Driver salary identification by hypothesis testing

This paper outlines comparing two independent samples, particularly average driver salary in Slovakia and the other EU countries. Significant differences between salaries are pointed out through two examples dealing with hypothesis testing. For Western European Union countries, the level of salaries in road transport is higher in absolute terms compared to the salaries of drivers in the central and eastern parts of the European Union. Different salaries and requirements of wage regulations create a discriminatory environment among entrepreneurs however harmonization of salaries in the field of road transport is specific because carriers offer transport throughout the whole European Union market.


Introduction
In practice, we often need to compare the basic characteristics of a random variable in two basic sets based on the knowledge of the sample data only. For example, we want to compare average salaries of drivers and their variability in individual countries or the annual tax on motor vehicles and the like. If the comparison is based on the results of the sample research, the statistical induction methods must be reused for the interval estimates of the difference or share of the comparison parameters of the basic files or for verifying the hypotheses of their compliance.

Comparing Two Independent Samples
In the two-parameter match tests, independent choices is considered in which the collection of statistical units from a single base file does not depend on removing units from the second base file. Based on independent selections, the hypotheses of matching the parameters of two basic sets are verified, using selection characteristics with known probability distributions. These distributions often depend on whether the variance in the basic files is the same or not.

Testing a Difference of Two Variances
The random variable X is monitored in two basic files. Its probability distribution is normal for both files. It is assumed that in both the basic files, the random variable X has the same variability, which is expressed by the hypothesis : ( 1 ) The test characteristic is used as the test criterion for its test. If the null hypothesis is true, the test is ( 2 ) and Fisher's probability distribution F(n 1 -1; n 2 -1) which has number of degrees of freedom v 1 = n 1 -1 and v 2 = n 2 -1. An overview of critical regions for various alternative hypotheses is given in Table 1 [5], [6].

Testing a Difference between Two Means -Normal Distribution
If we know the variances σ 1 2 , σ 2 2 of these distributions and test the null hypothesis : 26 Poliak (et al.): Driver Salary Identification by Hypothesis Testing against any alternative, as a test criterion we can use a characteristic assuming the null hypothesis is valid (4) and normal probability distribution N (0; 1). Selected level of significance α and critical regions for different types of alternative hypotheses are given in Table 2. If we do not know the variances σ 1 2 , σ 2 2 but we can assume its equality σ 1 2 = σ 2 2 , the null hypothesis H 1 : μ 1 =μ 2 is tested against any alternative by the test characteristic, which is in the case of null hypothesis ( 5 ) and distribution t(n 1 +n 2 -2).
Critical values are determined at the chosen level of significance α as the appropriate fractil of Student's distribution t (n 1 + n 2 -2) according to Table 3 [5], [6]. If we do not know the variances σ 1 2 , σ 2 2 and we cannot assume its equality, a suitable test criterion is the characteristic which, in the case of the null hypothesis validity and in the case of validity of the null hypothesis μ 1 =μ 2 is ( 6 ) and the division t(v r ).
Critical regions in various alternative hypotheses are given in Table 3 with the difference that we are looking for the fractile of t-distribution at degrees of freedom v r .

Testing a Difference between Two Means -Other or Unknown Distribution
The assumptions about the consistency of the mean values of the two base files in which the random variable X does not have normal but any distribution or unknown probability distribution is tested if the sample files have a large range, n 1 →+∞, n 2 →+∞. The formula for the test statistic in this case is ( 7 ) and normal probability distribution N(0;1).
The normality of its distribution is a consequence of the validity of the central limiting theorem and the use of selective variances instead of the variances of the base sets is due to the consistency of point estimates [9], [10].

Testing a Difference between Two Means -Alternative Distribution
This is a special case of the average value match test of two arbitrary distributions. If X has a distribution A (π1) in the first one and A (π2) in the second base file, the formula for null hypothesis about the consistency of mean values is : ( 8 ) An appropriate test criterion in its test against any alternative is the characteristic, which is simplified in the case of the validity of the null hypothesis (9) Assuming n_1→∞, n_2→∞, the Z -test has normalized normal distribution and the assumption or rejection of the null hypothesis H 0 : π 1 =π 2 is decided based on Table 2 [12], [13].

