Se midieron dos variables en 10 hombres:
Objetivo: evaluar, con \(\alpha = 0.05\), si existe asociación lineal (equivalente a probar \(H_0:\rho=0\) vs \(H_1:\rho\neq0\)) usando la t de Pearson.
| Sujeto | Colesterol | PesoTalla |
|---|---|---|
| 1 | 254 | 2.71 |
| 2 | 240 | 2.96 |
| 3 | 279 | 2.62 |
| 4 | 284 | 2.19 |
| 5 | 315 | 2.68 |
| 6 | 250 | 2.64 |
| 7 | 298 | 2.37 |
| 8 | 384 | 2.61 |
| 9 | 310 | 2.12 |
| 10 | 337 | 1.94 |
| n | Media.y1 | SD.y1 | Media.y2 | SD.y2 | Cov.y1.y2. | r |
|---|---|---|---|---|---|---|
| 10 | 295.10 | 44.00 | 2.484 | 0.317 | -6.103 | -0.4379 |
La t para contrastar \(H_0:\rho=0\) es: \[ t = \dfrac{r\sqrt{n-2}}{\sqrt{1-r^2}}, \qquad df=n-2. \]
| r | t_obs | df | X.t..crit..dos.colas. | p.valor | Decisión..α… |
|---|---|---|---|---|---|
| -0.4379 | -1.378 | 8 | ±2.306 | 0.206 | 0.05): No rechazar H0 |
| IC95..inferior | IC95..superior |
|---|---|
| -0.837 | 0.265 |
Con \(t_{\text{obs}} = -1.378\),
\(df=8\) y \(p\)-valor = 0.206, al nivel \(\alpha = 0.05\) NO se
rechaza \(H_0\).
El tamaño del efecto muestral es \(r =
-0.4379\).
El IC95% para \(\rho\)
es \([-0.837, 0.265]\).
##
## Pearson's product-moment correlation
##
## data: y1 and y2
## t = -1.3778, df = 8, p-value = 0.2056
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
## -0.8368148 0.2646875
## sample estimates:
## cor
## -0.43792