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 Assessment of the reliability of the relationship of selfesteem and professional orientation of students  psychologists

The task of correlation analysis is to establish the direction (positive or negative) and form (linear, nonlinear) of the relationship between the varying features, measure its tightness, and, finally, to check the significance level of the obtained correlation coefficients.
The criterion for selecting “sufficiently strong” correlations can be both the absolute value of the correlation coefficient itself (from 0.7 to 1), and the relative value of this coefficient, determined by the level of statistical significance (from 0.01 to 0.1), depending on sample size. In small samples, for further interpretation, it is more correct to select strong correlations based on the level of statistical significance. For studies conducted on large samples, it is better to use the absolute values of the correlation coefficients.
Since the diagnostic method of T. D. Dubovitskaya, aimed at identifying the level of professional orientation, involves the expression of the level of professional orientation in a direct scale (that is, the more points, the higher the level of professional orientation), and the methodology of B. I. Dodonov’s selfassessment involves the expression the level of selfesteem in the reverse scale (the higher the score, the lower selfesteem), then in the case of a statistically significant relationship between selfesteem and the level of professional orientation, the rank coefficient Spearman's correlation should be negative. We rank the experimental data obtained according to the methods of T. D. Dubovitskaya and B. I. Dodonov (Appendixes 2 and 4, respectively) in Table 4.
Table 4
Experimental data ranking
Check the correctness of the ranking. Sum of ranks is checked by the formula
In our case
which coincides with the total amount obtained in the table of 4 ranks and is a confirmation of the ranking.
We calculate the Spearman rank correlation coefficient according to the formula
In our case
In this case, the negative value of Spearman's rank correlation coefficient, as already mentioned, means the presence of a direct linear correlation between the examined parameters.
The obtained value of the correlation coefficient equal to 0.736 means that there is a strong (on the Chaddock scale) direct correlation relationship between selfesteem and the level of professional orientation.
To assess the statistical significance of Spearman's rank correlation coefficient, we verify for each series of ranked data the fulfillment of two conditions:
 the normality of each ranked distribution;
 equality of their variances.
Check the first condition.
We consider the first ranked row. We put forward the hypothesis H0:
 the frequencies of the ranks of the first row obey the normal distribution.
Alternative hypothesis H1:
 the frequencies of the ranks of the first row are different from the normal distribution.
We set the significance level? = 0.05. We calculate the criterion? 2 Pearson for the first row.
We compile the calculation table 5, in which we calculate the frequency of the ranks of the first row of the distribution and the rest of the data.
Table 5
Calculation of the criterion? 2 Pearson for the first row
The average value of the rank is determined using Excel, it is equal to
The standard deviation of the rank is determined using Excel, it is equal to
The normalized rank is calculated by the formula
The calculation results are presented in the fourth column of table 5.
? (ui) is the local Laplace function of the variable ui. Its values are tabulated [16]. Enter the corresponding values in the fifth column of table 5.
Theoretical frequencies are calculated according to the formula [16, C. 251]
Here h is the step between the ranks. The average step is h = 5. We enter the calculations in the column of the sixth table 5. We calculate the indicators of column 7, determine the sum of column 7, which is the observed value of the criterion? 2abl.
In our case,? 2set. = 10.95
According to the table of critical distribution points? 2 [16, P. 393] for a given level of significance? and the number of degrees of freedom k = s1r = 1112 = 8 (s is the number of partitioning groups, r is the number of estimated parameters, in our case we estimate two parameters of the normal distribution: the average value and the standard deviation) we find the critical point? 2kr (?; k):
? 2 cr (0.05; 8) = 15.5
Since? 2set. = 10.95 <? 2 cr (0.05; 8) = 15.5, there is no reason to reject the hypothesis of a normal distribution of the first series of data.
We consider the second ranked row. We put forward the hypothesis H0:
 the frequencies of the ranks of the second row obey the normal distribution.
Alternative hypothesis H1:
 the frequencies of the ranks of the second row are different from the normal distribution.
We set the significance level? = 0.05. We calculate the criterion? 2 Pearson for the second row.
We compile the calculation table 6, in which we calculate the frequency of the ranks of the second row of the distribution and the rest of the data.
In this case, we use the method of enlarging the intervals, due to the fact that many ranks occur once or twice (we combine such ranks into groups).
The average value of the rank is determined using Excel, it is equal to
The standard deviation of the rank is determined using Excel, it is equal to
Table 6
Calculation of the criterion? 2 Pearson for the second row
In our case,? 2set. = 6.21
According to the table of critical distribution points? 2 [16, P. 393] for a given level of significance? and the number of degrees of freedom k = s1r = 1012 = 7 we find the critical point? 2cr (?; k):
? 2 cr (0.05; 7) = 14.1
Since? 2set. = 6.21 <? 2 cr (0.05; 7) = 14.1, there is no reason to reject the hypothesis of a normal distribution of the second row of data.
We conclude that the requirement of normality of the series is provided.
Check the second condition.
We put forward the hypothesis H0:
 variances of the normal series of distribution of ranked data are equal.
Alternative hypothesis H1:
 variances of the normal series of distribution of ranked data are different.
We set the significance level? = 0.05. We calculate F  Fisher's criterion according to the formula
Since the root mean square deviations have already been calculated above, we calculate the observed value of F  the Fisher test
We determine the critical point of the Fisher distribution (k1 = s11; k2 = s21  the number of degrees of freedom)
Fcr (? / 2; k1; k2) = Fcr (0.25; 10; 9) = 3.13
Since Fnabl = 0.988
After checking these requirements, we evaluate the statistical significance of the Spearman rank correlation coefficient using the t  student criterion.
We put forward the hypothesis H0:
 Spearman's rank correlation coefficient is nonzero.
Alternative hypothesis H1:
 Spearman's rank correlation coefficient is zero.
We set the significance level? = 0.05.
We use the formula t  criterion
In our case
We determine the critical point t  Student's distribution (k1 = s11; k2 = s21  the number of degrees of freedom)
tcr (1 ?; n2) = tcr (0.95; 53) = 2.01
Since tabl = 7.91> tcr (0.95; 53) = 2.01, the null hypothesis is rejected in favor of the alternative. Consequently, the Spearman rank correlation coefficient is nonzero, that is, statistically significant. Thus, the obtained value of Spearman's rank correlation coefficient, which characterizes the relationship between selfesteem and the level of professional orientation as direct and strong, is reliable, and the revealed relationship is statistically significant.
The correlation correlation found at a rather high level of significance between the indicators of selfesteem and the level of professional orientation testifies to the confirmation of our hypothesis.  << Previous   Next >>  = Skip to textbook content = 
 Assessment of the reliability of the relationship of selfesteem and professional orientation of students  psychologists
  Graduate work. The relationship of selfesteem and professional orientation of student psychologists, 2012
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