ACCA F2 Management Accounting - 2010 - Study text

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Chapter 4: Business mathematics and computer spreadsheets

Errors (= outcome) Probability

x p px

0 0.7 0.0

1 0.2 0.2

2 0.1 0.2

Expected value

0.4

The expected value is 0.4 errors per entry, so if there are 10,000 entries, the expected

total number of errors per month is 10,000 × 0.4 = 400 errors.

1.3 How are expected values used?

Expected values are used to make forecasts or estimates, and to help management

with decision-making.

Expected values and decision-making

A decision is choice between two or more alternatives. Companies are often faced

with making decisions when the outcome is uncertain. There are two types of

decision where EVs might be used.

 Making a choice between doing something and not doing it. For example,

should we invest in this new equipment? The choice is between investing and

not investing.

 Making a choice between two or more different possible options. For example,

when a company has decided that it needs to replace an item of equipment,

there might be a choice between buying a smaller model or a larger model.

The following rules apply to the use of expected values for decision-making, when

there is risk or uncertainty and probabilities of different possible outcomes are

estimated.

1 When the choice is between doing something and not doing it.

- If the expected value of profit is negative, the decision should be NO (do

not do it).

- If the expected value of profit is positive, the decision should be YES (do

it).

2 If there are two or more options and each has a positive expected value of

profit, the option with the highest expected value of profit should be chosen in

preference to the others.

Example

The management accountant of a company has estimated that if the company tries to

sell a new product, Product X, there is a 60% probability that the product will make

an annual profit of $25,000 and a 40% probability that it will make a loss of $15,000.

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The EV of annual profit from the product is:

Profit ($) Probability

x p px

$ $

25,000 0.60 15,000

(15,000) 0.40 (6,000)

Expected value of annual profit

9,000

The expected value of profit is $9,000 per year. If the company makes its decisions

on the basis of expected value, it will go ahead and try to sell the new product.

Example

A company is considering investing in one of two projects, A and B. It cannot invest

in both of them. The profits and associated probabilities of each project are as

follows.

Project A

Project B

Monthly

profit

Probability

Monthly

profit

Probability

$ $

15,000 0.1 (18,000) 0.05

25,000 0.2 16,000 0.30

65,000 0.4 80,000 0.35

75,000 0.3 95,000 0.30

Required

Using the expected value technique, decide which project should be undertaken and

explain why.

Answer

The choice is between Option A and Option B. Using expected values, the option

with the higher expected value of profit should be chosen. If the expected value for

oth options shows a loss, neither project should be undertaken.

Option A Option B

Profit per

month

Probability Profit per

month

Probability

x p px x p px

$ $ $ $

15,000 0.1 1,500

(18,000)

0.05 (900)

25,000 0.2 5,000

16,000

0.30 4,800

65,000 0.4 26,000

80,000

0.35 28,000

75,000 0.3 22,500

95,000

0.30 28,500

Expected value of profit

55,000

EV of unit cost 60,400

Chapter 4: Business mathematics and computer spreadsheets

If EV is used to make the decision, Project B should be undertaken because it has a

higher EV of profit.

However, Option B is more risky, because there is some chance (a 5% chance) of

making a loss.

Note that the expected value is not equal to any of the possible monthly profit

figures. The EV is a weighted average value, not a value that is actually expected to

occur.

Exercise 1

There is a 70% chance that a company will make a profit of $800,000 next year and a

30% chance that it will make a loss of $150,000.

Required

Calculate the expected profit or loss for next year.

Exercise 2

A company that makes glassware products is considering a proposal to make and

sell a batch of 10,000 special drinking glasses, to commemorate the 100

anniversary

of the country’s independence.

The sales director has estimated that there is a 60% probability that the glasses will

sell for $5 each, and a 10% probability that a sales price of $6 can be achieved.

However, there is a 30% probability that the selling price will be just $4.

The management accountant has estimated that for a batch of 10,000 units, there is a

75% probability that the unit cost will be $3.60 and a 35% probability that it will be

$4.20.

Required

(a) Calculate the expected value of he sales price.

(b) Calculate the expected value of the unit cost.

recommend whether the company should make and sell the batch of glasses.

1.4 Limitations of the expected value method for decision-making

There are a number of limitations associated with using the expected value

technique in decision-making situations.

 The probabilities used in calculating expected values are estimates. If these

estimated probabilities are not reliable, then the expected values calculated from

the probabilities will also be unreliable.

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 An expected value might not be an ‘actual value’ of a possible outcome. For

example if there is a 0.50 probability that the cost will be $2,000 and a 0.50

probability that the cost will be $5,000, the EV of cost is $3,500. However a cost of

$3,500 will not happen: the cost will be either $2,000 or $5,000.

 Expected values don’t normally take into account the attitude to risk of the

person taking the decision. This will obviously vary from person to person. For

example, a project might have an EV of profit of $50,000, but there might be a

20% probability of making a loss of $20,000. The company might be unwilling to

take the risk of a $20,000 loss, even though the EV shows a profit.

Example

A company is considering investing in one of two projects, X and Y. It cannot invest

in both of them. The profits and associated probabilities of each project are as

follows.

