Tema 5: Inferencia Estadística II. Estimación por Intervalos de Confianza

class: center, middle, inverse, title-slide

# Tema 5: Inferencia Estadística II. Estimación por Intervalos de Confianza
### Estadística. Grado en Ciencias del Mar

---

# Intervalo de confianza

.resalta[
Sean:

* `$X$` una variable aleatoria cuya distribución queda caracterizada por un parámetro desconocido `$\theta$`

* `$\omega=\left\{ X_{1},X_{2},\dots,X_{n}\right\}$` una muestra aleatoria de observaciones de dicha variable.

Entonces `$\left[{\theta_{1}\left(\omega\right),\theta_{2}\left(\omega\right)}\right]$`,
donde `$\theta_{1}\left(\omega\right)$` y `$\theta_{2}\left(\omega\right)$` son variables aleatorias que dependen de la muestra, es un .blue[ __intervalo de confianza__] a nivel `$1-\alpha$` para el parámetro `$\theta$` si:

`$$P\left({\theta\in\left[{\theta_{1}\left(\omega\right),\theta_{2}\left(\omega\right)}\right]}\right)=1-\alpha$$`
]

---

## .blue[Ejemplo:]

Ya hemos visto que si `$X\approx N\left(\mu,\sigma\right)$`, y `$\bar{X}$` es la media aritmética de una muestra aleatoria de observaciones `$\omega=\left\{X_1,X_2,\dots,X_n\right\}$` de `$X$`:

`$$\frac{\bar{X}-\mu}{s/\sqrt{n}}\approx t_{n-1}$$`
--

y de aquí es posible deducir que _antes de tomar la muestra_:

`$$P\left(\mu \in \left[\bar{X}-t_{n-1,\alpha/2}\frac{s}{\sqrt{n}},\; \bar{X}+t_{n-1,\alpha/2}\frac{s}{\sqrt{n}}\right]\right)=1-\alpha$$`
--

Obviamente, los extremos del intervalo  `$\theta_{1}\left(\omega\right)=\bar{X}-t_{n-1,\alpha/2}\frac{s}{\sqrt{n}}\;$` y `$\;\theta_{2}\left(\omega\right)=\bar{X}+t_{n-1,\alpha/2}\frac{s}{\sqrt{n}}\;$` .ilumina[.blue[son variables aleatorias]] porque su valor no se conoce hasta que no se haya tomado la muestra.

---

## Interpretación del término "confianza"

* .blue[ __Antes de tomar la muestra__] tiene sentido hablar de la probabilidad de que el intervalo, una vez que se construya, contenga al parámetro desconocido.

* .blue[ __Después de tomar la muestra y calcular el intervalo__] éste contendrá o no contendrá al parámetro, pero no tiene sentido hablar de probabilidad porque el experimento ya se ha realizado.

* En la práctica, si la probabilidad .blue[ _a priori_] de capturar el parámetro en el intervalo es `$1-\alpha$`, si tomáramos muchas muestras de tamaño `$n$` y para cada una construyéramos el intervalo `$\left[\theta_1(\omega), \theta_2(\omega)\right]$`, podríamos esperar que el `$100(1-\alpha)\%$` de dichos intervalos contuviese al parámetro `$\theta$`.

* Pero como en realidad .ilumina[solo tenemos una muestra y construimos un único intervalo] .red[ __no sabemos__] si este intervalo contiene o no a `$\theta$`, pero _confiamos_ en que sea uno de entre el `$100\left(1-\alpha\right)\%$` de intervalos que contienen al parámetro. De ahí que valoremos nuestra confianza en `$1-\alpha$`

---

## .blue[Ejercicio]

* Podemos simular la obtención de una muestra de tamaño `$n=40$` de una variable `$X\approx N\left(\mu=80,\sigma=10\right)$` mediante el siguiente código R:

```r
muestra=rnorm(40,80,10)
```

* Podemos también obtener un intervalo de confianza para `$\mu$` a partir de esta muestra mediante:

```r
t.test(muestra)$conf.int
```

```
## [1] 79.08081 85.72290
## attr(,"conf.level")
## [1] 0.95
```

* Como vemos, en este caso el valor del parámetro `$\mu$` ha quedado contenido en el intervalo.

