Forecasting One Mark

COF1STAT3386 – ASSIGNMENT + SOLUTION

Question 1

1 points Save

Given an actual demand of 61, a previous forecast of 58, and an > of .3, what would the forecast for the next period be using simple exponential smoothing?

45.5

57.1

58.9

61.0

65.5

Answer: 58.9

Explanation:

The given information in the problem can be represented as follows

The actual demand for the previous period = 61

Forecast for the previous period = 58

Smoothing constant = 0.3

The forecast for the next period is calculated using simple exponential smoothing is given by

F t=Ft−1+α∗( At−1−Ft−1 )

Where

Ft = forecast for period t

Ft-1 = forecast for the previous period = 58

α = smoothing constant = 0.3

At−1= actual demand for the previous period = 61

The new forecast for the next period from the given information is given by

F t=Ft−1+α∗( At−1−Ft−1 )

= 58 + 0.3 * (61 – 58)

= 58.9

Hence the forecast for the next period is 58.9

Question 2

1 points Save

Which of the following values of alpha would cause exponential smoothing to respond the most slowly to forecast errors?

0.10

0.20

0.40

0.80

cannot be determined

Answer: 0.10

Explanation:

The quickness of forecast adjustment to error is determined by the smoothing constant, α.

The closer its value is to zero, the slower the forecast will be to adjust to forecast errors.

Thus the smoothing constant α = 0.10 is the closest value to zero responds the most slowly to forecast errors.

Hence the value of alpha is 0.10

Question 3

1 points Save

A forecasting method has produced the following over the past five months. What is the mean absolute deviation?

>

-0.2

-1.0

0.0

1.2

8.6

Answer: 1.2

Explanation:

The mean absolute deviation (MAD) is given by

MAD =

∑|e|n

Where

e is the forecast error = Actualt – Forecastt

n = number of the month = 5

Using the given information in the above table, mean absolute deviation (MAD) is given by

MAD =

∑|e|n

=

1+2+2+0+15

=

65

= 1.2

Hence the mean absolute deviation is 1.2

Question 4

1 points Save

Given forecast errors of -1, 4, 8, and -3, what is the mean absolute deviation?

2

3

4

8

16

Answer: 4

Explanation:

The given information in the problem can be represented as follows

The forecast errors are – 1, 4, 8, – 3


MAD =

∑|e|n

Where

e is the forecast error = Actualt – Forecastt = –1, 4, 8, –3

|e| = 1, 4, 8, 3

n = number of the period = 4


MAD =

∑|e|n

=

1+4+8+34

=

164

= 4

Hence the mean absolute deviation is 4

Question 5

1 points Save

The last four months of sales were 8, 10, 15, and 9 units. The last four forecasts were 5, 6, 11, and 12 units. The Mean Absolute Deviation

(MAD) is

2

-10

3.5

9

10.5

Answer: 3.5

Explanation:

The given information in the problem can be calculated as follows

The calculation of the forecast error to find the mean absolute deviation (MAD) is given in the following table:

Period Actual(sales) Forecast Error (A – F) |Error|1 8 5 3 32 10 6 4 43 15 11 4 44 9 12 –3 3TOTAL 14

From the above table we calculate MAD,


MAD =

∑|e|n

Where

e is the forecast error = Actualt – Forecastt

n = number of the month = 4


MAD =

∑|e|n

=

144

= 3.5

Hence the mean absolute deviation is 3.5

Question 6

1 points Save

A time series trend equation is 25.3 + 2.1 X. What is your forecast for period 7?

23.2

25.3

27.4

40.0

cannot be determined

Answer: 40.0

Explanation:

A linear trend equation has the form

Ft = a + b * t

Where

t = specified number of time periods from t = 0

Ft = forecast for period t

a = value of Ft at t = 0

b = slope of the line

The time series trend equation is given by

Ft = 25.3 +2.1 * X

Where

X = specified number of time period

The forecast for period 7 that is X = 7 is given by

Ft = 25.3 +2.1 * X

= 25.3 + 2.1 * 7

= 40.0

Hence, the forecast for period 7 is 40.0

Question 7

1 points Sav

e

For a given product demand, the time series trend equation is 53 - 4 X. The negative sign on the slope of the equation

is a mathematical impossibility

is an indication that the forecast is biased, with forecast values lower than actual values is an indication that product demand is declining

implies that the coefficient of determination will also be negative

implies that the RSFE will be negative

Answer: is an indication that product demand is declining

Explanation:

The negative sign on the slope of the equation indicates there is a inverse relationship between the two variables X and Y.

