Data Analysis and Linear Programming

This document covers the topic of Data Analysis and Linear Programming. It includes measures of location and spread, correlation coefficient, regression output, coefficient of determination, and maximization problem expression. The subject is not mentioned. The document type is an essay, and the type of assignment is not mentioned. The course code, course name, and college/university are not mentioned.

Data Analysis and Linear Programming 1
DATA ANALYSIS AND LINEAR PROGRAMMING
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Data Analysis and Linear Programming 2
Data Analysis and Linear Programming
Task 1
Part (a)
Measures of Location (i.e. measures of central tendency) will be calculated in excel and presented here as
follows:
Advertising('000s
)
Sales ('000s)
Mean 10.59 48.891375
Mode N/A N/A
Median 9.915 47.55
Measures of Spread (i.e. measures of dispersion
Advertising('000s) Sales ('000s)
1st Quartile 9.1325 44.3515
3rd Quartile 11.645 52.80675
Interquartile Range 2.5125 8.45525
Range 5.26 24.966
Variance 3.8728 64.87032855
Standard Deviation 1.967943089 8.054211852
Part (b)

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The Product-moment correlation coefficient (i.e. Pearson correlation coefficient) is indicated in the tables
below. According to the result, there is a very strong positive relationship (0.947159) between advertising
expenditure and sales revenue. Therefore, there follow the same pattern if one is increasing the other is also
increasing; likewise, if one is decreasing the other is also decreasing.
Advertising('000s) Sales ('000s)
Advertising('000s) 1
Sales ('000s) 0.947159454 1
Part (c)
From the regression output below, we can formulate the relation between advertising expenditure (x) and sales
revenue (y) as follows:
y=7.8398+3.8764(x )
Where a=7.8398 and b=3.8764
We can interpret “a” to represent the intercept; therefore, when advertising expenditure is £0, the sales revenue
will be £7,839.80. On the other hand, “ b” represents the slope and as a result an increment in advertising
expenditure by 1 unit (£1,000) will increase sales revenue by £3,876.40 (Montgomery 2013).

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7 8 9 10 11 12 13 14 15
35
40
45
50
55
60
65
70
f(x) = 3.87644487561602 x + 7.83982376722639
ScatterGraph
Sales ('000s)
Linear ( Sales ('000s))
Advertising Expenditure
Sales Revenue
SUMMARY OUTPUT
Regression Statistics
Multiple R 0.947159
R Square 0.897111
Adjusted R Square 0.879963
Standard Error 2.790492
Observations 8
ANOVA
df SS MS F Significance F
Regression 1 407.3712 407.3712 52.31529 0.000354
Residual 6 46.72109 7.786848
Total 7 454.0923
CoefficientsStandard Error t Stat P-value Lower 95% Upper 95%Lower 95.0%Upper 95.0%
Intercept 7.839824 5.760756 1.360902 0.22244 -6.25624 21.93589 -6.25624 21.93589
Advertising('000s) 3.876445 0.535944 7.232931 0.000354 2.565038 5.187852 2.565038 5.187852
Part (d)
Using the formula y=7.8398+ 3.8764( x ) we can be able to estimate the sales revenue as follows
Advertising Expenditure Sales Revenue
(i) £ 11,000.00 £ 50,480.72
-£ 50,520.00 -£ 187,998.17
(ii) £ 13,500.00 £ 60,171.83

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-£ 60,220.00 -£ 225,599.69
(iii) £ 15,000.00 £ 65,986.50
-£ 66,040.00 -£ 248,160.60
Part (e)
The coefficient of determination also referred to as R-squared is 0.89711. The coefficient indicates that 89.71%
of the change in sales revenue (dependent variable) can be explained by advertising expenditure (independent
variable).
Part (f)
Yes, the relationship is sufficiently strong because the coefficient of determination is considerably close to 1. A
value of 1 indicates that 100% of the change in the dependent variable is explained by the independent variable.
(NB: the value of significance F (0.000354) in the regression output above is considerably smaller than alpha of
0.05. indicating that the model is statistically significant)
Task 2
Maximization of Laptop Sales
The research question here is:
How many laptops and PCs should be sold in order to maximize profits?
The variables in this problem are
(i) Quantity of PCs i.e. X

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(ii) Quantity of Laptops i.e. Y
Maximization Problem Expression
Max z=400 x +700 y where x=quantity o f PCs∧ y=quantity of laptops
Constraints Expression
1000 x+1500 y ≤100,000
x ≥ 15
x ≤ 80
y ≥0
y ≤ 40
(Sultan 2014)
The Graphical Solution
Using an online graphing tool the objective function z is maximized (Z=$44,000) at the extreme point (40, 40)
hence x=40, and y=40.
Hence quantity of laptops that should be sold to maximize profit is 40.

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References
Montgomery, C., 2013. Introduction to Linear Regression Analysis. Hoboken: John Wiley & Sons.
Sultan, A., 2014. Linear Programming: An Introduction with Applications. Amsterdam: Elsevier.