* Vamos a repetir el proceso.

---

## .blue[Ejercicio]

* El siguiente código repite 100 veces el proceso de generar una muestra de tamaño 40 y construir el intervalo de confianza correspondiente; los intervalos resultantes se guardan en el objeto `intervalos`:

```r
intervalos <- replicate(100,t.test(rnorm(40,80,10))$conf.int)
```

* Podemos ver los primeros intervalos:

```r
intervalos[,1:6]
```

```
##          [,1]     [,2]     [,3]     [,4]     [,5]     [,6]
## [1,] 77.81020 78.95325 75.62774 78.74789 77.05240 76.54074
## [2,] 85.56283 85.63792 82.90436 84.47362 83.33463 82.39630
```

* O usar la siguiente sintaxis para verlos todos de una forma más cómoda:

```r
array(sprintf("[%.2f, %.2f]", intervalos[1,], intervalos[2,]),
      dim=c(20,5))
```

---

## .blue[Ejercicio]

Los 100 intervalos correspondientes a las 100 muestras de tamaño 40 son los siguientes. ¿Contienen todos a la media poblacional `$\mu=80$`?:

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83.92]</td><td class="cl-00e04a38">[76.32, 82.70]</td><td class="cl-00e04a38">[75.36, 81.92]</td><td class="cl-00e04a38">[78.05, 84.15]</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-00e04a42">[78.75, 84.47]</td><td class="cl-00e04a42">[75.47, 81.86]</td><td class="cl-00e04a42">[76.11, 82.33]</td><td class="cl-00e04a42">[75.22, 81.65]</td><td class="cl-00e04a42">[78.75, 85.43]</td><td class="cl-00e04a42">[76.03, 81.27]</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-00e04a38">[77.05, 83.33]</td><td class="cl-00e04a38">[75.44, 81.27]</td><td class="cl-00e04a38">[75.30, 82.36]</td><td class="cl-00e04a38">[74.22, 80.83]</td><td class="cl-00e04a38">[75.85, 82.27]</td><td class="cl-00e04a38">[79.01, 86.39]</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-00e04a42">[76.54, 82.40]</td><td class="cl-00e04a42">[73.07, 78.58]</td><td class="cl-00e04a42">[78.33, 83.57]</td><td class="cl-00e04a42">[76.15, 81.94]</td><td class="cl-00e04a42">[74.86, 81.09]</td><td class="cl-00e04a42">[77.21, 84.06]</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-00e04a38">[77.81, 84.30]</td><td class="cl-00e04a38">[77.24, 83.31]</td><td class="cl-00e04a38">[76.21, 83.06]</td><td class="cl-00e04a38">[77.51, 83.16]</td><td class="cl-00e04a38">[77.50, 83.24]</td><td class="cl-00e04a38">[77.57, 83.79]</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-00e04a42">[78.39, 84.66]</td><td class="cl-00e04a42">[75.46, 81.43]</td><td