Statistical Analysis to Support Decision Making - Assignment 2

This assignment solution demonstrates the application of statistical analysis techniques to support decision-making. It includes analysis of survey data, identifying variable types, and choosing appropriate graphical displays like histograms and bar graphs. The solution involves calculating descriptive statistics, performing hypothesis tests (z-tests and t-tests), and interpreting the results to draw conclusions. The assignment covers topics like stratified sampling, discrete and continuous variables, and the comparison of different datasets. The analysis includes examples of paired t-tests and the interpretation of p-values in relation to hypothesis acceptance or rejection. Furthermore, the solution addresses the relationship between income levels and consumer behavior, as well as the application of statistical methods to compare the effectiveness of different packaging methods.

Assignment 2
STATISTICAL ANALYSIS TO SUPPORT DECISION
MAKING
Question 1 (10 marks)
a)
The sampling survey was based on stratified sampling, where the sample number of cards
had part of the 400 respondents of the group. The stratified sampling enabled every
respondent from the group to be represented in the sample.
b)
Discrete variable was the type of variable that represented the cards, since the number
of cards were exact integer values between 0 and 6, and was not any other numbers
between 0 and 6 which could take decimal points.
c)
The appropriate graphical display to be used for the above discrete variables was a
frequency histogram.
d)
0 1 2 3 4 5 6
0
20
40
60
80
100
120
140
HISTOGRAM
# Cards
Frequency
e)
The above histogram clearly showed that more respondents were likely to carry 3
credit/debit/ATM cards every day, this was depicted with a high respondent frequency of 129,
in comparison zero respondents were likely to carry 4 credit/debit/ATM cards every day, this

was depicted with a low respondent frequency of 0
The calculated standard deviation from excel was equivalent to 48.38535788.

Question 2 (20 marks)
.
a)
Asset Liability Ratio variable would be represented as continuous variable since the
variables takes values that would be between two exact or specific values.
Operation variables would be represented as Ordinal variables since it they have two
categories which could be ranked as most or least levels but from this ranking no values
can be placed, it would remain only Yes or No.
b)
Interval
s
Bins Frequency
0.1 - 0.2 0.2 2
0.3 - 0.4 0.4 3
0.5 - 0.6 0.6 14
0.7 - 0.8 0.8 7
0.9 - 1.0 1 5
1.1 - 1.2 1.2 1
1.3 - 1.4 1.4 8
1.5 - 1.6 1.6 8
1.7 - 1.8 1.8 17
1.9 - 2.0 2 13
2.1 - 2.2 2.2 5
2.3 - 2.4 2.4 2
0.1 -
0.2 0.3 -
0.4 0.5 -
0.6 0.7 -
0.8 0.9 -
1.0 1.1 -
1.2 1.3 -
1.4 1.5 -
1.6 1.7 -
1.8 1.9 -
2.0 2.1 -
2.2 2.3 -
2.4
0
2
4
6
8
10
12
14
16
18
Frequency
Intervals
Frequency
From the excel histogram the highest number of Asset Liability Ratio was between the range
of 1.7 – 1.8, which had a frequency of 17, while the least number of Asset Liability Ratio was

between the range of 1.1 – 1.2, which had a frequency of 1.
c)
Bar graph
Bar graph was appropriate graphical presentation that would be used to compare the Asset
Liability Ratio for the now-closed and still operational small businesses, since it could
easily track the variable change over time
Interval
s
Bins Frequency for
Operational
Frequency for Now
_ closed
0.1 - 0.2 0.2 0 2
0.3 - 0.4 0.4 0 3
0.5 - 0.6 0.6 0 14
0.7 - 0.8 0.8 0 7
0.9 - 1.0 1 0 5
1.1 - 1.2 1.2 0 1
1.3 - 1.4 1.4 8 0
1.5 - 1.6 1.6 8 0
1.7 - 1.8 1.8 17 0
1.9 - 2.0 2 13 0
2.1 - 2.2 2.2 5 0
2.3 - 2.4 2.4 2 0
0.1 - 0.2
0.3 - 0.4
0.5 - 0.6
0.7 - 0.8
0.9 - 1.0
1.1 - 1.2
1.3 - 1.4
1.5 - 1.6
1.7 - 1.8
1.9 - 2.0
2.1 - 2.2
2.3 - 2.4
0 2 4 6 8 10 12 14 16 18
Frequency for Now _
closed
Frequency for Operational
Friquency
Intervals
d)
Now-closed small businesses.
Total 18.66 1.4970875
Count 32 32
mean 0.583125
Variance 0.048293145
Standard deviation 0.219757014

