0.725
Finally: Suppose a student signed up for three courses next semester, all of which are
taught by professors with good evaluation. What is the probability he enjoys all three?
 P  A | B     0.725   0.381
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