class="cl-00e04a42">[74.39, 81.75]</td><td class="cl-00e04a42">[76.77, 82.01]</td><td class="cl-00e04a42">[75.30, 80.62]</td><td class="cl-00e04a42">[77.47, 83.62]</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-00e04a38">[79.03, 84.93]</td><td class="cl-00e04a38">[74.26, 80.54]</td><td class="cl-00e04a38">[78.80, 83.82]</td><td class="cl-00e04a38">[77.78, 84.06]</td><td class="cl-00e04a38">[76.80, 83.57]</td><td class="cl-00e04a38">[78.35, 84.56]</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-00e04a42">[78.27, 83.95]</td><td class="cl-00e04a42">[76.74, 83.59]</td><td class="cl-00e04a42">[75.31, 83.09]</td><td class="cl-00e04a42">[78.44, 84.28]</td><td class="cl-00e04a42">[74.55, 81.48]</td><td class="cl-00e04a42">[76.11, 82.16]</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-00e04a38">[75.74, 81.84]</td><td class="cl-00e04a38">[77.24, 84.10]</td><td class="cl-00e04a38">[77.97, 83.94]</td><td class="cl-00e04a38">[74.49, 79.41]</td><td class="cl-00e04a38">[75.53, 82.28]</td><td class="cl-00e04a38">[74.72, 80.67]</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-00e04a42">[74.33, 81.12]</td><td class="cl-00e04a42">[78.19, 84.19]</td><td class="cl-00e04a42">[75.40, 81.51]</td><td class="cl-00e04a42">[79.60, 84.84]</td><td class="cl-00e04a42">[76.37, 82.74]</td><td class="cl-00e04a42">[77.36, 83.88]</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-00e04a38">[79.51, 86.42]</td><td class="cl-00e04a38">[77.10, 83.23]</td><td class="cl-00e04a38">[76.78, 83.91]</td><td class="cl-00e04a38">[78.95, 86.15]</td><td class="cl-00e04a38">[77.03, 83.00]</td><td class="cl-00e04a38">[80.46, 85.49]</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-00e04a42">[76.88, 82.80]</td><td class="cl-00e04a42">[76.07, 82.03]</td><td class="cl-00e04a42">[77.26, 84.45]</td><td class="cl-00e04a42">[76.28, 82.48]</td><td class="cl-00e04a42">[77.81, 83.99]</td><td class="cl-00e04a42">[78.82, 84.94]</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-00e04a38">[77.43, 82.42]</td><td class="cl-00e04a38">[75.63, 83.67]</td><td class="cl-00e04a38">[77.32, 82.51]</td><td class="cl-00e04a38">[75.59, 81.35]</td><td class="cl-00e04a38">[77.86, 82.65]</td><td class="cl-00e04a38">[72.89, 79.37]</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-00e04a42">[76.96, 84.06]</td><td class="cl-00e04a42">[74.10, 79.13]</td><td class="cl-00e04a42">[77.09, 83.56]</td><td 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---

