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Basic Valuation Analysis Assignment

Basic Valuation Analysis For Financial Markets Instruments

Learning Outcome 1: Apply quantitative tools in financial markets
Learning Outcome 2: Critically evaluate the impact of changes to financial models
Learning Outcome 3: Compute basic valuation analysis for financial markets instruments
Learning Outcome 4: Evaluation of the inputs variables and their outputs in the use of quantitative tools in financial markets
Learning Outcome 5: Critically evaluate the relative importance of the key elements in successful performance measurement and control

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Definition

Regression is a statistical method that is very important for the analysis in any business sector in today's modern world. This enables the identification and characteristics of the relationships among two or more variables or multiple variables (Duncan, and Klebanov, 1994). This method also helps to identify the relevant business risk factors the businesses face and enable them to calculate the risk scores for an individual business or sector or department of the business. It is also a graphical method that describes the average association among the two or more variable. Some of the methods applied in regression are many and do vary from business to business or sector.

However, when there is either a perfect positive or negative relationship among the variables, two regression lines will coincide, that is to say, we will have one line. The farther two regression lines are from each other, the lesser is the degree of correlation and the nearer the two lines the higher the degree of correlation. The variables are independent, where r is assumed to be zero and regression lines are a right angle, that is, they are parallel to X-axis. Some of the regression models used explains the important considerations that are put before the analysis is done or performed and how the outcome of the analysis done will look like and will be interpreted to fit the business need of the company (Duncan, and Klebanov, 1994). The beneficiaries of the analysis outcomes should be able to charge or gauge as to whether the methods which has been used are good and correct, and interpreted outcome or result are suitable to the needs of the company.

It must be noted that regression lines cuts each other at the point of average, if from the point where both lines cut each other, we get the mean value of and from the point a horizontal lines are drawn. It may not pass through all the points drawn, but it will lie on some place near or adjacent to drawn points or in the midst of them. Then from there the collection of points are seen and it will slope in the direction as suggested by the said points

Purpose of Regression

The purpose of this statistical estimation of data is over and over againpronounced as the associationamong two variables or among many or several ones (Christian, and Bryant, 1998). If for example, the company wants to know whether the sales in the coming month or year will increase by a certain percentage, but also whether there will be a likelihood of high volume of sales in the near future, it will be influenced by customer satisfaction and efficient services offered to customers (Christian, and Bryant, 1998).In most cases, regression permits multiple independent variable, with the adjustment of the coefficient to confound the effects among them. It also permits evaluation and determine the optional values.

If the equation "a" and "b" are two unknown constant which determine the position of the line completely. These constants are usually parameters of the line. These parameters determine the level of the best fit, that is, distance of the line directly above or below the origin. Parameter "b" usually determines the slope of the line which shows the change in X over the unit change in Y

The variable can be explained in terms of sales so called dependent variables or sometimes called response variable. The one that gives an explanation of sales are independent variable or forecaster variables. The one that measures the association provides an initial impression to the extent at which the statistical dependence that exists between the variables (Christian, and Bryant, 1998). If the depended and independent variables are of a continuous nature such as sales, there must be a correlation coefficient must be calculated to help in measuring the strength of the relationship that exist among themselves(Elder H, 1974). It must be noted that regression lines cuts each other at the point of average, if from the point where both lines cut each other, we get the mean value of and from the point a horizontal lines are drawn.

Scatter diagram is also used in regression whereby its main purpose is to create a straight-line relationship and a connection between which it supports the suggestion. For example, where r is above 0.4 0r 0.5, Then we say that there is a linear correlation between the two variable and their other errors of measurement and even observations which functions to get us a scatter point long the line as an alternative of exactly on it (Elder H, 1974).

To determine such relationship that exist between x and y, we have to understand what straight line need to be drawn through the meeting points of the scatter diagram (Elder H, 1974). It may not pass through all the points drawn, but it will lie on some place near or adjacent to drawn points or in the midst of them. Then from therethe collection of points are seen and it will slope in the direction as suggested by the said points. Then they form a regression line which can forecast the future of the sales of the company.

As we look at company's output on a monthly basis where y is taken to be total monthly costs of the manufacturing firm, the scatter diagram bases its records on last year's records. Line drawn through this points is clearly the one we think that is of best circumstances and many researchers do refer to regression lines as lines of best fit

Shortcomings of regression

Interpretation and performance of linear regression are topic to a variety of drawbacks which can be debated in details late on. It has both opportunities and limitation of use to the most business but it help the managers of the large companies to make economic decisions for their businesses (Elder H, 1974).

