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Managerial report Essay

I                                             MANAGERIAL REPORT

          Roughly examining the data we can say that the relationship between the sales and the advertising are looking to be related as the increase in the advertising expenses relevantly increased the sales but not in a manner. The graphical representation of the data may precisely help us to compare. Bar chart is a simple graphical tool to represent and compare data from which we can say about the relationship between two data. We can also say about the shape of the distribution. The representation of the sales and advertising expenditure data is as follows.

                                      Fig 1.1 Bar chart

          From the bar graph we can approximately say that there is a proportional increase in sales as the advertising expenditure increases. So we recommend increasing the advertising expenditures in a steady manner to increase the level of sales.  At the same time it will cause a reverse result also. If the sales are affected by any other reasons like economic crisis, the increase in the advertising expenditure will not help. So a precautionary plan must be done to increase the advertising expenditure. It is better to increase it while the purchasing power is high in the market.

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Also a bar graph is not sufficient tool to conclude, and let us look the other descriptive tools.

          The other similar and powerful graph tool is a frequency polygon which shows the shape of the distribution as primary. The frequency polygon for the sale and advertising data is as follows:

                                      Fig1.2 Frequency polygon

          From the frequency polygon we agree with the results of the bar graph and one more observation is that there is a steady increase in the advertising expenditure but an unsteady increase in the sales. This shows that the advertising expenditure may not contribute for a consistent sale here. Now let us move on the descriptive data analysis where we can interpret the results more precisely and accurately.

          The descriptive statistics is the type of statistics where the data are analyzed by the two main measures namely the measure of central tendency and the measure of variation. The important measures of central tendencies are the mean, median, mode and quartiles. The important measures of variations are the variance, standard deviation and quartile deviation. The drawback of these measures is that we cannot get any information about the relationship between two data. For that we have to perform the tasks like correlation and regression. First we do the individual analysis and then relationship analysis.

          Let us find the important measure of central tendencies Mean as the other measures may not help this analysis.

          Formula: “  = ” (Weisstein, Eric W)

          Mean is the average of the data which is found by dividing the sum of the data with the number of the data.

          Let us tabulate the data so as to perform the calculations simple.

Year
Sales

($1000s)
Advertising Expenditure ($1000s)
2004-05
19.0
1.0
2005-06
44.0
4.0
2006-07
40.0
6.0
2007-08
52.0
10.0
2008-09
53.0
14.0
Total
208
35

          Mean Sales = 208 / 5 = 41.6 ($1000s)

          Mean advertising expenditure = 35 / 5 = 7($1000s)

          The mean sales over the period 2004 – 2009 is 41.6 ($1000s) and the mean advertising expenditure is 7($1000s). There will be no information about the growth or decline in the mean. We can decide whether to maintain the average advertising expenditure to maintain the average sales. The other major drawback of the measure mean is that it does not give the details about individual data and is not robust of extreme values. To overcome these problems the measures of variation will help us.

          The measure of variation is the measure of variation of individual data from the mean. The important measure of the variation is the Standard deviation which is the square root of variance.

          Standard deviation:

          Formula = s = “SN  = ” (Weisstein, Eric W)

Sales (X)

($1000s)
Advertising

Expenditure(Y)

($1000s)

2004-05
19
1
-22.6
510.76
-6
36
2005-06
44
4
2.4
5.76
-3
9
2006-07
40
6
-1.6
2.56
-1
1
2007-08
52
10
10.4
108.16
3
9
2008-09
53
14
11.4
129.96
7
49
Total
208
35

757.2

104

          Standard deviation of Sales =

          Standard deviation of Advertising expenses =

          This show that the average deviation of sales is 12.306($1000s) and that of advertising expenditure is 4.56($1000s). This may tell us the individual property but not the relationship. The standard deviation of sales tells that the sales over one year to next year may deviate about 12.306($1000s) about the mean sales. The standard deviation of advertising expenses tells that the expenses of each year on average, deviates about 4.56 ($1000s) from the mean expenses. The other important thing we have to regard is that we can control or increase the amount of advertising expenses whereas we can only predict the sales. The standard deviation of expenses is higher, that is more than half of the mean which shows the gradual increase but the standard deviation of sales is less high that it is about one fourth of the mean, which shows that although there is gradual increase in the advertising expenses, may not lead to the same growth in the sales.

