# Developing a Model of Risk Factors of Injury in Track and Field Athletes
## 3 Results  
Forward stepwise selection was used to choose seven predictors for the model: previous injuries, sleep, age, blood, training load, atmospheric conditions, and competition. Table 2 shows that according to S and [PERSON] statistics, the other predictors should be removed from the model. This model will be referred to as M. Table 2 presents the last step of this method, confirming the choice of model M.  
Factors included in model M were classified into three groups introduced in Table 1:  
* age-demographics,
* sleep, blood, previous injuries, health/regeneration,
* training load, atmospheric conditions, training/competition. In our studies, the HL test value was 7.36, which produced a \(p\)-value equal to 0.5. The \(p\)-value was higher than the standard significance level of \(p\) = 0.05. This means there was no evidence to reject the hypothesis that the analyzed model of logistic regression was well fitted to the data. This was also confirmed by the high _AUC_ value of 0.88. Figure 3 presents the receiver operating characteristic (ROC) curve and area under the curve (AUC) value. Figure 2: Logistic regression—the pattern of approximate relationship between a dependent variable (y) and an independent (x) variables. The odds ratio for model M calculated from the classification matrix (Table 3) was 19.53. It was much higher than one and indicated that the classification was much better than that expected \"at random\". \begin{table}
\begin{tabular}{c c c c c c c} \hline \hline  & \multicolumn{5}{c}{**Model Creation**} \\
**Step 9** & \multicolumn{2}{c}{**Distribution: BINOMAIL, LINK Function: LOGI**} & \multicolumn{2}{c}{**Model Creation**} \\  & \multicolumn{5}{c}{**TModeled Probability of Injure = 1**} & \\ \hline Effect & df & Wald Stat. & Wald p & S. pkt Stat. & S. pkt p & \\ Previous injuries & 3 & 40.66 & 0.00 & & In model \\ Age & 4 & 18.30 & 0.00 & & In model \\ Sleep & 1 & 10.27 & 0.00 & & In model \\ Blood & 1 & 2.55 & 0.11 & & In model \\ Competition & 3 & 4.71 & 0.19 & & In model \\ Training Load & 4 & 9.67 & 0.04 & & In model \\ At. cond & 3 & 9.08 & 0.04 & & In model \\ Obj. cond & 4 & & 0.27 & 0.99 & Out \\ Warm-up & 3 & & 0.99 & 0.80 & Out \\ Experience & 1 & & 0.42 & 0.51 & Out \\ BMI & 4 & & 5.17 & 0.27 & Out \\ Level of performance & 4 & & 2.21 & 0.57 & Out \\ Sex & 1 & & 0.00 & 0.96 & Out \\ Diet & 1 & & 2.21 & 0.14 & Out \\ Biomechanics & 1 & & 0.45 & 0.50 & Out \\ Supplementation & 3 & & 0.54 & 0.46 & Out \\ Regeneration & 2 & & 0.78 & 0.85 & Out \\ Physiotherapy & 3 & & 0.74 & 0.69 & Out \\ Wellness & 2 & & 2.43 & 0.49 & Out \\ Illness & 1 & & 1.92 & 0.16 & Out \\ Summer performance & 1 & & 0.00 & 0.98 & Out \\ Winter performance & 1 & & 0.59 & 0.44 & Out \\ Injuries & 4 & & 6.50 & 0.16 & Out \\ Drinks & 2 & & 1.76 & 0.44 & Out \\ \hline \hline \end{tabular}
\end{table} 
Table 2: Modeled probability of injury using binomial and link distribution and logit function. Figure 3: Receiver operating characteristic (ROC) curve for model M.  
Next, WoE coefficients for all model M variables were calculated. The strongest relationship was observed between the factor _prv.inj_ (number of previous injuries) and the risk of injury, reflected by the highest value of IV = 0.993. Analysis of the graph presented in Figure 4 and WoE coefficients showed that the risk of injury increased with the category. The higher the category, the greater the number of previous injuries and the lower the WoE coefficient. The lower the WoE coefficient, the higher the risk of injury. The WoE coefficient gradually decreased (Figure 4) across categories 1 and 2 (positive WoE, risk of injury is relatively small for elements included in these classes) to categories 3 and 4 (negative WoE, increase in the risk of injury). The training period was the second predictor creating the strongest relationship with the risk of injury, for which the IV value was 0.326. The IV statistic of _age_ was frequently related to the training period was only slightly lower (IV = 0.282).