Hence the correct choice is “is an indication that product demand is declining”

Question 8

1 points Save

In trend-adjusted exponential smoothing, the forecast including trend (FIT) consists of

an exponentially smoothed forecast and an estimated trend value

an exponentially smoothed forecast and a smoothed trend factor the old forecast adjusted by a trend factor

the old forecast and a smoothed trend factor

a moving average and a trend factor

Answer: an exponentially smoothed forecast and a smoothed trend factor

Explanation:

The trend adjusted forecast (TAF) is composed of two elements: smoothed error and a trend factor

TAFt+1 = St + Tt

Where

St = previous forecast plus smoothed error

Tt = current trend estimate

Question 9

1 points Save

Which of the following is true regarding the two smoothing constants of the Forecast Including Trend (FIT) model?

One constant is positive, while the other is negative.

They are called MAD and RSFE.

Alpha is always smaller than beta.

One constant smoothes the regression intercept, whereas the other smoothes the regression slope. Their values are determined independently.

Answer: One constant smoothes the regression intercept, whereas the other smoothes the regression slope

Explanation:

For the the two smoothing constants of the Forecast Including Trend (FIT) model, it is true that one constant smoothes the regression intercept, whereas the other smoothes the regression slope.

Question 10 1 points

Demand for a certain product is forecast to be 800 units per month, averaged over all 12 months of the year. The product follows a seasonal pattern, for which the January monthly index is 1.25. What is the seasonally-adjusted sales forecast for January?

640 units

798.75 units

800 units

1000 units

cannot be calculated with the information given

Answer: 1000 units

Explanation:

Seasonally-adjusted sales forecast for January = 800 * 1.25 = 1000 units.

Question 11

1 points Save

A seasonal index for a monthly series is about to be calculated on the basis of three years' accumulation of data. The three previous July values were 110, 150, and 130. The average over all months is 190. The approximate seasonal index for July is

0.487

0.684

1.462

2.053

cannot be calculated with the information given

Answer: 0.684

Explanation:

The three year average of July month =

110+150+1303

=

3903

= 130

The seasonal index of the July moth is =

July moth average Over all month average

=

130190

=0.684

Hence, the seasonal index of the July moth is 0.684

Question 12

1 points

The percent of variation in the dependent variable that is explained by the regression equation is measured by the

mean absolute deviation

slope

coefficient of determination

correlation coefficient

intercept

Answer: coefficient of determination

Explanation:

The coefficient of determination (R-square) value indicates that the percent of variation in the

dependent variable that is explained by the regression equation.

Question 13

1 points

If two variables were perfectly correlated, the correlation coefficient r would equal

0

less than 1

exactly 1

-1 or +1

greater than 1

Answer: - 1 or +1

Explanation:

Since, the correlation coefficient r lies between ranges from -1 to +1. If the r value is -1 then

there is a perfect negative correlation exists, if the r value is +1 indicates that there is a perfect

positive correlation exist between two variables.

Question 14

1 points

The last four weekly values of sales were 80, 100, 105, and 90 units. The last four forecasts were 60, 80, 95, and 75 units. These forecasts illustrate

qualitative methods

adaptive smoothing

slope

bias

trend projection


>

Use the three-month moving-average method to forecast sales for June.

Fewer than or equal to 20 units

Greater than 20 but fewer than or equal to 22 units


Greater than 24 units

Answer: Greater than 24 units

Explanation:

The Three-month moving total carried out as follows:

The June month moving total is given by

22 +28+24 = 74

Three month moving averages for month June =

Three month total3

=

743

= 24.67

Calculation Three Year Moving average:

MonthActual Sales

3-Month Moving Total3-Month Moving

Average

January 23

February 18

March 22

April 28 63 21.00

May 24 68 22.67

June 74 24.67

From the above table we see that the 3-month moving average for June month is 24.67


>

What is the forecast for July with the two-month moving-average method and June sales of 40 units?





