Descriptive Statistics

This article covers various tasks related to descriptive statistics, including expenditure on stand mixers, correlation matrix, critical path, and break-even analysis.

Descriptive statistics

TABLE OF CONTENTS
PART 1............................................................................................................................................4
Task 1...........................................................................................................................................4
a)..................................................................................................................................................4
b)..................................................................................................................................................4
Task 2...........................................................................................................................................4
a)..................................................................................................................................................4
b)..................................................................................................................................................5
c)..................................................................................................................................................6
Task 3...........................................................................................................................................6
Task 4...........................................................................................................................................6
a) Network diagram.....................................................................................................................6
b) critical path..............................................................................................................................6
c) Difference between critical and non-critical activities............................................................7
Task 5...........................................................................................................................................7
a) correlation matrix....................................................................................................................7
b)..................................................................................................................................................7
c)..................................................................................................................................................8
d)..................................................................................................................................................8
e)..................................................................................................................................................8
Task 6...........................................................................................................................................8
a)..................................................................................................................................................8
b)..................................................................................................................................................8
Task 7...........................................................................................................................................9
PART 2............................................................................................................................................9

a)..................................................................................................................................................9
b)..................................................................................................................................................9
c)................................................................................................................................................10
d)................................................................................................................................................10
e)................................................................................................................................................10
f).................................................................................................................................................10
REFERENCES..............................................................................................................................11

PART 1
Task 1
a)
Expenditure (£) on Stand
Mixers
Mean
224.0
0
Standard Error 6.58
Median
224.0
0
Mode
224.0
0
Standard Deviation 75.00
Sample Variance 5625
Kurtosis -0.49
Skewness -0.32
Range
301.0
0
Minimum 68.00
Maximum
369.0
0
Sum
2912
0
Count 130
Coefficient of
Variation 33.5%
b)
The average number from the descriptive statistics entails that there are 224 customers
who are willingly pay for a stand mixer, retailer of kitchen appliance. Whereas, standard
deviation reflect that there is a high standard deviation which indicate that the data is more
spread.
Task 2
a)
Table 1
Expenditure
(£) Frequency
Frequency
(%)
Under 100 14 11%
100 and under 35 27%

200
200 and under
300 65 50%
300 and over 16 12%
Total: 130 100%
Table 2
Expenditure
(£) Frequency
Cumulative
Frequency
Cumulat
ive
Frequen
cy (%)
Under 100 14 14 10%
Under 200 35 49 34%
Under 300 65 100 68%
Under 400 16 81 55%
Total: 130 146 100%
Table 3
Expenditure
(£)
Frequency
(f)
midpoint
(x) fx (x-mean) (x-mean)2
f(x-
mean)2
Under 100 14 50 700
-
163.84615
38
26845.56
213
375837.8
698
100 and under
200 35 150 5250
-
63.846153
85
4076.331
361
142671.5
976
200 and under
300 65 250 16250
36.153846
15
1307.100
592
84961.53
846
300 and over 16 350 5600
136.15384
62
18537.86
982
296605.9
172
Total: 130 27800
900076.9
231
Mean
213.84615
38
Variance
6923.6686
39
Standard
Deviation 83.2

b)
There is a difference between the mean value and standard deviation of grouped and
ungrouped data and this is because of a class interval (K Zenner and Teimouri, 2021). Also,
ungrouped data will be more accurate because there is no midpoint theory applied within the
dataset.
c)
25% is the first quartile = ¼ (n+1)
= ¼ (4+1)
= 5/4 = 1.25 th term
= 14
Thus, it has been reflected that the estimated minimum amount that consumer needs to
spend on the stand mixer is £14.
Task 3
Time series data focuses on same variable over the period of time whereas cross sectional
data focus upon the several variable during a same time (Liu, Wang and Xu, 2021). Therefore, in
the context of retailer’s information needs, time series data is only useful in order to gain
understandings pertaining to demographic profile as well as lifestyle characteristic of the
customers. Thus, focusing upon single company at single time will help the business to improve
the customer need and meet their demand as well.
Task 4
a) Network diagram