In most cases, regression permits multiple independent variable, with the adjustment of the coefficient to confound the effects among them. It also permits evaluation and determine the optional values.Some of the regression models used explains that have no important considerations which are put before the analysis is done or performed and how the outcome of the analysis done will look like and will be interpreted to fit the business need of the company (Duncan, and Klebanov, 1994).

In addition, if there are many inappropriate variables are included in the model, over adjustment is likely to be the outcome, that is to say, irrelevant variables are found to an effect and have a purely change. Some of the variables are usually have random effects (Elder H, 1974). The inappropriate variable allow better fit with the data that is set under the study but since they have random effect, their outcome will get generally applicable outside the data that is set.

The causation connection
The causation's theory has been discussed by many research and for many years in the past. The concerns that have been discussed are many, however, is aimed at the academic need approach for the anticipated research method. It defines the cause of the event as anything which is perceived to be at some varying degree which is in controlfor the event which will not be said to exist but in the real sense it merely assumed to exist. Causation involved from ordinary knowledge looking to benefit and create generative methods and variables.

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QUESTION 1

a. Conduct a hypothesis test and determine if Sarah is right in her claim that on average female students performed better in economics courses compared to men. Use a significance level of 0.1.

Solution: Calculation of hypothesis testing. Given information:
Male students mean (x1) = 83 Sample size (x1) = 160 Standard deviation (S1) = 5
Female students mean (x2) = 84 Sample size (x2) = 272 Standard deviation (S1) = 7
Level of significance = 1% or 99% confidence
Critical values of z values are -2.58 and 2.58, therefore critical region z ≤ -2.58 or 2.58≥

Null hypothesis (Ho): U1 - U2 = 0, that is to say, Sarah's sample claim was drawn from same population.
Alternative Hypothesis (H1): u1 - u2 ≠ 0, which means, Sarah's claim was not drawn from same population.

The standard error of difference between the two means is given as:
∂(x1 - x2) = √(S^2 1)/n1+(S^2 2)/n2
√5^2/160+7^2/272
√0.15625 + 18015
√0.33 = 0.574
Z = (x1-x2)/(∂ (x1-x2)) = (83-84)/0.574= -1.74

Therefore Ho is accepted since the calculated Z value falls or lies within the critical region i.e. -1.74 is greater than -2.58 hence Sarah's claim is right or true.

b. Describe any additional assumptions required to validate the results of the hypothesis test in part a.

solution: Additional assumptions required to validate the hypothesis test above.

There are usually two assumptions
i) Sample analyzed using large sample, meaning that sample selected is assumed to be more than 30 items to be counted. Where the sample is selected the hypothesis more
ii) Sample analyzed using small sample, meaning that sample selected had equal likely chances of unselected ones. When they are drawn each element must be represented in an equal way.

QUESTION 2
a. Compute the least squares regression equation.

solution: a) Computing least square regression

X                     Y                     X2               Y2                          XY

336                  415                  112896            172225                        139440           

14                    21                    196                  441                  294

35                    38                    1225                1444                1330

80                    95                    6400                9025                7600

113                  152                  12769              23104              17176

146                  174                  21316              30276              25404

130                  144                  16900              20736              18720

288                  321                  82944              103041                        92448

496                  541                  246016            292681                        268336

173                  203                  29929              41209              35119

26                    39                    676                  1521                1014

∑X=1,837    ∑Y= 2143   ∑X2=531,267    ∑Y2= 695,703∑XY=606,881

Regression equation of X on Y is given as
X = a + bY

The normal equations for regression are
∑X = Na + b∑Y
∑XY = a∑Y + b∑Y^2

Substituting with the above provided values, we get
1,837 = 11a + 2,143b............(i)
606,881 = 2,143a + 695,703b...(ii)

Solving equations (i) and (ii) using elimination method, we multiply 194.8 with (i) equation above to make values of (a) equal, we get by subtracting equations (i) with (ii)

357,848 = 2,143a + 417,456b
606,881 = 2,143a + 695,703b
249,033 = 0a + 278,247b
Value of b = 278,247/249,033 = 1.2
Substituting b values in equation (i) will be
1,837 = 11a + (2,143 x 1.2)
11a = 2572 - 1,837, a = 735/11 = 66.8
Therefore the required regression equation of X on Y is gives as
X = 66.8 + 1.2Y
Regression equation of Y on X is calculated as: Y = a + bX
And normal equations are:
∑Y = Na + b∑X
∑XY = a∑X + b∑X^2
We substitute the figures with the formula
2,143 = 11a + 1,837b.............(i)
606,881 = 1,837a + 531,267b....(ii)
Solving equations (i) and (ii) using elimination method, we multiply 167 with (i) equation above to make values of (a) equal, we get by subtracting equations (i) with (ii).
357,881 = 1,837a + 306,779
606,881 = 1,837a + 531,267b
-249,000 = 0 + 224,488b
Therefore b = 1.1