Answer: Greater than 30 but fewer than or equal to 35 units

Explanation:

The Three-month moving total carried out as follows:

The July month moving total (Two months) is given by

24 + 40 = 64

Two month moving averages for month July =

Two month total2

=

642

= 32

Calculation Two Year Moving average:

MonthActual Sales

2-Month Moving Total2-Month Moving

Average

January 23

February 18

March 22 41 20.50

April 28 40 20.00

May 24 50 25.00

June 40 52 26.00

July 64 32.00

From the above table we see that the 2-month moving average for July month is 32.00

Question 17

1 points

The forecasting equation for a three-month weighted moving average is:Ft = W1Dt + W2Dt-1 + W3Dt-2

If the sales for June were 40 units and the weights are W1= 1/2, W2 = 1/3, and W3 = 1/6, what is the forecast for July?Assume Dt = June Demand = 40.






Explanation:

The formula of forecasting equation for a three-month weighted moving average is given by

Ft = W1Dt + W2Dt-1 + W3Dt-2

The forecasted value of July month is given by

=

12∗40+ 1

3∗24+ 1

6∗28

= 0.5*40 + 0.333*24 + 0.1667*28

= 20 +7.9999 + 4.667

= 32.667

Weighted Three- month moving average:

MonthDemand

(Dt)3-Month Weighted Moving Average

January 23

February 18

March 22

April 28 20.83

May 24 24.33

June 40 25.00

July 32.67

Hence, the forecasted value of July month is 32.67

Question 18

1 points

>

Using the 4-month weighted moving-average technique and the following weights, what is the forecasted demand for November?

>




More than 280 units


Explanation:

The formula of forecasting equation for a Four-month weighted moving average is given by

Ft = W1Dt + W2Dt-1 + W3Dt-2+ W4Dt-3

Where:

W1 = most recent month = 50% = 0.5

W2 = one month ago = 20% = 0.2

W3 = Two month ago = 20% = 0.2

W4 = three month ago = 10% = 0.1

The forecasted value of December month is given by

= 0.5*250 + 0.2*280 + 0.2*310 +0.1*240

=125 +56+62+24

= 267

Hence, the forecasted value of December month is 267

Question 19

1 points

>

Use an exponential smoothing model with a smoothing parameter of 0.30 and an April forecast of 525 to determine what the forecast sales would have been for June.

Fewer than or equal to 535

Greater than 535 but fewer than or equal to 545


Greater than 555


Explanation:

The Exponential smoothing forecast model is given below:

F t+1=αY t+(1−α )F t

Where:

F t+1 = forecast of the time series for period (t + 1)

Y t = actual value of the time series in period t

Ft = forecast of the time series for period t

α = smoothing constant (0 ≤α ≤ 1)

In our problem we see that the forecast value of April month is 525 and α = 0.3

Therefore, the forecasted value of May month is given by

F = 0.3* 550 + (1-0.30)*525

= 165 +367.5

= 532.5

The forecasted value of May month is 532.5

Similarly, the forecasted value of June month is given by

F = 0.3*590 + (1-0.30)*532.5

= 177 +372.75

= 549.75

Hence, the forecasted value of June month is 549.75


>

Use the exponential smoothing method with = 0.5 and a February forecast of 500 to forecast the sales for May.

Fewer than or equal to 530



Greater than 550


Explanation:



In our problem we see that the forecast value (F) of February month is 500 and α = 0.5

Therefore, the forecasted value of March month is given by

F = 0.5* 520 + (1-0.50)*500

= 260 +250

= 510

The forecasted value of March month is 510

Similarly, the forecasted value of April month is given by

F = 0.5* 535 + (1-0.50)*510

= 267.5 +255

= 522.5

The forecasted value of April month is 522.5

The forecasted value of May month is given by

F = 0.5* 550 + (1-0.50)*522.5

= 275 +261.25

= 536.25

The forecasted value of May month is 536.25


TOMBOW is a small manufacturer of pencils and has had the following

sales record for the most recent five months:

>

Use an exponential smoothing model to forecast sales in months 2, 3, 4, and 5. Let the smoothing parameter > equal 0.6; select F1 = 150 to get the forecast started.