          Next we move on the measures which determine the relationship between the data namely the correlation and regression.

          Correlation:

          “Correlation, measured with correlation coefficient indicates the strength and direction of the relationship between two random variables.” (Kenney J F and Keeping E S, 1962, pp. 252 – 285)

          The formula for finding the correlation coefficient is given by

            ”

          Tabulating the required parameters for this formula,

(X)

($1000s)
Advertising

Expenditure(Y)

($1000s)

( )( )

19
1
-22.6
-6
135.6
510.76
36
44
4
2.4
-3
-7.2
5.76
9
40
6
-1.6
-1
1.6
2.56
1
52
10
10.4
3
31.2
108.16
9
53
14
11.4
7
79.8
129.96
49
208
35

241
757.2

104
In general the correlation coefficient lies between -1 and 1. The zero shows that there is no relation, a positive show that there is a strong relation in same direction and a negative show that there is a strong relation in opposite direction. Our data obtained 0.858 which is positive and we conclude that there is a strong relation between sales and advertising expenditure in the same direction. So we need not worry to increase the advertising expenditure so that it would increase the sales.

          We have seen the property of relationship and now we are in the position to predict the relationship in a linear form. For that the regression helps us. Regression analysis is a statistical technique in the form of equation; interpret the relation of a dependent variable with an independent variable. Here the dependent variable is sales and the independent variable is advertising expenditure as sale is dependent on the advertisements to an extent.

          The regression equation takes the form,

          “X = a + b Y” (Kenney J F and Keeping E S, 1962, pp. 252 – 285) where X is assumed to be dependent and Y is independent in our case.

          The values of a and b are obtained from the normal equations,

          3 x = n a + b 3 y and

          3 xy = a 3 y + b 3 y2

          Tabulating the required parameters

X
Y
XY
Y2
19
1
19
1
44
4
176
16
40
6
240
36
52
10
520
100
53
14
742
196
208
35
1697
349

          Total

          208 = 5a + 35b…..(1)

          1697 = 35a + 349b…..(2)

          Solving the equations by any algebraic method, we can obtain the values of a and b.

          (1) ´ 7 Þ 1456 = 35a + 245b

          (2) ´ 1 Þ 1697 = 35a + 349b

          (2) – (1) = 241 = 104b

          Þ b = 2.37 Þ a = 24.85

          Hence the regression equation is

          X = 24.85 + 2.37 Y

          where X is the sales and Y is the advertising expenditure.

          We can predict the sales using this equation given the advertising expenditure.

          If the advertising expenditure is 16($1000) in 2009-10 then the sales is

          X = 24.85 + 2.37(16)

          X = 62.77

          Hence the sales will be 62.77($1000) in 2009-10 if the advertising expenditure will be 16($1000).

          This shows and increasing trend of the sales as the expenditure on the advertising increases.

          CONCLUSION:

          We advice Karen to keep improving the advertising campaigns in better methods year by year to keep the sales higher. The statistical techniques used are deterministic and hence descriptive. A better forecast can be used by inferential statistics. “Shortcomings, weaknesses, and limitations are admitted when the arguments are presented. Dealing with both the positive and the negative suggests objectivity” (Kenneth F.Harling, 2009)

II.      DATA: The table is the data collected from 25 professional including their designation, company and email addresses on the number of tele-calls they

