Assignment 3
ADM 2304 - Applications of Statistical Methods in Business
Due Date and Time: 16 March 2025 before 11:59 pm
Instructions for Submission
1. You may use StatCrunch, MS Excel Analysis ToolPak for calculations that require
technology. However, you must show your manual calculations when asked.
2. For questions that indicate StatCrunch, you must include your StatCrunch output
(e.g., screenshots of tables, plots).
3. If you are performing a hypothesis test, make sure you state the hypotheses, the
level of significance, the decision criteria/region, the test statistic, and/or the
critical value or p-value, if requested), your decision (whether to reject or not to
reject the null hypothesis), and a conclusion in managerial terms that answers
the question posed. These steps must be completed in addition to any software
output.
4. The data for this assignment can be found in the file Assignment 3 Data.xlsx.
5. Your assignment must be typed and uploaded to Brightspace in a single pdf file.
The most recent submission prior to the deadline will be graded. Start each
question on a different page and answer the questions in order.
6. Late submissions will be accepted according to the late submission policy
discussed in class and posted in the course outline on Brightspace.
7. Please do not submit a Declaration of Absence for this assignment. Declaration
of Absences apply for exams.
8. Remember to include your integrity statement. Assignments submitted without
a signed integrity statement will not be graded.
9. You are not permitted to use AI Large Language Models (LLMs) such as ChatGPT,
Claude, Gemini, to complete the assignment.
10. Students who fail to follow any of the instructions above will be penalized with
10% of the marks (for example, if the assignment is marked out of 50, the penalty
will be 5 marks).
11. IMPORTANT ADVICE FOR CALCULATIONS: Please keep 3 decimal places for all
intermediate steps and final result.
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Q1: [26]
A business analyst would like to study the effect of the age of the car owner on the size of the
cash offer for a used car by randomly selecting individuals in three age groups (young, middle,
elderly). A 10-year-old car was selected for the study, and cash offers (in hundred dollars)
received from different dealers are contained in the data set Assignment 3 data.xlsx.
Young Middle Elderly
20
18
32
25
26
35
15
30
35
30
29
40
22
26
28
20
20
29
15
22
10
(a) [2.5] Identify the experimental design, experimental unit/study subject or participant,
factor, level, and response (or output) variable.
The experimental design seemed to be completely randomized design because the
samples of the car owner in each group are random selected.
The experimental unit is car owner.
The factor is the age group of car owners
The levels are young, middle and elderly
The response variable is the cash offer for the used car
(b) [3] Graph the data using side-by-side box plots. Are the assumptions (i.e., normality
and equal variance) of ANOVA met here? Explain.
2
Box plots look symmetric for each group, with no extreme outliers.
3
P>=0.05, the assumption of equal variance is also met
(c) [5.5] Perform an appropriate ANOVA test to determine whether there is a difference in
average cash offers among 3 different age groups using both critical value and p-value
approach. Use α = 0.05 and describe all necessary steps of test of hypothesis (i.e.,
hypotheses, test statistic, critical value/p-value, decision rule, and decision with
justification). Note: You can use directly test statistic and p-value from STATCRUNCH
output. (STATCRUNCH)
H0: μYoung=μMiddle=μElderly
H1: at least one μ is different.
4
F statistic =7.9952
Df(between group_=2
Df (within group)=18
Significance level: α=0.05
Using df(2,18), α=0.05
5
F critical =3.55
Decision rule:（critical value approach）
If Fstatistic >Fcritical, reject H0
IF Fstatistic<= Fcritical, fail to reject H0
Since 7.9952>3.55,we reject H0
Decision rule:（p value approach）
·
From the ANOVA output, p-value = 0.0033.
·
Decision rule：
·
If p-value< a(0.05) reject H0
·
If p-value>=a(0.05),we fail to reject H0
6
·
Since 0.0033 <0.05,we reject H0
·
In conclusion we reject H0, which means at a=0.05 at least one μ is different
(d) [1.5] Calculate the pooled variance using standard deviations of samples obtained from
STATCRUNCH. Compare your pooled variance with the ANOVA output on STATCRUNCH.
Does your calculation agree with the STATCRUNCH output? (Manual Calculation and
STATCRUNCH)
Sp^2=[(n1-1)s1^2+(n2-1)s2^2+(n3-1)s3^2]/(n1+n2+n3-3)
=（7*39.69+5*23.37+6*34.4）/18
=33.36
From the Anova table in statcrunch ,the mean square error is also 33.6 which match my
manual calculate result.