The forecast for month 2 is:

fewer than or equal to 120 units.

greater than 120 but fewer than or equal to 140 units.


greater than 160 units.


Explanation:



Where:

F t+1 = forecast of the time series for period (t + 1)

Y t = actual value of the time series in period t

Ft = forecast of the time series for period t

α = smoothing constant (0 ≤α ≤ 1)

The forecast value of first month is (F1) = 150 and α = 0.6

The 2nd second moth (t + 1 = 1 + 1), the forecast (F2) is calculated as follows:

F2 =F1+1=αY 1+(1−α )F1

Where; α = smoothing constant = 0.6

⇒ F2=0. 6∗150+(1−0 .6 )∗150

F2 = 90 + 0.4*150

= 90+ 60

= 150

Hence, the forecasted value of 2nd month is 150

Question 22

1 points

TOMBOW is a small manufacturer of pencils and has had the following sales record for the most recent five months:

>

Use an exponential smoothing model to forecast sales in months 2, 3, 4, and 5. Let the smoothing parameter > equal 0.6; select F1 = 150 to get the forecast started.



greater than 140 but fewer than or equal to 150.




Explanation:

The forecast value of second month is (F2) = 150

The 3rd second moth (t + 1 = 2 + 1), the forecast (F3) is calculated as follows:

F3 =F1+2=αY 2+(1−α )F2


⇒ F3=0. 6∗145+(1−0 . 6)∗150

= 87 + 60

= 147

Hence, the forecasted value of 3rd month is 147

Similarly, the forecasted value of 4th month is given by

F4 =F1+3=αY 3+(1−α)F3


F3 = the forecasted value of 3rd month = 147

⇒ F4=0 . 6∗160+(1−0. 6 )∗147

= 96+58.8

= 154.8

Hence, the forecasted value of 4th month is 154 .8

Question 23

1 points


>

Use an exponential smoothing model to forecast sales in months 2, 3, 4, and 5. Let the smoothing parameter = 0.6; select F1 = 150 to get the forecast started.







Explanation:

The forecast value of second month is (F4) = 154.8 (by referring above answer)

The 5th moth (t + 4 = 4 + 1), the forecast (F5) is calculated as follows:

F5 =αY 4+(1−α )F4


⇒ F3=0. 6∗180+(1−0 . 6)∗154 . 8

= 180 + 61.92

= 169.92

Hence, the forecasted value of 5th month is 169.92

Question 24

1 points Save


>

Use an exponential smoothing model to forecast sales in months 2, 3, 4, and 5. Let the smoothing parameter = 0.6; select F1 = 150 to get the forecast started.

The cumulative sum of errors CFE from months 2 through 5 is:

fewer than or equal to 80.



greater than 90.

Answer: Greater than 80 but fewer than or equal to 85

Explanation:

From the actual and the forecasted value obtained in the previous question, the cumulative sum

of errors can be calculated as follows:

Error = Actual – Forecast

Month

Units Sold(Actual)

ForecastFt

Error =Actual – Forecast

1 150 150.00 0.002 145 150.00 -5.003 160 147.00 13.004 180 154.80 25.205 220 169.92 50.08 Sum = 83.28

From the above table, it can be clearly seen that the cumulative for months 2 through 5 is 83.28

which is greater than 80 but fewer than or equal to 85.

Question 25

1 points Save


Use an exponential smoothing model to forecast sales in months 2, 3, 4, and 5. Let the smoothing parameter equal 0.6; select F1 = 150 to get the forecast started.

What is the MAD for months 2 through 5?

Less than or equal to 20

Greater than 20 but less than or equal to 25

Greater than 25 but less than or equal to 30

Greater than 30

Answer: Greater than 20 but less than or equal to 25

Explanation:

From the Cumulative sum of errors CFE obtained in the previous question, the Mean Absolute

Deviation (MAD) can be calculated as follows:

MAD =

∑ ( Actual− Forecast )n

Month Units Sold Ft Error | Error|

1 150 150.00 0.00 02 145 150.00 -5.00 53 160 147.00 13.00 134 180 154.80 25.20 25.25 220 169.92 50.08 50.08 83.28 93.28

MAD = 23.32

From the above table, we have

MAD =

∑ ( Actual− Forecast )n =

93 .284 = 23.32

Thus, the MAD for months 2 through 5 is 23.32 which is greater than 20 but less than or

equal to 25.