Name
Designation
Company/Institution
Email
Call received
Ravi chandran
Technical support
Symantec
[email protected]
7
Frank Carmody
Senior Manager
Edrise
[email protected]
1
Jegadeesh
Senior Techinician
Titan
[email protected]
0
Nilesh
Instructor
Actuarial Society
[email protected]
2
Rajaram
Senior Engineer
Titan
[email protected]
1
Kalaiyarasan
Officer
Titan
[email protected]
1
Jothi santhan
Officer
Titan
[email protected]
3
Vijayakumar
Hardware Engineer
Titan
[email protected]
0
Vadivel
Engineer
Titan
[email protected]
0
Rajavelpandian
Senior Techinician
titan
[email protected]
4
Lavgupta
Commercial Head
BSNL
[email protected]
8
Chawla
Chairman
Indian institute of Engineers
[email protected]
5
Jerald
Lecturer
Govt. college
[email protected]
2
Muthu
Officer
Titan
[email protected]
0
Sreejith
Engineer
Titan
[email protected]
4
Sunil
Engineer
Titan
[email protected]
1
Sathan
Executive
Blue cross
[email protected]
6
Inbavanav
Technical support
IRCTC
[email protected]
11
Jayaram
CRM
Titan
[email protected]
2
Syam
Admin
Ukmaths
[email protected]
5
Niroop
Customer care
ICICI bank
[email protected]
10
Rajakkannu
Consultant
Sathyam
[email protected]
1
Subba rao
Customer care
ICICI prulife
[email protected]
11
Rengasamy
IT consultant
ISG nova soft
[email protected]
5
Jaffer
Service Engineer
Olympus
[email protected]
7

1)      THE UNGROUPED FREQUENCY DISTRIBUTION TABLE:

          The ungrouped frequency distribution table of the unwanted calls is as follows

X (Number of Calls)
F(Number of Victims)
0
4
1
5
2
3
3
1
4
2
5
3
6
1
7
2
8
1
9
0
10
1
11
2

2)      THE PERCENTAGE PIE CHART:

          The percentage pie chart is drawn by the help of the following table which converts the percentage frequencies into angles.

X (Number of Calls)
F(Number of Victims)
Percentage

= X/25
Angle

= Percentage * 360
0
4
0.16
57.6
1
5
0.2
72
2
3
0.12
43.2
3
1
0.04
14.4
4
2
0.08
28.8
5
3
0.12
43.2
6
1
0.04
14.4
7
2
0.08
28.8
8
1
0.04
14.4
9
0
0
0
10
1
0.04
14.4
11
2
0.08
28.8

          Constructing the Percentage pie chart from the data above,

                                      Fig.2.1 Pie chart

3)      CALCULATION OF MEAN AND STANDARD DEVIATION

x
f
fx
x –
(x – )2
f(x – )2
0
4
0
-3.88
15.0544
60.2176
1
5
5
-2.88
8.2944
41.472
2
3
6
-1.88
3.5344
10.6032
3
1
3
-0.88
0.7744
0.7744
4
2
8
0.12
0.0144
0.0288
5
3
15
1.12
1.2544
3.7632
6
1
6
2.12
4.4944
4.4944
7
2
14
3.12
9.7344
19.4688
8
1
8
4.12
16.9744
16.9744
9
0
0
5.12
26.2144
0
10
1
10
6.12
37.4544
37.4544
11
2
22
7.12
50.6944
101.3888
Total
25
97

296.64

          The Mean for the frequency distributed data is  = 97/25 = 3.88

          The standard deviation is  =

          The mean unwanted call is 3.88 calls per person with standard deviation of 3.44. The variation is high but the mean suggests that it will be a problem as the average is 3 calls per day for a person.

BIBLIOGRAPHY

          Kenneth F.Harling(2009), Writing a Managerial report,   The Maple leaf conference, [Online]. Available at http://info.wlu.ca/~wwwsbe/MapleLeaf/Report_Writing_c.html [Accessed: 18th April, 2009]

          Kenney J F and Keeping E S, “Linear Regression and Correlation”. Ch.15 in “Mathematics of Statistics”, Pt.1, 3rd ed. Princeton, NJ: Van Nostrand, pp. 252 – 285, 1962.

          Weisstein, Eric W., “Mean” From Mathworld – A Wolfram Web resourse. [Online]. Available at http://mathworld.wolfram.com/Mean.html [Accessed : 19th April 2009]

          Weisstein, Eric W. “Standard Deviation.” From Math world –    A Wolfram Web Resource. [Online] Available at http://mathworld.com/StandardDeviation.html [Accessed: 19th April,2009]

 

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