median 0.53
First Quartile 0.43
Third Quartile 0.755
Interquartile
Range
0.325
Operational small businesses
Total 91.65 3.050169811
Counts 53 53
Mean 1.729245283
Variance 0.058657112
Standard deviation 0.242192303
median 1.75
First Quartile 1.55
Third Quartile 1.87
Interquartile
Range
0.32
e)
The frequency for still operational small business in the bar chat were higher as compared to
those of Now _ closed small business, this would mean that asset liability ratio for still
Operation small business that were recorded were higher as compared to still Now _ closed
small business.
The number of counts for Operational small business was 53 while that for Now _ closed
small business was 32, the Operational small business had a high mean of 1.729245, while that
for Now _ closed small business had a mean of 0.583125, the standard deviation for
Operational small business was high with a value of 0.242192 as compared to Now _ closed
small business which had 0.219757, the interquartile range for Now _ closed small business
was high with a value of 0.325 as compared to Operational small business having a value of
0.32.
f)
1. Mean = 1.73
S.d = 0.24
(Z > 1.97) = 1.97−1.73
0.033 = 0.49
From the z – table
P(Z > 0.49) = 1 - 0.6879 = 0.3121
2.
Since 0.3121 > 0.05, we accept the hypothesis at 5% significant level, since the z – score
had a high z- score than 5% significance level.
g)
Mean = 1.73 and 0.58

S.d = 0.24 and 0.22
t =
x 1−x 2
√ σ 2 1
n 1 + σ 2 2
n 2
t =
1.73−0.58
√ 0.242
53 + 0.222
32
= 1.15/0.051
= 22.56
Criterion; reject if t < -tα/2(n1 + n2 – 2)
Or
t > tα/2(n1 + n2 – 2)
t0.025(53 + 32 – 2) = t0.025(83)
80 = 1.990
83 = ?
100 = 1.984
Interpolating
20 = 0.006
3 = ?

= 3∗0.006
20
= 0.0009
Therefore for df of 83 = 1.9891
Reject H0 t < 1.9891 or t > 1.9891
H0:u1 = u2
H1:u1 ≠ u2
Conclusion
Since 22.56 > 1.9891, H0 would be rejected and H1 would be accepted, therefore there were
significant difference in mean Asset Liability Ratio, for small businesses which were still
operating five years later and those that were not, of at least 1.

Question 3 (10 marks)
a)
H0:u1 = u2
H1:u1 ≠ u2
Criterion; reject if t < -tα/2(n1 + n2 – 2)
Or
t > tα/2(n1 + n2 – 2)
t0.025(10 + 10 – 2) = t0.025(18)
Therefore for df of 18 = 2.101
Mean = 15.6 and 11.5
S.d = 6.26 and 5.21
t =
x 1−x 2
√ σ 2 1
n 1 + σ 2 2
n 2

t =
15.6−11.5
√ 6.262
10 + 5.212
10
= 4.1/2.58
= 1.59
Conclusion
Since 1.59 < 2.101, H0 would be accepted and H1 would be rejected, therefore there were
significant difference in mean between the abilities of the two different kinds of packaging to
protect the cheese from mould.
b)
H0:u1 - u2 = 1
H1:u1 −¿ u2 < 1
Criterion; reject if t < -tα/2(n1 + n2 – 2)
Or
t > tα/2(n1 + n2 – 2)
t0.025(10 + 10 – 2) = t0.025(18)

Therefore for df of 18 = 2.101
Mean = 15.6 and 11.5
S.d = 6.26 and 5.21
t =
( x ¿1−x 2)−1
√ σ2 1
n1 + σ2 2
n 2
¿
t =
( 15.6−11.5 ) −1
√ 6.262
10 + 5.212
10
= 3.1/2.58
= 1.20
Conclusion
Since 1.20 < 2.101, H0 would be accepted and H1 would be rejected, therefore there were
significant difference in mean days for mould to appear would surely not be more than one day
Question 4 (10 marks)
Income
< $50K (“low”) ≥ $50K (“high”) Total
LookUp Yes 322 338 660
No 1031 677 1708
Total 1353 1015 2368
a)
Proportion of yes = 660/2368 = 0.2787
In percentage
Proportion of yes = 660/2368 = 0.2787 * 100 = 27.87%
Proportion of No = 1708/2368 = 0.7213
In percentage
Proportion of No = 1708/2368 = 0.7213 * 100 = 72.13%
b)

< $50K (“low”)
≥ $50K (“high”)
Income
0 200 400 600 800 1000 1200
LookUp Yes; 338
LookUp No; 677
Responces by income levels
Look Up
The cluster bar showed that for the two income levels of low and high income than 50K,
majority of the respondents did not check in the past 30 days and looked up for the price of a
product while they were in a store.
The cluster Bar graph was appropriate graphical presentation that would be used to compare
the level of incomes and respondent of looking up the prices of products while they were still
in store, since it could easily track the variable change over time
c)
t-Test: Paired Two Sample for Means
< $50K
(“low”)
≥ $50K
(“high”)
Mean 676.5 507.5
Variance 251340.5 57460.5
Observations 2 2
Pearson Correlation 1
Hypothesized Mean
Difference
0
df 1
t Stat 0.913514
P(T<=t) one-tail 0.264377
t Critical one-tail 6.313752

P(T<=t) two-tail 0.528754
t Critical two-tail 12.7062
The p value = 0.264377 which is greater than 0.05, therefore we accept the null
hypothesis that Income is related to LookUp