Driver Salary Identification
Regarding the remuneration of drivers, we conducted a research of drivers' compensation in selected transport companies. The research was conducted from January to March 2017 and the drivers of transport companies from 10 different EU states participated on it. Overall, drivers from 310 transport companies took part in the research. The countries were chosen so that in the research were the states which uphold the introduction of a minimum salary for the work of international road transport drivers (Germany and Austria), Western European countries which incline to a certain protection of the transport market (Belgium, Denmark, the Netherlands, Luxembourg, Italy) and the countries of the central and eastern part of the European Union which are against the unification of the minimum salary across the EU (Czech Republic, Poland, Slovakia).
In the research, the drivers responded to the form of remuneration and the amount of remuneration which appertain to them. The drivers were divided into groups according to whether they are remunerated with a monthly or hourly rate. On average, the highest hourly salary rate is reached by drivers in Luxembourg, it is 15.875 € / hour (Table 4). By the rates above € 10 / hour are also remunerated drivers in Germany and Austria. Significantly lower rates are in the Czech Republic and Slovakia, where they are below 5 € / hour. In several countries, drivers are not remunerated with € / hour, but with monthly salaries. The highest average salary for drivers is in Luxembourg, Germany and the Netherlands and it is above 2000 € / month. At the level of significance α = 0.05, we assume that average salaries are the same in both countries. We want to verify the hypothesis H 0 : μ 1 = μ 2 versus the alternative H 1 : μ 1 ≠ μ 2 . The sample size is large with an unknown variances σ 1 2 , σ 2 2 . In order to be able to choose a test criterion correctly, we must first verify the hypothesis H 0 : σ 1 2 = σ 2 2 against the alternative H 1 : σ 1  Then we determine critical values of F-distribution with degrees of freedom α = 0.05: The value of the test criterion does not belong to the critical region (Fig. 1.) 0, ∪ , ∞ In Excel, using the TINV function, we find the 0.05 critical values for a T-distribution with 145 degrees of freedom to be −1.98 and 1.98. So to accept the null hypothesis, we need out test statistic to satisfy t < 1.98 [12], [13].

Figure 2. Distribution Plot -t-test
The value of the test criterion does not belong to the critical area (Figure 2.), therefore we assume the null hypothesis H 0 : μ 1 = μ 2 for the significance level α = 0.05. At the level of significance α = 0.05 we assume that in the SR and the Czech Republic the average salary of drivers is not significantly different.

Example 2 -Salary Comparison of Slovak driver and drivers from the other EU countries
From research we know the average salaries of drivers in SR and the other EU countries. In the SR we determined the average salaries of drivers with random sample of size 116 (the same salaries as example 1 are considered).
In selected EU countries (Belgium, Denmark, Luxembourg, Netherlands, Norway, Poland, Austria, and Italy) we determined the following average salaries of drivers with random sample of size 60 (all entries are in euro): At the level of significance α = 0.05, we assume that average Slovak salaries are the same as salaries in selected EU countries.
We want to verify the hypothesis H0: μ 1 = μ 2 versus the alternative H1: μ 1 ≠ μ 2 . The sample size is large with an un-known variances σ 1   The value of the test criterion does not belong to the critical region (Fig. 3.)

3.4595
In Excel, using the TINV function, we find α 0.05 critical values for a T-distribution with 174 degrees of freedom to be −1.97 and 1.97 [12], [13].
Since t = -3.4595 is inside of our critical region (Fig. 4), we accept the null hypothesis and we can conclude at α 0.05 significance level that that the average salary of the driver in the Slovak Republic and in the other analysed EU countries is significantly different. Since t = -3.4595 is inside of our critical region (Fig. 4.), we accept the null hypothesis and we can conclude at α 0.05 significance level that that the average salary of the driver in the Slovak Republic and in the other analysed EU countries is significantly different.

Conclusions
On the basis of testing hypotheses, we can conclude that there are differences between driver salaries in the selected EU countries. Testing has proven that in the Slovak Republic and the Czech Republic the average salary of drivers is not significantly different at the level of significance α 0.05 but on the other hand there are significant differences between the average salary of the driver in the Slovak Republic and in the other analysed countries, mostly Western European Union countries. For Western European Union countries, the level of salaries in road transport is higher in absolute terms compared to the salaries of drivers in the central and eastern parts of the European Union. Therefore, it can be expected that in the western part of the EU drivers are less interested in this profession, and companies are often looking for workers from the eastern part of the EU or from outside the EU. The harmonization of salaries in the field of road transport is specific because carriers offer transport throughout the whole European Union market. Every carrier established in the European Union and with a license issued by the Community has the same access to this market. Individual states put pressure on increase of salaries in international road transport with the use of national regulations in order to reduce the competitive pressure of lower prices of carriers located in the central and eastern parts of the European Union.