## .blue[Ejercicio]

Podemos comprobar que no todos los intervalos contienen al parámetro `$\mu=80$`:

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class="cl-00ea7f76">[77.05, 83.33]</td><td class="cl-00ea7f76">[75.44, 81.27]</td><td class="cl-00ea7f76">[75.30, 82.36]</td><td class="cl-00ea7f76">[74.22, 80.83]</td><td class="cl-00ea7f76">[75.85, 82.27]</td><td class="cl-00ea7f76">[79.01, 86.39]</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-00ea7f77">[76.54, 82.40]</td><td class="cl-00ea7f80">[73.07, 78.58]</td><td class="cl-00ea7f77">[78.33, 83.57]</td><td class="cl-00ea7f77">[76.15, 81.94]</td><td class="cl-00ea7f77">[74.86, 81.09]</td><td class="cl-00ea7f77">[77.21, 84.06]</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-00ea7f76">[77.81, 84.30]</td><td class="cl-00ea7f76">[77.24, 83.31]</td><td class="cl-00ea7f76">[76.21, 83.06]</td><td class="cl-00ea7f76">[77.51, 83.16]</td><td class="cl-00ea7f76">[77.50, 83.24]</td><td class="cl-00ea7f76">[77.57, 83.79]</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-00ea7f77">[78.39, 84.66]</td><td class="cl-00ea7f77">[75.46, 81.43]</td><td class="cl-00ea7f77">[74.39, 81.75]</td><td class="cl-00ea7f77">[76.77, 82.01]</td><td class="cl-00ea7f77">[75.30, 80.62]</td><td class="cl-00ea7f77">[77.47, 83.62]</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-00ea7f76">[79.03, 84.93]</td><td class="cl-00ea7f76">[74.26, 80.54]</td><td class="cl-00ea7f76">[78.80, 83.82]</td><td class="cl-00ea7f76">[77.78, 84.06]</td><td class="cl-00ea7f76">[76.80, 83.57]</td><td class="cl-00ea7f76">[78.35, 84.56]</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-00ea7f77">[78.27, 83.95]</td><td class="cl-00ea7f77">[76.74, 83.59]</td><td class="cl-00ea7f77">[75.31, 83.09]</td><td class="cl-00ea7f77">[78.44, 84.28]</td><td class="cl-00ea7f77">[74.55, 81.48]</td><td class="cl-00ea7f77">[76.11, 82.16]</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-00ea7f76">[75.74, 81.84]</td><td class="cl-00ea7f76">[77.24, 84.10]</td><td class="cl-00ea7f76">[77.97, 83.94]</td><td class="cl-00ea7f80">[74.49, 79.41]</td><td class="cl-00ea7f76">[75.53, 82.28]</td><td class="cl-00ea7f76">[74.72, 80.67]</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-00ea7f77">[74.33, 81.12]</td><td class="cl-00ea7f77">[78.19, 84.19]</td><td class="cl-00ea7f77">[75.40, 81.51]</td><td class="cl-00ea7f77">[79.60, 84.84]</td><td class="cl-00ea7f77">[76.37, 82.74]</td><td class="cl-00ea7f77">[77.36, 83.88]</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-00ea7f76">[79.51, 86.42]</td><td class="cl-00ea7f76">[77.10, 83.23]</td><td class="cl-00ea7f76">[76.78, 83.91]</td><td class="cl-00ea7f76">[78.95, 86.15]</td><td class="cl-00ea7f76">[77.03, 83.00]</td><td class="cl-00ea7f80">[80.46, 85.49]</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-00ea7f77">[76.88, 82.80]</td><td class="cl-00ea7f77">[76.07, 82.03]</td><td class="cl-00ea7f77">[77.26, 84.45]</td><td class="cl-00ea7f77">[76.28, 82.48]</td><td class="cl-00ea7f77">[77.81, 83.99]</td><td class="cl-00ea7f77">[78.82, 84.94]</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-00ea7f76">[77.43, 82.42]</td><td class="cl-00ea7f76">[75.63, 83.67]</td><td class="cl-00ea7f76">[77.32, 82.51]</td><td class="cl-00ea7f76">[75.59, 81.35]</td><td class="cl-00ea7f76">[77.86, 82.65]</td><td class="cl-00ea7f80">[72.89, 79.37]</td></tr><tr style="overflow-wrap:break-word;"><td class="cl-00ea7f77">[76.96, 84.06]</td><td class="cl-00ea7f80">[74.10, 79.13]</td><td class="cl-00ea7f77">[77.09, 83.56]</td><td class="cl-00ea7f77">[78.73, 84.71]</td><td class="cl-00ea7f77">[77.30, 82.81]</td><td class="cl-00ea7f77"></td></tr><tr style="overflow-wrap:break-word;"><td class="cl-00ea7f76">[77.18, 84.75]</td><td class="cl-00ea7f76">[75.17, 81.63]</td><td class="cl-00ea7f76">[74.78, 81.09]</td><td class="cl-00ea7f76">[75.72, 81.49]</td><td class="cl-00ea7f76">[77.54, 83.29]</td><td class="cl-00ea7f76"></td></tr></tbody></table></div></template>
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---

## .blue[Ejercicio]

* R puede utilizarse para contar "de forma automática" cuantos de estos intervalos contienen al parámetro, en este caso, al valor de la media poblacional `$\mu=80$`:

```r
contieneMu <- which(intervalos[1,]<=80 & 80<=intervalos[2,])
length(contieneMu)
```

```
## [1] 94
```

* En este ejemplo, 94 de 100 intervalos (el 94%) contienen al verdadero valor del parámetro.