Regression equation of Y on X is calculated as: Y = a + bX
And normal equations are:
∑Y = Na + b∑X
∑XY = a∑X + b∑X^2
We substitute the figures with the formula
2,143 = 11a + 1,837b.............(i)
606,881 = 1,837a + 531,267b....(ii)
Solving equations (i) and (ii) using elimination method, we multiply 167 with (i) equation above to make values of (a) equal, we get by subtracting equations (i) with (ii).
357,881 = 1,837a + 306,779
606,881 = 1,837a + 531,267b
-249,000 = 0 + 224,488b
Therefore b = 1.1

Substituting b values in equation (i) will be
2,143 = 11a + (1,837 x 1.1)
11a = 2021 - 2,143 = -122
Therefore, a = -122/11 = 11
Hence, the required regression equation of X on Y is gives as
Y = 11 + 1.1X

b. If the computer-generated account balance was 100, what would you expect to be the actual account balance as verified by the
accountant?

solution: If computer generated balance was 100, then the verified value confirmed by accountant will be
Y =11 + 1.1(100) = 121

c. The computer-generated balance for Oliver Buxton is listed as 100 in the computer-generated account record. Calculate a 95% interval estimate for Mr. Buxton's actual account balance.

solution: Calculation of the 95% interval estimate will be

P = 100/121 = 0.8
Q= 1-p (1 - 0.8) = 0.2
N= 121
∂p = √pq/n = √((0.8 x 0.2))/121 = √0.16/121 = 0.4/11 = 0.036
Population proportion = p± 1.96 (∂p)
0.8 ± 1.96 (0.036)
0.8 ± 0.07
0.73 to 0.87 which is between 73% to 87%
d. Calculate the P-value of the regression and interpret it. Given the sample of the data, what is the chance that a relationship between
the actual and computer-generated account balance doesn't exist?

Calculating p- values and interpret it

Z= 0.8- 0.2 / √ 0.2 x (1 - 0.2) /121
= 0.6 / √ 1.32
= 0.6/1.15
= 0.14
Using Z tables, 0.14 under 0.05 significance level, P-value is 0.0735
Therefore Ho is accepted since the calculated P-Value is less than 0.05 hence hypothesis in this case is strongly supported right or true because it falls within the accepted region.

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QUESTION 3


  X Y X2

1

2005

329670

4020025

2

2136

332980

4562496

3

1897

326810

3530641

4

2245

361240

5040025

5

2053

347380

4214809

6

1920

309990

3686400

7

2989

453420

8934121

8

2952

444960

8714304

9

2353

372900

5536609

10

1860

302840

3459600

11

2028

320600

4112784

12

2040

328800

4161600

13

2029

341550

4116841

14

2233

366630

4986289

15

1932

305920

3732624

16

1872

303720

3504384

17

2653

422510

7038409

18

2005

337920

4020025

19

2509

394020

6295081

20

1993

333410

3972049


69631

7037270

97639116

a. Produce a regression equation to predict the selling price for residences using a model of the following form:

solution:

Regression equation of X on Y is given as
X = a + bY
The normal equations for regression are
∑X = Na + b∑Y
∑XY = a∑Y + b∑Y^2
Substituting with the above provided values, we get
43,704 = 20 a + 7037270 b............(i)
606,881,675 = 7037270 a + 695,703,854b...(ii)
Solving equations (i) and (ii) using elimination method, we multiply 351863.5 with (i) equation above to make values of (a) equal, we get by subtracting equations (i) with (ii)

43704 = 7037270a + 4,468,456b
606,881,675 = 7037270a + 695,703b
606837971 = 0a + 3772753b
Value of b = 606837971 / 3772753 = 16
Substituting b values in equation (i) will be
43704 = 20a + (4,468,456 x 16)
11a = 7149296 - 43704, a = 1392620 / 20 = 69631
Therefore the required regression equation is gives as
Price = 69631-3630 type + 90.4 Square Feet

b. What effect would an increase of the square footage by 100 square feet have on the expected price? And what is the effect on the price if the property on the market is a single-family home?

