Time: 9:00am-12:00pm
Date: April 11, 2015
Place: CC (Performance Gym A&B 25-28)
FACULTY OF ADMINISTRATION
UNIVERSITY OF NEW BRUNSWICK
ADM2623: Business Statistics
Final Examination
Instructions:
1. Please justify your answers by providing the methods and/or the procedures for your final
results. You will not get any credit for a single answer without proper explanation.
2. The exam is closed book and notes.
3. Formulas and tables are provided along with the exam papers at the end of this booklet.
4. Scratch papers will be supplied upon request.
5. There are 13 questions in total.
6. The total points are 35.
Name:_________________________________ID_______________________________
Question 1 (5/35 points) Short explanation (be brief and long explanations may reduce your
mark):
(a) (1 point) Explain the Central Limit Theorem
ANS: The sample mean as a random variable under SRR approximately follows a
normal distribution when the sample size is sufficiently large.
(b) (1 point) Explain the Empirical Rule.
ANS:
68% within the one standard deviation away from its mean
89% within the two standard deviations away from its mean
99.7% within the three standard deviations away from its mean
(c) (1 point) Explain the Chebychev’s Theorem.
𝟏
ANS: The interval [𝝁 − 𝒌𝝈, 𝝁 + 𝒌𝝈 ] contains at least 𝟏 − 𝒌𝟐 percentage of the
entire data.
(d) (1 point) What is a random experiment?
ANS: A random experiment is a process satisfying the following properties:
1.
2.
3.
4.
the process can be repeated as many trials as you want
the outcome of any trial is uncertain
well-defined set of possible outcomes
each outcome has a probability associated with it
Page 1 of 12
Time: 9:00am-12:00pm
Date: April 11, 2015
Place: CC (Performance Gym A&B 25-28)
(e) (1 point) What is inferential statistics?
ANS: estimating attributes and drawing inference from a sample about the entire
population
Question 2 (5/35 points) Short calculations:
(a) (1 point) Suppose you invested in a stock with annual return of 7%. How many years
will it take to double your wealth (round your final answer up to the closest
integer)?
ANS: Suppose you invested $1 at the beginning, and after n years your wealth will
be $2 (doubled). 2=1(1+0.07)^n. So n=log2/log(1.07)= 10.24477. So approximately
10 years.
(b) (1 point) For the following grouped data as a sample, calculate the standard deviation
(keep two digits after the decimal):
Class
[-2,4)
[4,10)
[10,16)
[16,22)
[22,28]
Frequency
(f)
3
2
5
10
1
21
Midpoint
(x)
1
7
13
19
25
fx
fx^2
3
14
65
190
25
297
3
98
845
3610
625
5181
ANS: Var=(5181-297^2/21)/(21-1)= 49.02857. So SD=7.00.
(c) (1 point) In a math class of 30 students, 17 are boys and 13 are girls. On a unit test, 4
boys and 5 girls made an A grade. If a student is chosen at random from the class, what is
the probability of choosing a girl or an A student?
ANS:
P(girl or A) = P(girl) + P(A) - P(girl and A)
=
13
30
=
17
30
+
9
30
5
30
(d) (1 point) Find z such that 𝑃(𝑍 > 𝑧)=0.07 by using the normal table at the end of this
exam paper.
Page 2 of 12
Time: 9:00am-12:00pm
Date: April 11, 2015
Place: CC (Performance Gym A&B 25-28)
ANS: 𝑷(𝒁 > 𝒛) = 𝟎. 𝟎𝟕 is equivalent to 𝑷(𝟎 < 𝒁 < 𝒛) = 𝟎. 𝟒𝟑 . Form the normal
table, we have z=1.48
(e) (1 point) X is a normally distributed random variable with E(X) = -10 and SD(X) = 2.
Find a number x such that P(-x-20<X<x ) = 0.84.
ANS: P(-x<X<x ) = 0.84 is equivalent to
−𝒙−𝟏𝟎
𝑷(
𝟐
<
𝑿−𝝁
𝝈
<
𝒙+𝟏𝟎
𝟐
) = 𝟎. 𝟖𝟒.
Let
z=
𝒙+𝟏𝟎
𝝈
.
Then
𝑷(−𝒛 < 𝒁 < 𝒛) = 𝟎. 𝟖𝟒.
Then
𝑷(𝟎 < 𝒁 < 𝒛) = 𝟎. 𝟒𝟐
From the normal table, we have z=1.41.