* Si repetimos el proceso con otras muestras distintas veremos que siempre obtenemos valores de cobertura (proporción de intervalos que contienen al parámetro) próximos al 95%

---

# .blue[Intervalos de confianza más habituales]

### Esperanza `$\mu$` de una variable `$X$` con distribución normal
.resalta[
`$$\mu \in \left[{\bar{X}-\frac{S}{{\sqrt{n}}}t_{n-1,\alpha/2},\bar{X}+\frac{S}{{\sqrt{n}}}t_{n-1,\alpha/2}}\right]$$`
]

### Varianza `$\sigma^{2}$` de una variable `$X$` con distribución normal
.resalta[
`$$\sigma^2 \in \left[\frac{\left(n-1\right)S^{2}}{\chi_{n-1,\alpha/2}^{2}},\,\frac{\left(n-1\right)S^{2}}{\chi_{n-1,1-\alpha/2}^{2}}\right]$$`
]

---

### Cociente de varianzas `$\sigma_{1}^{2}/\sigma_{2}^{2}$` de variables normales.
.resalta[
`$$\frac{\sigma_1^2}{\sigma_2^2} \in \left[\frac{S_{1}^{2}/S_{2}^{2}}{F_{n_{1}-1,n_{2}-1,\alpha/2}},\frac{S_{1}^{2}/S_{2}^{2}}{F_{n_{1}-1,n_{2}-1,1-\alpha/2}}\right]$$`
]

---

### Diferencia de medias de variables normales .red[independientes]

.resalta[
`$$\mu_1-\mu_2\in\left[\left({\bar{X}_{1}-\bar{X}_{2}}\right)\pm t_{n,}{}_{\alpha/2}\sqrt{\frac{{s_{1}^{2}}}{{n_{1}}}+\frac{{s_{2}^{2}}}{{n_{2}}}}\right]$$`
]

`$$s_{p}=\sqrt{\frac{{\left({n_{1}-1}\right)s_{1}^{2}+\left({n_{2}-1}\right)s_{2}^{2}}}{{n_{1}+n_{2}-2}}}\;\qquad \; n=\frac{{\left({\frac{{s_{1}^{2}}}{{n_{1}}}+\frac{{s_{2}^{2}}}{{n_{2}}}}\right)^{2}}}{{\left({\frac{{s_{1}^{2}}}{{n_{1}}}}\right)^{2}\frac{1}{{n_{1}-1}}+\left({\frac{{s_{2}^{2}}}{{n_{2}}}}\right)^{2}\frac{1}{{n_{2}-1}}}}$$`
---

### Diferencia de medias de variables normales .red[emparejadas]

.resalta[
`$$\mu_1-\mu_2\in\left[\left({\bar{X}_{1}-\bar{X}_{2}}\right)\pm t_{n-1,\alpha/2}\frac{{S_{D}}}{{\sqrt{n}}}\right]$$`
]

donde:

`$S_{D}=\sqrt{S_{1}^{2}+S_{2}^{2}-2S_{12}}=\sqrt{S_{1}^{2}+S_{2}^{2}-2rS_{1}S_{2}}$`

---

### Diferencia de medias de variables no normales .red[independientes] (sólo si `$\small n_{1}\ge30,n_{2}\ge30$`).

.resalta[
`$$\mu_1-\mu_2\in\left[\left({\bar{X}_{1}-\bar{X}_{2}}\right)\pm z{}_{\alpha/2}\sqrt{\frac{{s_{1}^{2}}}{{n_{1}}}+\frac{{s_{2}^{2}}}{{n_{2}}}}\right]$$`
]

### Diferencia de medias de variables no normales .red[emparejadas] (sólo si `$\small n\ge 60$`)

.resalta[
`$$\mu_1-\mu_2\in\left[\left({\bar{X}_{1}-\bar{X}_{2}}\right)\pm z_{\alpha/2}\frac{{S_{D}}}{{\sqrt{n}}}\right]$$`
]

siendo `$\small S_{D}=\sqrt{S_{1}^{2}+S_{2}^{2}-2S_{12}}=\sqrt{S_{1}^{2}+S_{2}^{2}-2rS_{1}S_{2}}$`

---

### Intervalo de confianza para la proporción `$\pi$` de éxitos en una `$B(n,\pi)$`.

Sean `$n$` el tamaño de la muestra y `$N_{E}$` el número de éxitos observados.