solution: If the condominium is the residence, the price is likely to decrease on average by $3630. With increase of one unit in square feet the price is likely to increase by $100 which causes an average of $90.4

c. Calculate P-values and a 95% confidence interval for both slopes. What is the probability that the observed effect of the property being a condominium and of the square footage really influence the price?

solution: Calculation of the 95% interval estimate will be

P-value = 16/20 = 0.822
Q= 1-p (1 - 0.8) = 0.2
N= 20
∂p = √pq/n = √((0.8 x 0.2))/20 = 0.4/4.47 =
Population proportion = p± 1.96 (∂p)
0.8 ± 1.96 (0.089)
0.8 ± 0.17
0.63 to 97 which is between 63% to 97%

Z= 0.8- 0.2 / √ 0.2 x (1 - 0.2) /20
= 0.6 / √ 0.008
= 0.6/0.089
= 6.74
Using Z tables 6.74 under 0.05 significance level, we find it is outside the recommended region
Since the p-value of 6.74 is greater than level of significance of 0.05, we reject the null hypothesis and we have to conclude that the relationship that is existing between the selling price and the square footage is not different from condominiums and single family homes.

d. Produce an equation that describes the relationship between the selling price and the square footage of (1) condominiums and (2) single-family homes.

solution:

d) i) The condominiums equation from (a) above will give us, Price = 66,001 + 90.4 Square Feet
ii) The single family equation is also given as Price = 69,631 + 90.4 Square Feet

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QUESTION 4

a. Calculate the multiple linear regression equation using quarterly data as dummy variables.

solution: The calculation of the multiple regression analysis using dummy variables to determine the revenue.

Year                Quarter            Revenue          t           Q1       Q2       Q3

2015                1                      492                  1          1          0          0

                        2                      457                  2          0          1          0

                        3                      485                  3          0          0          1

                        4                      578                  4          0          0          0

2016                1                      554                  5          1          0          0

                        2                      559                  6          0          1          0

                        3                      574                  7          0          0          1

                        4                      630                  8          0          0          0         

2017                1                      587                  9          1          0          0

                        2                      651                  10        0          1          0

                        3                      645                  11        0          0          1

                        4                      728                  12        0          0          0

2018                1                      701                  13        1          0          0

                        2                      729                  14        0          1          0

                        3                      776                  15        0          0          0

                        4                      840                  16        0          0          0

Regression analysis

Overall fit

Multiple R                   0.89089

R Square                     0.9576

Adjusted R Square     0.98185

Standard error             0.53759

Observations               16

                                    Coeff               Std error          t stat

Intercept                      14.327             0.4488             30.73

T                                  0.6524             0.0139             16.59

Q1                               -3.9381            0.4367             -8.23

Q2                               -5.7375            0.4321             -13.488           

Q3                               -2.4938            0.4211             -5.922

Explanation

The data shown is for the two independent variables, that is, square footage and corresponding prices for each footage cost. So we are looking at the relationship in the form of
Sales = a + b1x the square feet + b2 x price. Therefore the above calculation shows the value for the interpretation and variables within the line of best fit identified as
Sales = 585.96 + (9.90 x 0.957)
The coefficient of correlation r = 0.957. That means, a very strong linear relationship exist. Then this is confirmed by determination coefficient, r2, which shows that 99.7%.
The adjusted r2 which removes the bias but is only slightly lesser that that calculated value.

b. Interpret each slope and the intercept in the multiple linear regression equation from part a. Calculate a 95% and 99% confidence interval for each variable slope and assess whether it is statistically significant.

solution: Calculation of the 95% and 99% interval estimate will be
Lower at 95% significance level
P = 14/20 = 0.7
Q= 1-p (1 - 0.7) = 0.3
N= 16
∂p = √pq/n = √((0.7 x 0.3))/16 = 0.115
Population proportion = p± 1.96 (∂p)
0.7 ± 1.96 (0.115)
0.7 ± 0.2254
0.47 to 0.93

Upper at 99% confidence
0.7 ± 2.58 (0.115)
0.7 ± 0.2967
0.4 to 0.99

c. Create a forecast for 2019 and 2020.

solution: c) Let us assume that sales will constantly increase in year 2019 and 2020 consecutively
Ending revenue as of Q4 2018 was 840 with 16 quarters (4 x 4 years).
By end of Q4 of 2019 sales will be: = 840 + (9.90 x 0.997 x 20 quarters)
840 + 189 = 1029.
By end of Q4 of 2019, sales forecast will be = 1029 + (9.90 x 0.957 x 24 quarters)
1029 + 227 = 1256