Question 26

1 points Save

A sales manager wants to forecast monthly sales of the machines the company makes using the following monthly sales data.

>

Use this information, and use the 3-month weighted moving-average method to calculate the forecast for month 9. The weights are 0.60, 0.30, and 0.10, where 0.60 refers to the most recent demand.

$3,916

$3,880

$3,396

$3,229

Answer: $ 3,916

Explanation:

The formula of forecasting equation for a three-month weighted moving average is given by

Ft+1 = W1Dt + W2Dt-1 + W3Dt-2

The forecasted (3-month weighted moving average) value of the 3rd month (t = 3) is calculated

as follows:

F3+1=W 1∗B3+W 2∗B2+W 3∗B1

F4 = 0.60 * 3469 + 0.30 * 2558 + 0.10 * 3803 = 3229.1

Similarly, the forecasted values for the remaining months (4th to 8th month) have been

calculated and are presented in the following table:

Month Balance Ft1 3803 2 2558 3 3469 4 3442 3229.15 2682 3361.76 3469 2988.77 4442 3230.28 3728 3974.1 9 3916.3

From the above table, we can see that the forecasted value for the month of 9 is 3,916.3 or

approximately 3,916.

Question 27

1 points Save

A sales manager wants to forecast monthly sales of the machines the company makes using the following monthly sales data.

>

Use this information, if the forecast for period 7 is $4,300, what is the forecast for period 9 using exponential smoothing with an alpha equal to 0.30?

$4,300

$4,342

$4,158

$3,957

Answer: $ 4,158

Explanation:

The forecast value of 7 month is (F7) is given as 4,300

Then, the forecast for the 8th month, F8 (t + 1 = 7 + 1 = 8) is calculated as follows:

F8 =αY 7+(1−α )F7

.where α = smoothing constant = 0.30

⇒ F8 =0.30∗4 ,442+(1−0. 3 )∗4 ,300

= 4342.60

Now, the forecast for the 9th month, F9 is calculated as follows:

F9 =αY 8+(1−α )F8

F9 = 0.30 * 3728 + (1 – 0.30) * 4342.60

= 4158.22 or 4158 (approximately)

Question 28

1 points Save

The management of an insurance company monitors the number of mistakes made by telephone service representatives for a company they have subcontracted with. The number of mistakes for the past several months appears in this table along with forecasts for errors made with three different forecasting techniques. The column labeled Exponential was created using exponential smoothing with an alpha of 0.30. The column labeled MA is forecast using a moving average of three periods. The column labeled WMA uses a 3-month weighted moving average with weights of 0.65, 0.25, and 0.10 for the most-to-least recent months.

>

Using this table, what is the MSE for months 6-10 for the exponential smoothing technique?

Less than 591

Greater than or equal to 591 but less than 595

Greater than or equal to 595 but less than 599

Greater than 599

Answer: Greater than 599

Explanation:

The formula for MSE (Mean Square Error) is given below:

MSE =

∑i=1

n

(Y i−Y i )2

n

Month

Mistakes Y i

ExponentialY

(Y i−Y )2

1 55 2 61 3 71 71 04 77 71 365 88 73 231.046 100 77 512.577 109 84 617.428 122 92 923.779 126 101 638.85

10 126 108 313.04

From the above table, the square of the deviation (Y i−Y )2 has been calculated for the month 3

through 10 and we have highlighted the values for the month 6 through 10.

Now, the sum of the square of the deviation for the month 6 through 10 has been calculated and

is given by

∑i=6

10

(Y i−Y i)2

= 512.57 + 617.42 + 923.77 + 638.85 + 313.04 = 3005.65

MSE =

∑i=1

n

(Y i−Y i )2

n =

3005 .625 = 601.131

Thus, the value of MSE for the month 6-10 is ‘601.131’ that is greater than 599.