So
x = -10 + 1.41*2 = -7.18
Question 3 (5/35 points) Thirty AA batteries were tested to determine how long they would
last. The results, to the nearest minute, were recorded as follows:
423, 369, 387, 411, 393, 394, 371, 377, 389, 409,
392, 408, 431, 401, 363, 391, 405, 382, 400, 381,
399, 415, 428, 422, 396, 372, 410, 419, 386, 390
Organize the above data into a frequency distribution table
(a) (1 point). Calculate the number classes:
ANS: 5
(b) (1 point). Calculate the class width:
ANS: 14
(c) (1 point). Calculate the lower limit of the first class
ANS: 362
(d) (1point) Fill up the following frequency distribution table using the information you
obtained from (a), (b) and (c).
class
frequency
[362,376)
[376,390)
4
6
Page 3 of 12
Time: 9:00am-12:00pm
Date: April 11, 2015
Place: CC (Performance Gym A&B 25-28)
[390,404)
[404,418)
[418,432]
9
6
5
(e) (1 point) Draw a cumulative frequency polygon (ogive).
Question 4 (3/35 points) Suppose you are a hospital manager, and you are considering the use
of a new method to diagnose a rare form of bowel syndrome. You know that only 0.1% of the
population suffers from that disease. You also know that if a person has the disease, the test has
99% of chance of turning out positive. If the person doesn’t have the disease, the test has a
98% chance of turning negative. How feasible is this diagnostics method? That is, given that a
test turned out positive, what are the chances of the person really having the disease?
ANS: Let’s say that event DIS is having the disease, and event POS is getting a positive
test. To solve the problem we want to find P(DIS|POS). The description of the problem
tells us that:
P(DIS) = 0.001 (population with the disease)
P(DIS') = 0.999 (population without the disease, so complement of event DIS)
P(POS|DIS) = 0.99 (positive test given patient has disease)
P(POS|DIS') = 0.01 (positive test given patient doesn't have disease)
P(POS'|DIS) = 0.02 (negative test given patient has disease)
P(POS'|DIS') = 0.98 (negative test given patient doesn't have the disease)
Bayes’ theorem is the following:
Page 4 of 12
Time: 9:00am-12:00pm
Date: April 11, 2015
Place: CC (Performance Gym A&B 25-28)
P(DIS|POS) = P(POS|DIS) P(DIS) / P(POS)
Note that P(POS) is equal to
P(POS) = P(DIS)P(POS|DIS) + P(DIS')P(POS|DIS')
P(POS) = 0.001 * 0.99 + 0.999 * 0.01
P(POS) = 0.01098
Now we can apply Bayes’ theorem:
P(DIS|POS) = 0.99 * 0.001 / 0.01098 = 0.09016393
In other words, given the test turned out to be positive, the person only has a chance of
4.7% of actually having the disease. Clearly this is not a feasible method for diagnosing
the rare disease.
Question 5 (2/35 points) Every day a stock trader either makes 50% with probability 0.6 or
loses 50% with probability 0.4. Suppose this trader traded 20 days.
(a) (1 point) How many days can he/she expect to make 50%?
ANS: Let X be the number of days this trader makes 50% over the 20 days. Then
X ~ B(20, 0.6)
So E[X]=20(0.6)=12
(b) (1 point) What is the probability that he/she will make 50% for at least seven days?
ANS:
P(X>=7) =1-P(X<=6)
=1-P(X=0)-P(X=1)-P(X=2) -P(X=3)-P(X=4)-P(X=5)-P(X=6)
=1-0.001-0.005=0.994
Question 6 (5/35 points) Given a joint probability as follows:
X
Y
-3
4
10
-5
-3
-2
0.50
0.02
0.10
0.62
0.05
0.12
0.06
0.23
0.10
0.02
0.03
0.15
Page 5 of 12
0.65
0.16
0.19
Time: 9:00am-12:00pm
Date: April 11, 2015
Place: CC (Performance Gym A&B 25-28)
(a) (2 points) Find the marginal distributions of X and Y respectively
ANS:
X
-5
-3
-2
P(X)
0.62
0.23
0.15
XP(X)
-3.1
-0.69
-0.3
Y
P(Y)
YP(Y)
-3
0.65
-1.95
4
0.16
0.64
10
0.19
1.9
(b) ( 1point) Compute the expectations of X and Y, respectively.
ANS: E[X]=-4.09 and E[Y]=0.59
(c) (1 mark) Are X and Y independent or not, justify your answer.
ANS: No because P(X=-5, Y=-3)=0.5 which is not equal to
P(X=-5)*P(Y=-3)=0.62*0.65=0.403
(d) (1 mark) Compute the expectation E [2+3X-4Y].
