Pearson's product-moment correlation
data: data$relative_strength_imtp and data$dragon_score
t = 3.5996, df = 83, p-value = 0.000541
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
0.1674558 0.5384226
sample estimates:
cor
0.3674655 General Linear Models
Does strength correlate with movement skill?
Competency in fundamental movement skills (FMS) are seen as the “building blocks” for lifelong physical activity and long-term athlete development (LTAD).
Strength, fitness and perceived movement skills competence theorized to be key moderators or mediates of that relationship.
Hypotheses
We collected strength (IMTP peak and relative peak force & CMJ height) and movement skill Dragon Challenge) for 85 youth athletes (10 - 15 years of age, 59 female):
- Strength would be associated with movement skill (Pearson’s correlations with lower 95% CI greater than 0.1)
- Sex and maturation may be important confounding variables in the analysis.
Code
A simple Pearson’s correlation will provide the strength of the relationship between predictor (strength) and outcome movement skill:
Plotting
Linear model
We can set up the same correlation using the linear model function.
Call:
lm(formula = dragon_score ~ relative_strength_imtp, data = data)
Residuals:
Min 1Q Median 3Q Max
-7.5601 -2.2497 -0.3844 2.5778 6.2955
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 39.22863 2.05312 19.11 < 2e-16 ***
relative_strength_imtp 0.26940 0.07484 3.60 0.000541 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 3.125 on 83 degrees of freedom
Multiple R-squared: 0.135, Adjusted R-squared: 0.1246
F-statistic: 12.96 on 1 and 83 DF, p-value: 0.000541Sex as co-variate
- We can then plot and see if a factor such as “sex” effects the relationship
ggplot(data) +
aes(
x = relative_strength_imtp,
y = time,
colour = sex
) +
geom_point(size = 2L) +
geom_smooth(method = "lm", se = TRUE) +
scale_color_manual(values = c(f = "#40004B", m = "#00441B")) +
labs(
x = "Relative Isometric Strength (N/kg)",
y = "Dragon Challenge Time (s)"
) +
theme_classic() +
theme(
axis.title.y = element_text(size = 13L,
face = "bold"),
axis.title.x = element_text(size = 13L,
face = "bold"),
axis.text.y = element_text(size = 13L),
axis.text.x = element_text(size = 13L),
legend.text = element_text(size = 14L),
legend.title = element_text(size = 14L)
) 
Sex as co-variate
Visually boys are scoring higher than girls but the slopes look the same?
Slopes
We can model this:
Call:
lm(formula = dragon_score ~ relative_strength_imtp + sex, data = data)
Residuals:
Min 1Q Median 3Q Max
-7.5211 -1.7629 -0.4793 2.4677 6.1480
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 38.26778 2.20441 17.360 < 2e-16 ***
relative_strength_imtp 0.31598 0.08447 3.741 0.000339 ***
sexm -0.97918 0.83032 -1.179 0.241698
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 3.118 on 82 degrees of freedom
Multiple R-squared: 0.1495, Adjusted R-squared: 0.1287
F-statistic: 7.204 on 2 and 82 DF, p-value: 0.001311Checking models
We can check models using the performance packages
# Comparison of Model Performance Indices
Name | Model | R2 | R2 (adj.) | RMSE | Sigma | AIC weights
-------------------------------------------------------------------
lm_addsex | lm | 0.149 | 0.129 | 3.062 | 3.118 | 0.429
lm_base | lm | 0.135 | 0.125 | 3.088 | 3.125 | 0.571
Name | AICc weights | BIC weights | Performance-Score
----------------------------------------------------------
lm_addsex | 0.404 | 0.181 | 57.14%
lm_base | 0.596 | 0.819 | 42.86%ANOVA
Let’s look at absolute strength
Simple Pearson’s correlation?
Pearson's product-moment correlation
data: data$peak_force_imtp and data$dragon_score
t = 0.97674, df = 83, p-value = 0.3315
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
-0.109000 0.312622
sample estimates:
cor
0.1066005 But let’s plot it by sex
Notice how we have opposing slopes.
Run models
Adding sex (keeping slopes fixed)
Adding force x sex interaction (modeling different slopes for boys and girls)
Model performance
# Comparison of Model Performance Indices
Name | Model | R2 | R2 (adj.) | RMSE | Sigma | AIC weights
-------------------------------------------------------------------------
lm_interaction | lm | 0.075 | 0.041 | 3.194 | 3.272 | 0.623
lm_base | lm | 0.011 | -5.476e-04 | 3.302 | 3.341 | 0.274
lm_addsex | lm | 0.012 | -0.012 | 3.301 | 3.361 | 0.102
Name | AICc weights | BIC weights | Performance-Score
---------------------------------------------------------------
lm_interaction | 0.574 | 0.151 | 87.12%
lm_base | 0.318 | 0.764 | 31.78%
lm_addsex | 0.107 | 0.084 | 0.17%Model ANOVA
Analysis of Variance Table
Model 1: dragon_score ~ peak_force_imtp + sex
Model 2: dragon_score ~ peak_force_imtp * sex
Res.Df RSS Df Sum of Sq F Pr(>F)
1 82 926.22
2 81 867.04 1 59.172 5.5279 0.02115 *
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Results from best model
Call:
lm(formula = dragon_score ~ peak_force_imtp * sex, data = data)
Residuals:
Min 1Q Median 3Q Max
-9.5935 -2.1603 0.0473 1.9788 6.0967
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 48.475926 1.921964 25.222 <2e-16 ***
peak_force_imtp -0.001702 0.001517 -1.122 0.2651
sexm -6.887493 3.118490 -2.209 0.0300 *
peak_force_imtp:sexm 0.004977 0.002117 2.351 0.0211 *
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 3.272 on 81 degrees of freedom
Multiple R-squared: 0.07488, Adjusted R-squared: 0.04062
F-statistic: 2.185 on 3 and 81 DF, p-value: 0.09607Plotting model outputs
First we generate a data set from the model
Plotting model outputs
# Predicted values of dragon_score
sex: f
peak_force_imtp | Predicted | 95% CI
------------------------------------------
700 | 47.28 | 45.46, 49.11
750 | 47.20 | 45.51, 48.89
800 | 47.11 | 45.55, 48.68
sex: m
peak_force_imtp | Predicted | 95% CI
------------------------------------------
700 | 43.88 | 40.93, 46.83
750 | 44.04 | 41.23, 46.86
800 | 44.21 | 41.52, 46.90Plotting model outputs
Then we can use ggplot:
plot <- ggplot(eff, aes(x = x, y = predicted, colour = group, fill = group)) +
geom_line(size = 1) +
geom_ribbon(aes(ymin = conf.low, ymax = conf.high), alpha = 0.2, colour = NA) +
labs(
x = "IMTP Relative Peak Force (N/kg)",
y = "Dragon Challenge Score (AU)",
colour = "Sex",
fill = "Sex"
) +
theme_classic()Plotting model outputs
Conclusions
Have a clear questions first!
Consider theory based hypotheses
Consider confounding factors
Quick visualisations can help inform model choices
Check models and go with the best ones