(e) [6] If necessary, determine which groups cash offerings are different from the others
and for which groups you cannot confirm, by using Bonferonni adjusted confidence
intervals. Assume the family level of significance α = 0.05. (Manual Calculation)
aadjusted=0.05/3=0.0167
a/2=0.00835
1-0.00835=0.99165
Comparison 1: Young vs. Middle
SE=√33.36*（1/6+1/8）=3.12
ME=（24.833-19.625-2.9618*3.12，24.833-19.625+2.9618*3.12）
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=(5.208-9.240816,5.208+9.240816)
=(-4.03,14.45)
Since the interval contains zero, we cannot confirm a significant difference between
Young and Middle groups.
Comparison 2: Young vs. Elderly:
SE:√33.36*(1/7+1/8)=2.99
ME;(31.571-19.625-2.96188*2.99，31.571-19.625+2.96188*2.99)
ME（3.09,20.8）
Since the interval does not contain zero, there is a significant difference
between Young and Elderly groups.
Comparison 3: middle vs elderly
SE;√33.36*(1/6+1/7)=3.21
ME：（31.571-24.833-2.9618*3.21+31.571-24.833+2.9618*3.21）
=(-2.77.16.25)
Since the interval contains zero, we cannot confirm a significant difference between
Middle and Elderly groups.
In summary:
Elderly car owners offer significantly higher cash values than young car owners.
We cannot confirm significant differences between Young and middle-aged car owners,
Middle-aged and elderly car owners.
(f) [5] Notwithstanding your answer to part (b), perform Kruskal Wallis nonparametric
test at level of significance α = 0.05 to determine whether there is difference among the
median cash offers among the 3 age groups using critical value approach. Be sure to
sate your hypotheses, test statistic, critical value, decision rule, rank calculation in
EXCEL, //and decision with justification. (Manual Calculation and EXCEL)
H0: μYoung=μMiddle=μElderly
8
H1: at least one μ is different.
α = 0.05
（Critical Value）
α = 0.05， df=2， critical value=5.991
decision value:
if H> critical value ,reject H0
if H<=critical value, fail to reject H0
H:(12/21*22)*(53^2/8+64^2/6+114^2/7)-3*22
=9.074
because H=9.07>5,991, we reject H0
·
conclusion： in the significance level of α = 0.05, there are sufficient evidence to prove
there are at least one μ is different
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(g) [2.5] Perform the Kruskal Wallis nonparametric test using STATCRUNCH, include your
output and give your decision using p-value approach. Note: You need to uncheck
Adjust for Ties in the STATCRUNCH application. (STATCRUNCH)
10
The output from statcrunch match my manual calculation and excel output.
Q2: [24]
The Aeroplan program is the frequent flyer program offered by Canada’s largest airline, Air
Canada. It allows travelers to earn and redeem points for various travel-related rewards
towards flights, hotel, car rental, and merchandise & gifts. The CEO of Air Canada is
concerned that many of its loyalty members have accumulated large quantities of free miles
and the airline may face difficulties if the redeem requests are submitted around the same
time. Hence, a business analyst of Aeroplan program conducted an experiment where four
methods of redeeming frequent flyer miles were offered to a random sample of 60 members
who had accumulated more than 80,000 frequent flyer miles. The frequent flyer members
were equally divided into three age groups. The number of miles redeemed for each frequent
flyer member during the 8-week experimentation period are contained in the data set
Assignment 3 data.xlsx and also given in the following table.
Flights Hotel Car Rental
Merchandise and Gifts
Under 25
25 to 50
Over 50
30000
0
30000
65000
0
55000
28000
12000
10000
35000
25000
0
40000
70000
0
20000
38000
65000
0
18000
55000
45000
0
40000
0
30000
15000
10000
0
45000
40000
0
35000
60000
25000
38000
38000
48000
30000
50000
50000
40000
45000
25000
20000
0
10000
10000
12000
25000
20000
0
0
30000
0
35000
42000
35000
0
0
Note: You can get the exact critical value from STATCRUNCH/EXCEL or approximate critical
value from the table.
(a) [4] Identify the experimental design, number of factors along with levels, number of
treatments, number of replications, and response (or outcome) variable.
