Question

find the correlation coefficient, r, of the data described below. owen keeps detailed information about his model airplane collection. the data includes not only the characteristics of each model, but also the time it took to assemble it. he is thinking of buying a new model airplane and decided to look at this information to understand how long assembly might take. from his data, owen found the number of pieces in each model, x, and how many minutes each took to assemble, y. pieces assembly time 45 44 53 53 61 46 67 50 92 56 99 55 round your answer to the nearest thousandth. r =

Explanation:

Explicación:

Paso 1: Calcular la media de \(x\) y \(y\)

Sean \(x_i\) los valores de la variable \(x\) (número de piezas) y \(y_i\) los valores de la variable \(y\) (tiempo de ensamblaje).
Tenemos \(n = 6\) pares de datos.
\(\bar{x}=\frac{45 + 53+61+67+92+99}{6}=\frac{417}{6}=69.5\)
\(\bar{y}=\frac{44 + 53+46+50+56+55}{6}=\frac{304}{6}\approx50.667\)

Paso 2: Calcular los productos y sumas necesarias

Calculemos \(\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})\), \(\sum_{i=1}^{n}(x_i-\bar{x})^2\) y \(\sum_{i = 1}^{n}(y_i-\bar{y})^2\)

Para \(i = 1\): \((x_1-\bar{x})(y_1 - \bar{y})=(45 - 69.5)(44-50.667)=(- 24.5)(-6.667)=163.3415\)
\((x_1-\bar{x})^2=(45 - 69.5)^2=(-24.5)^2 = 600.25\)
\((y_1-\bar{y})^2=(44 - 50.667)^2=(-6.667)^2 = 44.448889\)

Repitiendo para todos los \(i\) de \(1\) a \(6\):

\(\sum_{i = 1}^{6}(x_i-\bar{x})(y_i - \bar{y})=(-24.5)(-6.667)+(-16.5)(2.333)+(-8.5)(-4.667)+(-2.5)(-0.667)+(22.5)(5.333)+(29.5)(4.333)\)
\(=163.3415-38.4945 + 39.6695+1.6675+119.9925+127.8235=413.999\)

\(\sum_{i=1}^{6}(x_i-\bar{x})^2=600.25 + 272.25+72.25 + 6.25+506.25+870.25=2327.5\)

\(\sum_{i = 1}^{6}(y_i-\bar{y})^2=44.448889+5.443889+21.778889+0.444889+69.441889+18.776889 = 160.339334\)

Paso 3: Calcular el coeficiente de correlación \(r\)

La fórmula para el coeficiente de correlación de Pearson es \(r=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sqrt{\sum_{i=1}^{n}(x_i-\bar{x})^2\sum_{i = 1}^{n}(y_i-\bar{y})^2}}\)
Sustituyendo los valores calculados:
\(r=\frac{413.999}{\sqrt{2327.5\times160.339334}}=\frac{413.999}{\sqrt{373244.0934}}=\frac{413.999}{610.936}\approx0.678\)

Respuesta:

\(r\approx0.678\)

Answer:

Explicación:

Paso 1: Calcular la media de \(x\) y \(y\)

Sean \(x_i\) los valores de la variable \(x\) (número de piezas) y \(y_i\) los valores de la variable \(y\) (tiempo de ensamblaje).
Tenemos \(n = 6\) pares de datos.
\(\bar{x}=\frac{45 + 53+61+67+92+99}{6}=\frac{417}{6}=69.5\)
\(\bar{y}=\frac{44 + 53+46+50+56+55}{6}=\frac{304}{6}\approx50.667\)

Paso 2: Calcular los productos y sumas necesarias

Calculemos \(\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})\), \(\sum_{i=1}^{n}(x_i-\bar{x})^2\) y \(\sum_{i = 1}^{n}(y_i-\bar{y})^2\)

Para \(i = 1\): \((x_1-\bar{x})(y_1 - \bar{y})=(45 - 69.5)(44-50.667)=(- 24.5)(-6.667)=163.3415\)
\((x_1-\bar{x})^2=(45 - 69.5)^2=(-24.5)^2 = 600.25\)
\((y_1-\bar{y})^2=(44 - 50.667)^2=(-6.667)^2 = 44.448889\)

Repitiendo para todos los \(i\) de \(1\) a \(6\):

\(\sum_{i = 1}^{6}(x_i-\bar{x})(y_i - \bar{y})=(-24.5)(-6.667)+(-16.5)(2.333)+(-8.5)(-4.667)+(-2.5)(-0.667)+(22.5)(5.333)+(29.5)(4.333)\)
\(=163.3415-38.4945 + 39.6695+1.6675+119.9925+127.8235=413.999\)

\(\sum_{i=1}^{6}(x_i-\bar{x})^2=600.25 + 272.25+72.25 + 6.25+506.25+870.25=2327.5\)

\(\sum_{i = 1}^{6}(y_i-\bar{y})^2=44.448889+5.443889+21.778889+0.444889+69.441889+18.776889 = 160.339334\)

Paso 3: Calcular el coeficiente de correlación \(r\)

La fórmula para el coeficiente de correlación de Pearson es \(r=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sqrt{\sum_{i=1}^{n}(x_i-\bar{x})^2\sum_{i = 1}^{n}(y_i-\bar{y})^2}}\)
Sustituyendo los valores calculados:
\(r=\frac{413.999}{\sqrt{2327.5\times160.339334}}=\frac{413.999}{\sqrt{373244.0934}}=\frac{413.999}{610.936}\approx0.678\)

Respuesta:

\(r\approx0.678\)