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 Ttest. 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:
(N_{1}1) S²_{1} +(N_{2}1)
S²_{2}
N_{1}+N_{2}2
S² = .......equation 4.1
? (X1iA_{1}) ²+? (X2iA_{2}) ²
N_{1}+N_{2}2
= ....equation 4.2
Numerically, we have:
S²=[(51)*33.04²
+(51)*26.74²]/(5+51)............equation 4.3
S²=802.96
The difference in the two means is expressed as follows:
ì_{1}ì_{2 }=474325
ì_{1}ì_{2 }=149F
The difference in the two means is therefore normally distributed
with the following parameters:
A_{1}A_{2}~N [149, 802.96]
At this level, it is now possible to test the hypotheses, which
were stated as follows:
H_{o}: (Null hypothesis)
A_{1}A_{2}=0, there is no difference between the sample means
and both can be used for the same purposes.
H_{1}: (Alternative Hypothesis)
A_{1}A_{2}?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+n22 }to be calculated,
The decision will be thus depending on the results
obtained:
Reject Ho if, Tc>T_{á/2, n1+n22} or
Tc< T_{á/2, n1+n22}
In this case, we have a Twotailed test,
The level
of significance á=0.05
First sample size N1=5
Second sample size N2=5
The Degree of freedom F=N1+N22
F=5+52=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:
A1A2
Tc= ..........equation4.4
S (1/N1 +1/N2)^{1/2}
But S is the square root of S², S=802.96^{1/2}
474325
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+n22. }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.