Probability

Project X

profit/(loss)

Project Y

profit/(loss)

$ $

0.6 15,000 40,000

0.4 (10,000) (20,000)

Project X has an EV of profit of $5,000 and Project Y has an EV of profit of $16,000.

(Workings are not shown but you should be able to calculate these expected values

yourself.) Applying the normal rule, both projects would be worth undertaking.

However, since only one project can be selected, Project Y should be preferred

because it has a higher EV of profit.

The limitations of expected values should be apparent.

 Although both projects have an EV of profit, they each have a 40% probability of

making a loss.

 Although Project Y has a higher EV of profit, the risk is greater, because the loss

might be as high a $20,000, whereas with Project X it will not exceed $10,000.

It is important to remember that an expected value is a weighted average of the

different expected results. Like all averages it is only ‘accurate’ when it is used for a

large number of repetitive events or outcomes, where the actual results are likely to

be close to the weighted average. For ‘one off’ events, expected values should not be

used without giving careful consideration to the possible variations in the results

that might occur, and the risk involved.

Chapter 4: Business mathematics and computer spreadsheets

Regression analysis

 The purpose of linear regression analysis

 The linear regression formulae

 Applying the linear regression formulae

2 Regression analysis

2.1 The purpose of linear regression analysis

Linear regression analysis is a statistical technique for calculating a line of best fit

from a set of data:

y = a + bx

The data is in ‘pairs’, which means that there are a number of different values for x,

and for each value of x there is an associated value of y in the data.

Linear regression analysis can be used to estimate fixed costs and the variable cost

per unit from historical data for mixed costs. For example, there might be historical

data for the total monthly costs of operating a workshop. It might be assumed that

total costs each month vary with the number of direct labour hours worked.

Historical data can be collected about total monthly costs in the past (cost in each

month = y) and the associated number of direct labour hours worked in that month

(hours in each month = x). This data can then be used to estimate fixed costs per

month and variable costs per direct labour hour, which can be expressed in a

formula y = a + bx.

Linear regression analysis can also be used to predict future sales by projecting the

historical sales trend into the future (on the assumption that sales growth is rising at

a constant rate, in a ‘straight line’).

Regression analysis is an alternative technique to high-low analysis, which was

described in the previous chapter. It can be used for all the same purposes in cost

and management accounting that high-low analysis is used for.

Regression analysis and high-low analysis compared

Regression analysis and the high-low analysis are used for the same purposes.

Regression analysis is therefore an alternative technique to high-low analysis, which

was described in the previous chapter, but there are important differences between

them.

 High-low analysis uses just two sets of data for x and y, the highest value for x

and the lowest value for x. Regression analysis uses as many sets of data for x

and y as are available.

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 Because regression analysis calculates a line of best fit for all the available data, it

is likely to provide a more reliable estimate than high-low analysis for the values

of a and b.

 In addition, regression analysis can be used to assess the extent to which values

of y depend on values of x. For example, if a line of best fit is calculated that

estimates total costs for any volume of production, we can also calculate the

extent to which total costs do seem to be linked (or ‘correlated’) to the volume of

production. This is done by calculating a correlation co-efficient, which is

explained later.

 Regression analysis uses more complex arithmetic than high-low analysis, and a

calculator or small spreadsheet model is normally needed

In summary, linear regression analysis is a better technique than high-low analysis

because:

 it is more reliable and

 its reliability can be measured.

High-low analysis has the benefit of simplicity and speed/ease of calculation.

2.2 The regression analysis formulae

The formulae for estimating costs using regression analysis are provided in a

formulae sheet in your examination. You might be required to apply the formulae to

data provided in an examination question.

Linear regression analysis is used to calculate values for a and b in the linear cost

equation: y = a + bx.

The linear regression formulae for calculating a and b are shown below. The

number of pairs of data that are used in the calculation is n.

The regression analysis formulae are as follows. They will be given to you in your

examination:

a =

∑

−

∑

b =

∑

xy –

∑

n ∑ x

–(∑ x)

where:

n = the number of pairs of data used for x and y

x, y represent the values of x (volume of activity or output) and y (total cost)

The value of ‘b’ is an item in the formula for calculating ‘a’. It is therefore necessary

to calculate ‘b’ first before calculating ‘a’.

Understanding the formulae

These formulae might seem very complicated, and in your examination you might

be given a question that requires you to show your understanding of what the

different items in the formulae mean.

Chapter 4: Business mathematics and computer spreadsheets

 You should think of the pairs of data as a list of values set out in two columns,

with one column for the values of x and the second column for the associated

values of y.

 You should then prepare two more columns.

 There should be an additional column for x². Calculate the square of each value

of x in the x column and write the answer in the x² column.

 The fourth column is for xy. For each pair of data for x and y, multiply the value

of x by the value of y, and enter the answer in the xy column.

 You must then add up the totals for each of the four columns. ∑x is the total of

all the values in the x column. ∑y is the total of all the values in the y

column.∑x² is the total of all the values in the x² column and ∑xy is the total of

all the values in the xy column.

