Example No.11

 

Enter data:  

Glucose in blood serum (xi) : 2.5, 2.6, 2.8, 2.9, 3.0, 3.1, 3.2, 3.5, 3.9, 4.2, 4.5 [mmol.l-1]

Urine keto-substances (yi): 0, 0, 2, 1, 2, 3, 2, 4, 2, 4, 3 [units]

 

 

1) Calculation of mean values:

Results:  Glucose in blood serum: 3.2909 mmol.l-1

                Urine keto-substances: 2.0909  units

 

2) Calculation of coefficients in equation of linear regression (y=bx + a):

Result: Linear regression y = 1.52667x - 2.93322

 

 

3) Calculation of the correlation coefficient r:

 

 

4) Calculation of significance of the correlation coefficient:

                            DF = n-2 = 9

 

As calculated t > tcrit= 2.262 (DF=9, α=0.05) the correlation coefficient is statistically significant. 

 

Conclusion: between glucose level and keto-substances of urine the relation was found that is described by means of linear regression y=1.52667x - 2.93322 and correlation coefficient 0.7305 that is statistically significant (p<0.05).

Note:

Both variables monitored in this example are equally valuable, then its relation is reciprocal (correlative relation) and it is possible to evaluate it also by means of the method, based on the mutual exchange of data sets (X« Y).

 

 

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