Question 29

1 points Save

The management of an insurance company monitors the number of mistakes made by telephone service representatives for a company they have subcontracted with. The number of mistakes for the past several months appears in this table along with forecasts for errors made with three different forecasting techniques. The column labeled Exponential was created using exponential smoothing with an alpha of 0.30. The column labeled MA is forecast using a moving average of three periods. The column labeled WMA uses a 3-month weighted moving average with weights of 0.65, 0.25, and 0.10 for the most-to-least recent months.

>

Using this table, what is the order of the forecasting techniques from most accurate to least accurate based on their errors for months 6-10?

Exponential smoothing, weighted moving average, moving average Exponential smoothing, moving average, weighted moving average Moving average, exponential smoothing, weighted moving average Weighted moving average, moving average, exponential smoothing

Answer: Weighted moving average, moving average, exponential smoothing

Explanation:

The formula for MSE (Mean Square Error) is given below:

MSE =

∑i=1

n

(Y i−Y i )2

n

For the given data, the MSE has been calculated for the given three models and are shown

below:

Month Mistakes Exponential 1 55 2 61 3 71 71 04 77 71 365 88 73 231.046 100 77 512.577 109 84 617.428 122 92 923.779 126 101 638.85

10 126 108 313.04 SUM (6th to 10th) = 3005.65 MSE (6th to 10th) = 601.13

Month Mistakes MA 1 55 2 61 3 71 4 77 62 2255 88 70 3246 100 79 4417 109 88 4418 122 99 5299 126 110 256

10 126 119 49 SUM (6th to 10th) = 1716.00 MSE (6th to 10th) = 343.20

Month Mistakes WMA

1 55 2 61 3 71 4 77 67 1005 88 74 1966 100 84 2567 109 95 1968 122 105 2899 126 117 81

10 126 123 9 SUM (6th to 10th) = 831.00 MSE (6th to 10th) = 166.20

The forecasting techniques have been ordered from most accurate to least accurate based on their

MSE for months 6-10 and are presented below:

Model MSE RANKWMA 161 1STMA 323.57 2ND

EXPONENTIAL 409.09 3RD

Thus, the order of the forecasting models is:

Weighted moving average (WMA), Moving Average (MA), Exponential Smoothing

Question 30

1 points Save

The management of an insurance company monitors the number of mistakes made by telephone service representatives for a company they have subcontracted with. The number of mistakes for the past several months appears in this table along with forecasts for errors made with three different forecasting techniques. The column labeled Exponential was created using exponential smoothing with an alpha of 0.30. The column labeled MA is forecast using a moving average of

three periods. The column labeled WMA uses a 3-month weighted moving average with weights of 0.65, 0.25, and 0.10 for the most-to-least recent months.

>

Using this table, what is the mean absolute percent error for months 6-10 using the exponential smoothing forecasts?

Less than 22%

Greater than or equal to 22% but less than 24%

Greater than or equal to 24% but less than 26%

Greater than 26%

Answer: Less than 22%

Explanation:

The mean absolute percent error is given by the following formula:

MAPE = 100n

∑i=1

n

|Y i−F iY i

|

Month Mistakes Exponential

Y i−F iY i

1 55 2 61 3 71 71 4 77 71 5 88 73 6 100 77 0.237 109 84 0.238 122 92 0.259 126 101 0.20

10 126 108 0.14 SUM = 1.04 100/n = 100/5 20.00

MAPE = 20.8902


MAPE = 100n

∑i=1

n

|Y i−F iY i

| = 20.8902, which is less than 0.22 (or 22%).

QUESTION 31. A. B. C.

A. Compute descriptive statistics for each stock and the S&P 500. Comment on the results. Which stocks are the most volatile?

B. Compute the value of beta for each stock. Which of these stocks would you expect to perform best in an up market? Which would you expect to hold their value best in a down market?