b) critical path
1 + 3 + 5 + 9 +10
= 6 weeks + 4 weeks + 6 weeks + 5 weeks + 2 weeks
= 23 weeks
c) Difference between critical and non-critical activities
The difference between the critical and non-critical activities reflected that there is a high
difference between completing the overall work which is as mentioned below:
1 + 3 +4 +7 + 6 + 5 + 9 +10
= 6 weeks + 4 weeks + 2 weeks + 3 weeks + 5 weeks + 6 weeks + 5 weeks + 2 weeks
= 33 weeks
Task 5
a) correlation matrix
Sales Revenue
(£'000)
Total Costs
(£'000)
Average Order
Value (£)
Gross
Profit
(£'000)
Sales Revenue
(£'000) 1
Total Costs (£'000) 0.54003673 1
Average Order
Value (£) 0.974847027 0.464074002 1
Gross Profit (£'000) 0.423092657 0.839835895 0.338727951 1
b)
The best predictor identified from the correlation matrix is sales revenue because the
variable is highly associated with each other as compared to other ones.

c)
0 500 1000 1500 2000 2500 3000
0
200
400
600
800
1000
1200
1400
1600
f(x) = 0.246629484443074 x + 474.021153537997
R² = 0.179007396268785
Total Costs (£'000) Average Order Value (£)
Gross Profit (£'000) Linear (Gross Profit (£'000))
d)
The coefficient of determination from the graph reflected that there is 17% change in the
dependent variable when independent one changes. However, the coefficient of correlation
reflected that there is low association between the variables.
e)
The regression equation entails that y = 0.2466x + 474.02 which means 474.02 is an
intercept and the slop indicate the value 0.2466 that is slope upward.
Task 6
a)
For venture A For venture B
Coefficient of variation =
Standard deviation / net
expected value *100
30.08 / 47 *100 = 64 13.32 / 41.2 *100 = 32.33
b)
The retailer must choose venture A because it has a string association with the variables
and also provide effective results as compared to Project B. Even though the retailer is riskier,
but still project B must be rejected over A.

Task 7
Yes, I perfectly did the assignment, because with the help of tutor, I am able to generate
the best outcomes.
The most challenging aspect is findings formulas for the Task 6. It is so because I am not
able to get the answer and due to not-known terms, I face difficulties.
If the same assignment will use, I ensure to increase my excel and analytical skills so that
I can do the same in better manner without consuming high time.
PART 2
a)
High low method is used in the scenario in order to calculate the variable and fixed cost
of a product entity with mixed cost.
Highest Activity cost = 4,00,000
Highest activity Units = 36,000
Lowest Activity cost = 183,200
Lowest activity units = 13,320
Variable Cost per unit = (HAC – LAC)/ (HAU - LAU)
= (400000 – 183200) / (36000 - 13320)
= 216800 / 22680
= 9.56
b)
Fixed cost = Highest Activity Cost – (Variable cost per units * Highest Activity Units)
= 400000 – (9.56 * 36000)
= 400000 – 344160
= 55840

c)
Break event point = Fixed Cost / Selling price – variable cost
= 55840 / (15 – 9.56)
= 55840 / (5.44)
= 10264.70
Thus, Andreea Ltd has to produce and sell 10264.70
d)
Margin of safety = (Actual sales – Break Even Point) / Selling Price per Unit
= (36000 * 15 – 10264.70) / 15
= (54000 – 10264.70) / 15
= 529735.3 / 15
= 35315.68
e)
Through the above, it has been analyzed that margin of safety is 35315.68 which means it
is the forecasted sales for the future (Kadhim, 2021).
f)
Limitation of breakeven analysis
 It is difficult to separate the Fixed and variable cost in this situation
 Sales as well as variable cost does not increases in same proportion with the value of
production (Abdurofi and et.al., 2021).

REFERENCES
Books and Journals
Abdurofi, I. and et.al., 2021. Application of Cost-Benefit and Break-Even Analysis for the
Development of Stingless Bees Farming in Malaysia. International Journal of Business
and Society. 22(2). pp.846-861.
K Zenner, E. and Teimouri, M., 2021. Modeling in Forestry Using Mixture Models Fitted to
Grouped and Ungrouped Data. Forests. 12(9). p.1196.
Kadhim, Z. R., 2021. Margin of Safety of Hiring Decision of Agricultural Machinery Services
by Rice Farmers in Alnajaf Al-Ashraf Province. IRAQI JOURNAL OF
AGRICULTURAL SCIENCES. 52(3). pp.756-762.
Liu, L., Wang, Y. and Xu, Y., 2021. A practical guide to counterfactual estimators for causal
inference with time-series cross-sectional data. arXiv preprint arXiv:2107.00856.