ANS: E [2+3X-4Y]=2+3E[X]-4E[Y]=2+3*(-4.09)-4*(0.59)=-12.63
Question 7 (4/35 points) A research firm conducted a survey to determine the mean amount
steady smokers spend on cigarettes during a week. A sample of 49 steady smokers revealed
that sample mean of $20 with a sample standard deviation of $5.
(i) (2 points) Obtain a 97% confidence interval estimate for mean amount steady smokers
spend on cigarettes during a week.
ANS: 𝟏 − 𝜶 = 𝟎. 𝟗𝟕 so 𝜶 = 𝟎. 𝟎𝟑 and 𝒛𝜶/𝟐 = 𝒛𝟎.𝟎𝟑/𝟐 = 𝒛𝟎.𝟎𝟏𝟓 = 𝟐. 𝟏𝟕
So the 97% CI is
[20-2.17*5/7, 20+2.17*5/7]=[18.45, 21.55]
(ii) (2 points) If it is desired to obtain a 97% confidence interval estimate for the mean amount
steady smokers spend on cigarettes during a week with a margin of error as $0.5, what
should be the sample size?
𝟓
𝟐
ANS: 𝒏 ≥ (𝟐. 𝟏𝟕 ∗ 𝟎.𝟓) = 𝟒𝟕𝟎. 𝟖𝟗. So n is at least471.
Page 6 of 12
Time: 9:00am-12:00pm
Date: April 11, 2015
Place: CC (Performance Gym A&B 25-28)
Question 8 (4/35 points) Your investment advisor proposes you a monthly income investment
scheme which promises a variable return each month. You will invest in it only if you are
assured of exactly an average of a $180 monthly income. Your advisor also tells you that for
the past 300 months, the scheme had returns with an average value of $190 and standard
deviation of $75.
(a) (2 points) From this sample data, can we conclude, at 0.02 significance level, that the
you want to invest in this scheme?
ANS:
Step 1. The hypothesis is
𝐻0 : 𝜇 = 180
𝐻1 : 𝜇 ≠ 180
Step 2. 𝛼 = 0.02 so 𝑧𝛼 = 𝑧0.01 = 2.33
2
Step 3. Decision rule: reject the null hypothesis whenever |𝑧| ≥ 𝑧𝛼/2
190−180
Step 4. |𝑧| =
= 2.309401 < 2.33
75/√300
Step 5. So there is no enough evidence to reject the null hypothesis and hence you want to
invest in this scheme.
(b) (2 points) Suppose that you will invest in it only if you are assured of at least an
average of a $180 monthly income. From this sample data, can we conclude, at 0.02
significance level, that the you want to invest in this scheme?
ANS: This becomes an one-tailed test as follows:
Step 1.
𝐻0 : 𝜇 ≥ 180
𝐻1 : 𝜇 < 180
Step 2. 𝛼 = 0.02 so 𝑧𝛼 = 𝑧0.02 = 2.05
Step 3. Decision rule: reject the null hypothesis whenever |𝑧| ≥ 𝑧𝛼
190−180
Step 4. |𝑧| = 75/√300 = 2.309401 > 2.05
Step 5. So there is enough evidence to reject the null hypothesis and hence you do not want
to invest in this scheme.
Question 9 (2/35 points) Suppose you have a population of 4 persons with ages
18, 20, 25, 26
Obtain the sampling distribution of the sample mean age when a sample of two persons is
chosen using SRR.
ANS: The samples and their means are
Samples
Sample mean
18,18
18
18,20
19
20,18
19
18,25
21.5
Page 7 of 12
Time: 9:00am-12:00pm
Date: April 11, 2015
Place: CC (Performance Gym A&B 25-28)
25,18
18,26
26,18
20,20
20, 25
25, 20
20,26
26,20
25,25
25,26
26,25
26,26
21.5
22
22
20
22.5
22.5
23
23
25
25.5
25.5
26
So the sample distribution of the sample mean is given as follows
Sample mean
18
19
20
21.5
22
22.5
23
25
25.5
26
Prob.
1/16
2/16
1/16
2/16
2/16
2/16
2/16
1/16
2/16
1/16
Page 8 of 12
Time: 9:00am-12:00pm
Date: April 11, 2015
Place: CC (Performance Gym A&B 25-28)
Page 9 of 12
Time: 9:00am-12:00pm
Date: April 11, 2015
Place: CC (Performance Gym A&B 25-28)
Page 10 of 12
Time: 9:00am-12:00pm
Date: April 11, 2015
Place: CC (Performance Gym A&B 25-28)
Page 11 of 12
Time: 9:00am-12:00pm
Date: April 11, 2015
Place: CC (Performance Gym A&B 25-28)
Page 12 of 12