C. Comment on how much of the return for the individual stocks is explained by the market.

Solution:

(a) Descriptive Statistics

There are 8 stocks and S&P 500 which the the descriptive statistics has been calculated using

Excel tool (Data Data Analysis Descriptive Statistics) and the results are presented below:

Microsoft Exxon Mobil Caterpillar

Mean0.00502555

6 Mean0.01663722

2 Mean0.03009722

2

Standard Error 0.00756193 Standard Error0.00922334

8 Standard Error0.01142720

9Median 0.004 Median 0.012785 Median 0.040815Mode #N/A Mode #N/A Mode #N/AStandard Deviation

0.045371583

Standard Deviation

0.055340091

Standard Deviation

0.068563251

Sample Variance0.00205858

1 Sample Variance0.00306252

6 Sample Variance0.00470091

9Kurtosis -

0.93997177Kurtosis 6.58430206

8Kurtosis 0.28574371

9

3

Skewness 0.02437306 Skewness1.51829391

2 Skewness0.33059476

7Range 0.17084 Range 0.34863 Range 0.31907Minimum -0.08201 Minimum -0.11646 Minimum -0.1006Maximum 0.08883 Maximum 0.23217 Maximum 0.21847Sum 0.18092 Sum 0.59894 Sum 1.0835Count 36 Count 36 Count 36

For the Stock “Microsoft”, the mean return is 0.00503 and its standard deviation is 0.0454.

The coefficient of variation (CV) for the return of “Microsoft” is given by

CV = Standard deviation / Mean = 0.00756 / 0.0454 = 9.03 %

For the Stock “Exxon Mobil”, the mean return is 0.0166 and its standard deviation is 0.0553.

The coefficient of variation (CV) for the return of “Exxon Mobil” is given by


For the Stock “Caterpillar”, the mean return is 0.0301 and its standard deviation is 0.0686

The coefficient of variation (CV) for the return of “Caterpillar” is given by


Johnson & Johnson McDonald's Sandisk Mean 0.005295556 Mean 0.02447417 Mean 0.06926194Standard Error 0.005811 Standard Error 0.0113494 Standard Error 0.03256624Median -0.001475 Median 0.037015 Median 0.074135Mode #N/A Mode #N/A Mode #N/AStandard 0.034866 Standard 0.06809637 Standard 0.19539743

Deviation Deviation DeviationSample Variance 0.001215638 Sample Variance 0.00463712 Sample Variance 0.03818016Kurtosis 0.434491683 Kurtosis 0.224422 Kurtosis -0.1931613Skewness 0.595957212 Skewness 0.2101202 Skewness 0.30934715Range 0.16251 Range 0.297 Range 0.78496Minimum -0.05917 Minimum -0.11443 Minimum -0.28331Maximum 0.10334 Maximum 0.18257 Maximum 0.50165Sum 0.19064 Sum 0.88107 Sum 2.49343Count 36 Count 36 Count 36

For the Stock “Johnson & Johnson”, the mean return is 0.0053 and its standard deviation is

0.0349

The coefficient of variation (CV) for the return of “Johnson & Johnson” is given by


For the Stock “McDonald's”, the mean return is 0.0245 and its standard deviation is 0.0681

The coefficient of variation (CV) for the return of “McDonald's” is given by


For the Stock “Sandisk”, the mean return is 0.0693 and its standard deviation is 0.1954

The coefficient of variation (CV) for the return of “Sandisk” is given by


Qualcomm Procter & Gamble

Mean 0.02836028 Mean 0.01058856

Standard Error 0.01436449 Standard Error 0.00617781Median 0.038705 Median 0.0133255Mode #N/A Mode #N/AStandard Deviation 0.08618695

Standard Deviation 0.03706685

Sample Variance 0.00742819 Sample Variance 0.00137395Kurtosis -0.4890999 Kurtosis -0.5558073Skewness 0.00243987 Skewness 0.15628025Range 0.33225 Range 0.141483Minimum -0.1217 Minimum -0.05365Maximum 0.21055 Maximum 0.087833Sum 1.02097 Sum 0.381188Count 36 Count 36

For the Stock “Qualcomm”, the mean return is 0.0284 and its standard deviation is 0.0862

The coefficient of variation (CV) for the return of “Qualcomm” is given by


For the Stock “Procter & Gamble”, the mean return is 0.0106 and its standard deviation is

0.0371

The coefficient of variation (CV) for the return of “Procter & Gamble” is given by


The coefficient of variation for the 8 stocks is consolidated below:

Stock C V Microsoft 9.028 (most volatile)

Exxon Mobil 3.326

Caterpillar 2.278

Johnson & Johnson 6.584

McDonald's 2.782

Sandisk 2.821

Qualcomm 3.039

Procter & Gamble 3.501

From the above table, we can see that the coefficient of variation for the stock

“Microsoft” is 9.028 which is the highest value as compared to that of other stocks. Thus, the

most volatile stock is the Microsoft as it has highest coefficient of variation.