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Experimental design: two-way ANOVA design
Numbers of factors and levels:
Factor 1: Age group (3 levels: under 25,25 to 50, over 50)
Factor 2: Redemption Method (4 levels: Flights, Hotel, Car Rental, Merchandise and
Gifts)
Number of Treatments: 3*4=12 treatments
Number of Replications:
60/3=20 members per age group
20 ÷ 4 = 5 replications per treatment
Response variable:
Number of miles redeemed during the 8-week experimentation period
(b) [5] Do the data provide sufficient evidence to indicate an interaction between age
group and type of redemption? Conduct an appropriate test of hypothesis at 5%
level of significance using both critical value and p-value approach.Please provide
hypotheses, test statistic, critical value, p-value, decision rule, and decision with
justification. (STATCRUNCH)
Hypotheses:
H₀: There is no interaction between age group and type of redemption
H₁: There is an interaction between age group and type of redemption
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Decision rule: Reject H₀ if F > F-critical. Fail to reject H₀ if F <= F-critical
Base on the output ; F=1.2771823
critical（α = 0.05，df₁ = 6，df₂ = 48）= 2.30
Since F = 1.2771823 < 2.30, we fail to reject the null hypothesis.
decision rule；Reject H₀ if p-value < α (0.05) ,fail to reject H0 if p value >=a(0.05)
Since p-value = 0.2856 > 0.05, we fail to reject the null hypothesis.
Using both approaches, we reach the same conclusion: there is insufficient evidence at
the 5% significance level to conclude that there is an interaction between age group
and type of redemption.
(c) [3] Test at 5% level of significance, if the average number of redeemed miles differs
among the three age groups. Use the critical value approach. (STATCRUNCH)
Hypotheses:
H₀: μ₁ = μ₂ = μ₃
H₁: At least one mean is different
Significance level: α = 0.05
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F = 1.7639311
Critical F-value (α = 0.05, df₁ = 2, df₂ = 48) = 3.19
Decision Rule:
Reject H₀ if F > F-critical (3.19)
Fail to reject H₀ if F ≤ F-critical (3.19)
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Since F = 1.7639311 < 3.19, we fail to reject the null hypothesis.
Using the critical value approach at a 5% significance level, there is not sufficient
evidence to conclude that the average number of redeemed miles differs among the
three age groups.
(d) [3] Test at 5% level of significance, if the average number of redeemed miles differs
among the four types of redemption options. Use the critical value approach.
(STATCRUNCH)
Hypotheses；
H₀: μ₁ = μ₂ = μ₃ = μ₄
H₁: At least one mean is different
Significance level: α = 0.05
F = 6.1468565（from statcrunch）
Critical F-value (α = 0.05, df₁ = 3, df₂ = 48) = 2.80
Decision Rule:
Reject H₀ if F > F-critical (2.80)
Fail to reject H₀ if F <= F-critical (2.80)
Since F = 6.1468565 > 2.80, we reject the null hypothesis.
Using the critical value approach at a 5% significance level, there is sufficient
evidence to conclude that the average number of redeemed miles differs
among the four types of redemption options.
(e) [9] Calculate 95% Bonferroni margin
of error for the confidence intervals based on all the pairwise differences between the
average number of redeemed miles. Now, test the difference between the mean
redeemed miles for the following two treatments (Manual Calculation):
(i) (Under 25, Hotel) versus (Under 25, Car Rental).
g = number of pairwise comparisons = 12C2 = 66
α/2g = 0.05/(2×66) = 0.000379
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df = 48
MSE = 3.1255e8
n₁ = n₂ = 5
Finding t-critical value for α/(2g) = 0.000379 with df = 48:,t=3.94
Bonferroni
=3.94 × √(3.1255e8*2*（1/5）)
= 3.94 × √(1.2502e8)
=3.94 × 11180.9
=44053
(Under 25, Hotel) vs (Under 25, Car Rental): （41000 -44053，41000+44053）
= (-3,053, 85,053)
(ii) (Under 25, Hotel) vs (Over 50, Car Rental).
=（ 44000 - 44053 55600+44053）= (547, 89153)
Now, draw two interaction plots (Plot 1: Age (x axis) vs Type (y axis); Plot 2: Type (x
axis) vs Age (y axis)) using STATCRUNCH and verify whether you notice the same as
you concluded in the above two comparisons (STATCRUNCH).
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=
17
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Course Code:
Assignment No. / Title:
ADM 2304
ASSIGNMENT 1
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Signature:
Feb.5th
Yilei Wang
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