 You now have all the values you need to calculate a value for b in the formula y

= a + bx.

 The only other item you might not be sure about is (∑x)². This is not the same as

∑x². It is the square of the value for ∑x.

 Having calculated a value for b, you can then use the formula for calculating the

value of a: this includes the value for b, which is why you need to calculate b

first.

2.3 Applying the regression analysis formulae

The formulae might seem complicated, but they are actually fairly straightforward

provided that you understand the meaning of the ∑ items in the formulae, and

remember that n is the number of pairs of data used to make the estimate. Study the

following example carefully.

Example

The total costs and output volumes of a manufacturing company in the first six

months of the year have been as follows.

Month Output Total cost

000 units $000

January 5 146

February 7 152

March 6 148

April 5 142

May 8 164

June 6 151

Required

Using regression analysis:

 estimate a value for fixed costs each month and a variable cost per unit

 estimate the expected costs in September if output volume is expected to be

8,000 units.

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Answer

The starting point is to draw a table showing the values of x (output) and y (total

cost). For each value of x and y, you should calculate the value of:

 x

and

 xy.

For reasons to be explained later, the value of y

will also be calculated here.

There are six pairs of data, so there should be six different values for x, y, x

, xy and y

Add the figures in each column to obtain totals for:

 ∑x (= the sum of the values of x)

 ∑y (= the sum of the values of y)

 ∑x

 ∑xy

 ∑y

. (The reason for calculating ∑y

will be explained later.)

Month Output

Total cost

000 units

$000

xy y

January 5

146

730 21,316

February 7

152

1,064 23,104

March 6

148

888 21,904

April 5

142

710 20,164

May 8

164

1,312 26,896

June 6

151

906 22,801

903

235

5,610 136,185

∑x

= ∑y

= ∑x

= ∑xy

= ∑y

There are six pairs of data, so n = 6.

We now have all the figures we need to calculate values for ‘a’ and ‘b, starting with ‘b’.

b =

n ∑ xy – ∑ x∑ y

n ∑x

– ∑x

()

b =

6 (5,610)–(37)(903)

6 (235)–(37)

33,660 – 33,411

1,410 – 1,369

249

b = 6.07

This is the estimate of the variable cost per unit. The value of ‘b’ is used to calculate

a value for ‘a’, as follows.

Chapter 4: Business mathematics and computer spreadsheets

a =

∑ y

–

∑

a =

903

–

6.07(37)

= 150.5 – 37.4

a (in $000) = 113.1

Answer (a)

Estimated fixed costs each month = a = $113,100

Estimated variable cost per unit = b = $6.07

Therefore, linear cost function, y, is estimated to be:

y = 113,100 + 6.07x

For simplicity, this could be rounded to:

y = 113,000 + 6x.

Answer (b)

Using y = 113,100 + 6.07x, when monthly output is expected to be 8,000 units, the

expected total costs will be:

Fixed 113,100

Variable (8,000 × $6.07) 48,560

161,660

Linear regression analysis and your examination

As you might appreciate, calculating a linear equation using linear regression

analysis is a time-consuming process, unless you have a programmed calculator or

spreadsheet. You will therefore not be required to use the full technique in your

examination. However, you might be required to:

 comment on the meaning of items in the formulae, such as ∑x², ∑xy and (∑x)²,

 calculate a value for a or b, given values in the question for ∑x, ∑y, ∑x²

and∑xy.

Paper F2: Management accounting

Correlation and the correlation coefficient

 Correlation

 Degrees of correlation

 Correlation coefficient, r

 Coefficient of determination, r

3 Correlation and the correlation coefficient

3.1 Correlation

Linear regression analysis can be applied to any sets of data where the data is in

pairs (x and y). It could be used, for example, to calculate a line of best fit for total

weekly costs of an activity by taking pairs of data for total weekly costs (= y) and the

associated values for the activity. However, the activity level might be any of the

following:

 units of output each week, if it is assumed that weekly costs vary with the

volume of output produced

 direct labour hours each week, if it is assumed that weekly costs vary with the

number of direct labour hours worked

 machine hours operated each week, if it is assumed that weekly costs vary with

the number of machine hours.

Which activity is the best one to choose as ‘x’?

You could even use linear regression analysis to calculate a line of best fit between

total weekly costs and the air temperature on the factory floor at midday, if you

decided that total costs vary with the air temperature. (This might seem a ridiculous

idea, but the point that you should try to understand is that linear regression

analysis can be used to calculate a line of best fit for any two variables, given pairs

of data, even when there is no actual connection between them.)

Since a line of best fit can be calculated between any two variables using regression

analysis, important questions are:

 How can we assess the reliability of the line of best fit?’

 How do we know whether our choice of activity as ‘x’ is a good one?

The answer to both questions is that the reliability of the regression formula can be

assessed using a statistic called the coefficient of correlation.

Correlation is a measure of the strength of the relationship between two variables.

Variables are said to be correlated if a change in one variable results in a change in

the other variable.

ACCA F2 Management Accounting - 2010 - Study text - Emile Woolf Publishing

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