Question

find the correlation coefficient, r, of the data described below. anna owns a car wash and has noticed that her business fluctuates throughout the year. she is curious to know whether these fluctuations are related to changes in the local pigeon population. over several mornings, anna counted the number of pigeons that were sitting on power lines in her neighborhood each day, x, and the number of car washes that were purchased during that day, y. pigeons counted car washes 13 9 14 12 16 4 22 48 22 34 round your answer to the nearest thousandth. r =

Explanation:

Explicación:

Paso 1: Calcular las sumas básicas

Sean $x_i$ el número de palomas y $y_i$ el número de lavados de automóviles.
Tenemos $n = 5$ datos.
$\sum_{i = 1}^{n}x_i=13 + 14+16+22+22=87$
$\sum_{i = 1}^{n}y_i=9 + 12+4+48+34=107$
$\sum_{i = 1}^{n}x_i^2=13^2 + 14^2+16^2+22^2+22^2=169+196+256+484+484 = 1589$
$\sum_{i = 1}^{n}y_i^2=9^2 + 12^2+4^2+48^2+34^2=81+144+16+2304+1156 = 3691$
$\sum_{i = 1}^{n}x_iy_i=13\times9+14\times12 + 16\times4+22\times48+22\times34=117+168+64+1056+748 = 2153$

Paso 2: Calcular la covarianza

La fórmula de la covarianza $S_{xy}=\frac{\sum_{i = 1}^{n}x_iy_i-\frac{1}{n}(\sum_{i = 1}^{n}x_i)(\sum_{i = 1}^{n}y_i)}{n - 1}$
$S_{xy}=\frac{2153-\frac{1}{5}\times87\times107}{4}=\frac{2153 - \frac{9309}{5}}{4}=\frac{2153-1861.8}{4}=\frac{291.2}{4}=72.8$

Paso 3: Calcular la desviación estándar de $x$

La fórmula de la desviación estándar de $x$, $S_x=\sqrt{\frac{\sum_{i = 1}^{n}x_i^2-\frac{1}{n}(\sum_{i = 1}^{n}x_i)^2}{n - 1}}$
$S_x=\sqrt{\frac{1589-\frac{1}{5}\times87^2}{4}}=\sqrt{\frac{1589-\frac{7569}{5}}{4}}=\sqrt{\frac{1589 - 1513.8}{4}}=\sqrt{\frac{75.2}{4}}=\sqrt{18.8}\approx4.336$

Paso 4: Calcular la desviación estándar de $y$

La fórmula de la desviación estándar de $y$, $S_y=\sqrt{\frac{\sum_{i = 1}^{n}y_i^2-\frac{1}{n}(\sum_{i = 1}^{n}y_i)^2}{n - 1}}$
$S_y=\sqrt{\frac{3691-\frac{1}{5}\times107^2}{4}}=\sqrt{\frac{3691-\frac{11449}{5}}{4}}=\sqrt{\frac{3691 - 2289.8}{4}}=\sqrt{\frac{1401.2}{4}}=\sqrt{350.3}\approx18.716$

Paso 5: Calcular el coeficiente de correlación

La fórmula del coeficiente de correlación $r=\frac{S_{xy}}{S_xS_y}$
$r=\frac{72.8}{4.336\times18.716}=\frac{72.8}{81.17}\approx0.897$

Respuesta:

$r\approx0.897$

Answer:

Explicación:

Paso 1: Calcular las sumas básicas

Sean $x_i$ el número de palomas y $y_i$ el número de lavados de automóviles.
Tenemos $n = 5$ datos.
$\sum_{i = 1}^{n}x_i=13 + 14+16+22+22=87$
$\sum_{i = 1}^{n}y_i=9 + 12+4+48+34=107$
$\sum_{i = 1}^{n}x_i^2=13^2 + 14^2+16^2+22^2+22^2=169+196+256+484+484 = 1589$
$\sum_{i = 1}^{n}y_i^2=9^2 + 12^2+4^2+48^2+34^2=81+144+16+2304+1156 = 3691$
$\sum_{i = 1}^{n}x_iy_i=13\times9+14\times12 + 16\times4+22\times48+22\times34=117+168+64+1056+748 = 2153$

Paso 2: Calcular la covarianza

La fórmula de la covarianza $S_{xy}=\frac{\sum_{i = 1}^{n}x_iy_i-\frac{1}{n}(\sum_{i = 1}^{n}x_i)(\sum_{i = 1}^{n}y_i)}{n - 1}$
$S_{xy}=\frac{2153-\frac{1}{5}\times87\times107}{4}=\frac{2153 - \frac{9309}{5}}{4}=\frac{2153-1861.8}{4}=\frac{291.2}{4}=72.8$

Paso 3: Calcular la desviación estándar de $x$

La fórmula de la desviación estándar de $x$, $S_x=\sqrt{\frac{\sum_{i = 1}^{n}x_i^2-\frac{1}{n}(\sum_{i = 1}^{n}x_i)^2}{n - 1}}$
$S_x=\sqrt{\frac{1589-\frac{1}{5}\times87^2}{4}}=\sqrt{\frac{1589-\frac{7569}{5}}{4}}=\sqrt{\frac{1589 - 1513.8}{4}}=\sqrt{\frac{75.2}{4}}=\sqrt{18.8}\approx4.336$

Paso 4: Calcular la desviación estándar de $y$

La fórmula de la desviación estándar de $y$, $S_y=\sqrt{\frac{\sum_{i = 1}^{n}y_i^2-\frac{1}{n}(\sum_{i = 1}^{n}y_i)^2}{n - 1}}$
$S_y=\sqrt{\frac{3691-\frac{1}{5}\times107^2}{4}}=\sqrt{\frac{3691-\frac{11449}{5}}{4}}=\sqrt{\frac{3691 - 2289.8}{4}}=\sqrt{\frac{1401.2}{4}}=\sqrt{350.3}\approx18.716$

Paso 5: Calcular el coeficiente de correlación

La fórmula del coeficiente de correlación $r=\frac{S_{xy}}{S_xS_y}$
$r=\frac{72.8}{4.336\times18.716}=\frac{72.8}{81.17}\approx0.897$

Respuesta:

$r\approx0.897$