The use of job costing as a tool for the pricing and cost control decisions in the printing industry: the case of Société de Presse et d'Editions (SOPECAM)

VII. ANALYSIS OF THE DATA

In this part, our work will be focused on testing the hypotheses outlined in the first chapter; for this, we will use the student's T-test. This test is concerned with analysing the difference in means; in our case, the two means are the weekly unit cost obtained using absorption and marginal costing.

This test requires us to compute some elements of analysis such as:

· The pooled weighted variance S²

· The various standard deviations

· The difference in means

In the previous chapter, explanations have been given on the above concepts; we therefore do not need to dwell on it again. As such, we have the following:

(N₁-1) S²₁ +(N₂-1) S²₂

N₁+N₂-2

S² = .......equation 4.1

? (X1i-A₁) ²+? (X2i-A₂) ²

N₁+N₂-2

= ....equation 4.2

Numerically, we have:

S²=[(5-1)*33.04² +(5-1)*26.74²]/(5+5-1)............equation 4.3

S²=802.96

The difference in the two means is expressed as follows: ì₁-ì₂ =474-325

ì₁-ì₂ =149F

The difference in the two means is therefore normally distributed with the following parameters:

A₁-A₂~N [149, 802.96]

At this level, it is now possible to test the hypotheses, which were stated as follows:

H_o: (Null hypothesis) A₁-A₂=0, there is no difference between the sample means and both can be used for the same purposes.

H₁: (Alternative Hypothesis) A₁-A₂?0 the two sample means are different and cannot be used for the same purposes.

These hypotheses will be tested using the following rules:

Given the test statistic Tc and the critical value T_á/2,n1+n2-2 to be calculated,

The decision will be thus depending on the results obtained:

Reject Ho if, Tc>T_{á/2, n1+n2-2} or Tc<- T_{á/2, n1+n2-2}

In this case, we have a Two-tailed test,

The level of significance á=0.05

First sample size N1=5

Second sample size N2=5

The Degree of freedom F=N1+N2-2

F=5+5-2=8

From the statistical tables, we therefore have: t_{á/2 ,8} =2.306 or : t_{á/2 ,8=} -2.306 A1=474

A2=314

The test statistic is therefore equal to:

A1-A2

Tc= ..........equation4.4

S (1/N1 +1/N2)^1/2

But S is the square root of S², S=802.96^1/2

474-325

Tc= ..............equation 4.5

28.34 (1/5 +1/5)^1/2

Tc=8.312

The critical value previously obtained was 2.306, which is less than the test statistic obtained: Tc>T_{á/2, n1+n2-2.} This leads us to reject the null hypothesis stating that: there is no difference between the sample means and both can be used for the same purposes. And we accept the alternative hypothesis stating that: the two sample means are different and cannot be used for the same purposes.

This conclusion ends our presentation and analysis of the data collected at SOPECAM, and helps us to directly enter into the problem of the lessons and conclusion to draw from this study, which is the main concern of the next chapter.