* .blue[ __Método de Agresti-Coull__ ] (sólo si `$n>40$`, `$N_E\ge5$`, `$n-N_E\ge5$`)

Llamando: `$\tilde{N}_{E}=N_{E}+z_{\alpha/2}^2/2\;$`, `$\tilde{n}=n+\left(z_{\alpha/2}\right)^{2}$` y `$\tilde{\pi}=\tilde{N}_{E}/\tilde{n}$`:

.resalta[
`$$\pi \in \left[{\tilde{\pi}\pm z_{\alpha/2}\sqrt{\frac{{\tilde{\pi}\left({1-\tilde{\pi}}\right)}}{{\tilde{n}}}}}\right]$$`
]

* .blue[ __Método de Clopper y Pearson__] (cuando no se puede aplicar el anterior)

Llamando: `$F_{1}=F_{2(n-N_{E}+1),2N_{E},\alpha/2}$` y `$F_{2}=F_{2(N_{E}+1),2(n-N_{E}),\alpha/2}$`:

.resalta[
`$$\pi \in \left[\frac{N_{E}}{{(n-N_{E}+1)F_{1}+N_{E}}},\frac{{(N_{E}+1)F_{2}}}{{(n-N_{E})+(N_{E}+1)F_{2}}}\right]$$`
]

---

### Comparación de proporciones de dos variables binomiales `$X_1$` y `$X_2$`

Se han realizado `$n_1$` observaciones de `$X_1$` con `$N_{E_1}$` éxitos y `$n_2$` observaciones de `$X_2$` con `$N_{E_2}$` éxitos. Sean `$\hat{\pi_1}=\frac{N_{E_1}}{n_1}$` y `$\;\hat{\pi_2}=\frac{N_{E_2}}{n_2}$` las proporciones observadas de éxitos. Si `$n_{1}\ge30$`, `$n_{2}\ge30$`:

* .blue[ __Intervalo de confianza para la diferencia de proporciones__] 
.resalta[
`$$\left[{\left(\pi_{1}-\pi_{2}\right)\pm\left(z_{\alpha/2}\sqrt{\frac{{\hat{\pi}_{1}\left({1-\hat{\pi}_{1}}\right)}}{n_{1}}+\frac{{\hat{\pi}_{2}\left({1-\hat{\pi}_{2}}\right)}}{n_{2}}}+\frac{1}{2}\left(\frac{1}{n_{1}}+\frac{1}{n_{2}}\right)\right)}\right]$$`
]

* .blue[ __Intervalo de confianza para el cociente de proporciones__]
.resalta[
`$$\left(\frac{\pi_{1}}{\pi_{2}}\right)\in\left[\left(\frac{\hat{\pi}_{1}}{\hat{\pi}_{2}}\right)\cdot\exp\left(\pm z_{\alpha/2}\sqrt{\frac{\left(1-\hat{\pi}_{1}\right)}{n_{1}\hat{\pi}_{1}}+\frac{\left(1-\hat{\pi}_{2}\right)}{n_{2}\hat{\pi}_{2}}}\right)\right]$$`
]

---
### Intervalo de confianza para la esperanza `$\mu$` de una distribución exponencial

.resalta[
`$$\mu\in\left[\frac{2n\bar{X}}{\chi_{2n,\alpha/2}^{2}},\frac{2n\bar{X}}{\chi_{2n,1-\alpha/2}^{2}}\right]$$`
]

### Intervalo de confianza para el parámetro `$\lambda$` de una distribución de Poisson.
.resalta[
`$$\lambda \in \left[{\frac{1}{{2n}}\chi_{n_{1},1-\alpha/2}^{2},\frac{1}{{2n}}\chi_{n_{2},\alpha/2}^{2}}\right]$$`
]

siendo: `$n_{1}=2T,\,\,\,\,\,n_{2}=2(T+1),\,\,\,\,\,T=\sum_{i=1}^{n}X_{i}$`