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QUESTION 5

a) Calculating the best combination of asset stock, bonds and commodities in the financial market.

solution:

Asset case                   Stocks             Bonds              Commodities

Expected Return         100%               0%                   0%

                                    90%                 10%                 0%

                                    80%                 20%                 10%

                                    70%                 30%                 20%

                                    60%                 40%                 30%

                                    50%                 50%                 40%

                                    40%                 60%                 50%

                                    30%                 70%                 60%

                                    20%                 80%                 70%

                                    10%                 90%                 80%

                                    0%                   100%               90%

                                    0%                   0%                   100%

 

Standard deviation     100%               0%                   0%

                                    90%                 10%                 0%

                                    80%                 20%                 10%

                                    70%                 30%                 20%

                                    60%                 40%                 30%

                                    50%                 50%                 40%

                                    40%                 60%                 50%

                                    30%                 70%                 60%

                                    20%                 80%                 70%

                                    10%                 90%                 80%

                                    0%                   100%               90%

                                    0%                   0%                   100%

 

b)

Combination               Stocks                         Bonds                          Commodities

                                    ER       SD                   ER       SD                   ER       SD

                                    100%   100%               0%       0%                   0%       0%

                                    90%     90%                 10%     10%                 0%       0%

                                    80%     80%                 20%     20%                 10%     10%

                                    70%     70%                 30%     30%                 20%     20%

                                    60%     60%                 40%     40%                 30%     30%

                                    50%     50%                 50%     50%                 40%     40%

                                    40%     40%                 60%     60%                 50%     50%

                                    30%     30%                 70%     70%                 60%     60%

                                    20%     30%                 80%     80%                 70%     70%

                                    10%     20%                 90%     90%                 80%     80%

                                    0%       0%                   100%   100%               90%     90%

                                    0%       0%                   0%       0%                   100%   100%

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QUESTION 6

Find the multiple linear regression equation of X1 on X2 and X3 from the data relating to three variables given below

X1:      4          6          7          9          13        15

X2:      15        12        8          6          4          8

X3:      30        24        20        14        10        4

Calculate the regression equation above and estimate the probable revenue for the firm.

Solution:

Regression equation of X1 on X2 and X3 is given as
X1 = a + bX2 + bX3
The constants value are a, b1 and b2 are obtained by solving the following three normal equations as given below
∑X1 = Na + b∑X2 + b∑X3
∑X1X2 = a∑X3 + b∑X^2 2 +b∑X2X3
∑X1X3 = a∑X3 +b∑X2X3+ b∑X^2 3
Calculating the required values

X1       X2       X3       X1X2    X1X3     X2X3     X21      X22      X23   

4          15        30        60        120      450      225      900      16

6          12        24        72        144      288      144      576      36

7          8          20        56        140      160      64        400      49

9          6          10        54        126      84        36        196      81

13        4          10        52        130      40        16        100      169

15        3          4          45        60        12        9          12        225

54        48        102      339      720      1,034   494      2,188   576

Substituting the values in the normal equations, we get
6a + 48b + 102b = 54 ..........................(i)
48a + 494b + 1034b = 339 ...................(ii)
102a + 1034b + 2188b = 720 .................(iii)

Multiplying equation (i) by 8, we get
48a + 384b + 816b = 432 .....................(iv)
Subtracting equation (ii) from equation (iv), we get
110b + 218b = -93 ............................(v)
Multiplying equation (i) by 17, we get
102a + 816b + 1734b = 918 .................(vi)
Subtracting equation (iii) from equation (iv), we get
218b + 454b = -198 ...........................(vii)
Multiplying equation (v) by 109, we obtain
11990b + 23762b = -10137 .................(viii)
Multiply equation (vii) by 55, we get
11990b + 24070b = -10890 .................(ix)
Subtracting equation (viii) from equation (ix), we get
1208b = -753
B = -753 / 1208 = 0.623
Substituting this value of b in equation (v), we get
110b + 218 (-o.623) = -93
110b + 135.814 - 93
B = 42.814 / 110 = +0.389

Substituting the value of b3 and b2
110b + 218(-0.623) in equation (i), we get
6a + 48 (0.389) + 102 (-0.623) = 54
6a = 54 + 63.546 - 18.672 = 98.874
A = 16.479
The regression equation is given as
X1 = 16.479 + 0.389X2 - 0.623X3

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