(b) Beta coefficient:

The beta coefficient can be calculated by the following formula:

Beta =

Covariance(Stock Price, S&P Index )Variance(S&P Index )

Using Excel tool, the values of beta for the given 8 stocks have been calculated and are shown

below:

Beta (for Microsoft) =

Covariance(Stock Price, S&P Index )Variance(S&P Index ) =

0 . 00310 .000674 = 0.4584

Similarly, the beta for the remaining stocks has been calculated and is presented below:

Covariance Variance Beta

Microsoft 0.00031 0.4584

Less Volatile

Exxon Mobil 0.00049 0.7309

Less Volatile

Caterpillar 0.00101 1.4932 > 1

More Volatile

Johnson & Johnson 0.00001

0.0087Less Volatile

McDonald's 0.00101 1.5032 > 1

More Volatile

Sandisk 0.00176 2.6049 > 1

More Volatile

Qualcomm 0.00095 1.4138 > 1

More Volatile

Procter & Gamble 0.00034

0.5065Less Volatile

S&P 500 0.0006740

Going by the above table, we can see that the beta value for the stocks Caterpillar,

McDonald’s, Sandisk and Qualcomm are greater than 1 which indicates that those stocks are

Most Volatile.

(c) Comment on how much of the return for the individual stocks is explained by the market.

In order to calculate the amount of return for the individual stocks that is explained by the

market (S&P 500), we need to calculate the R-square value. Using Excel function, the

correlation coefficient has been calculated from which the R-square (square of correlation)

values for all the stocks have been calculated and are presented below:

S&P 500 r r^2 R^2 in %Microsoft 0.2660 0.070754 7.0754%

Exxon Mobil 0.3477 0.120929 12.0929%Caterpillar 0.5734 0.328805 32.8805%Johnson &

Johnson 0.0066 0.000044 0.0044%McDonald's 0.5812 0.337813 33.7813%

Sandisk 0.3510 0.123206 12.3206%Qualcomm 0.4319 0.186552 18.6552%

Procter & Gamble 0.3598 0.129460 12.9460%


R-square value for the pair “Microsoft and S&P 5000” = 7.08%

The above R-square value indicates that there is approximately 7.08% of the total variation in the

dependent variable “Microsoft” is explained by the market (S&P 500).

That is, there is approximately 7.08% of the return for the stock “Microsoft” is explained by the

market (S&P 500).

Similarly,

R-square value for the pair “Exxon Mobil and S&P 5000” = 12.09%

That is, there is approximately 12.09% of the return for the stock “Exxon Mobil” is explained by

the market (S&P 500).

R-square value for the pair “Caterpillar and S&P 5000” = 32.88%

That is, there is approximately 32.88% of the return for the stock “Caterpillar” is explained by


R-square value for the pair “Johnson & Johnson and S&P 5000” = 0.0044%

That is, there is approximately 0.004% of the return for the stock “Johnson & Johnson” is

explained by the market (S&P 500).

R-square value for the pair “McDonald's and S&P 5000” = 33.78%

That is, there is approximately 33.78% of the return for the stock “McDonald's” is explained by


R-square value for the pair “Sandisk and S&P 5000” = 12.32%

That is, there is approximately 12.32% of the return for the stock “Sandisk” is explained by the

market (S&P 500).

R-square value for the pair “Qualcomm and S&P 5000” = 18.66%

That is, there is approximately 18.66% of the return for the stock “Qualcomm” is explained by


R-square value for the pair “Procter & Gamble and S&P 5000” = 12.95%

That is, there is approximately 12.95% of the return for the stock “Procter & Gamble” is

explained by the market (S&P 500).

//---------------------------------//

Documents

Forecasting One Mark