Question

find the correlation coefficient, r, of the data described below. the counselor at georgetown high school suspects that students are signed up for too many extracurricular activities and thinks this negatively affects their academic performance. she decided to compile information about several students to find out. for each of the students, the counselor recorded the number of extracurricular activities, x, and the gpa (out of 4.0), y. extracurricular activities gpa 0 3.8 1 3.4 2 3.2 3 2.3 4 3.4 6 2.3 round your answer to the nearest thousandth. r =

Explanation:

Explicación:

Paso 1: Definir las sumas iniciales

Sean $n = 6$ los datos. Calculemos $\sum x$, $\sum y$, $\sum x^2$, $\sum y^2$ y $\sum xy$.
$\sum x=0 + 1+2 + 3+4 + 6=16$
$\sum y=3.8+3.4 + 3.2+2.3+3.4+2.3 = 18.4$
$\sum x^2=0^2+1^2 + 2^2+3^2+4^2+6^2=0 + 1+4 + 9+16+36 = 66$
$\sum y^2=3.8^2+3.4^2+3.2^2+2.3^2+3.4^2+2.3^2$
$=14.44+11.56+10.24+5.29+11.56+5.29 = 58.38$
$\sum xy=(0\times3.8)+(1\times3.4)+(2\times3.2)+(3\times2.3)+(4\times3.4)+(6\times2.3)$
$=0 + 3.4+6.4+6.9+13.6+13.8 = 44.1$

Paso 2: Aplicar la fórmula del coeficiente de correlación

La fórmula para el coeficiente de correlación $r$ es:
\[r=\frac{n\sum xy-\sum x\sum y}{\sqrt{n\sum x^2 - (\sum x)^2}\sqrt{n\sum y^2-(\sum y)^2}}\]
Sustituimos los valores:
\[n\sum xy=6\times44.1 = 264.6\]
\[\sum x\sum y=16\times18.4 = 294.4\]
\[n\sum x^2=6\times66 = 396\]
\[(\sum x)^2=16^2 = 256\]
\[n\sum y^2=6\times58.38 = 350.28\]
\[(\sum y)^2=18.4^2 = 338.56\]

\[r=\frac{264.6 - 294.4}{\sqrt{396 - 256}\sqrt{350.28-338.56}}\]
\[r=\frac{-29.8}{\sqrt{140}\sqrt{11.72}}\]
\[r=\frac{-29.8}{\sqrt{140\times11.72}}\]
\[r=\frac{-29.8}{\sqrt{1640.8}}\]
\[r=\frac{-29.8}{40.5068}\]
\[r\approx - 0.736\]

Respuesta:

$-0.736$

Answer:

Explicación:

Paso 1: Definir las sumas iniciales

Sean $n = 6$ los datos. Calculemos $\sum x$, $\sum y$, $\sum x^2$, $\sum y^2$ y $\sum xy$.
$\sum x=0 + 1+2 + 3+4 + 6=16$
$\sum y=3.8+3.4 + 3.2+2.3+3.4+2.3 = 18.4$
$\sum x^2=0^2+1^2 + 2^2+3^2+4^2+6^2=0 + 1+4 + 9+16+36 = 66$
$\sum y^2=3.8^2+3.4^2+3.2^2+2.3^2+3.4^2+2.3^2$
$=14.44+11.56+10.24+5.29+11.56+5.29 = 58.38$
$\sum xy=(0\times3.8)+(1\times3.4)+(2\times3.2)+(3\times2.3)+(4\times3.4)+(6\times2.3)$
$=0 + 3.4+6.4+6.9+13.6+13.8 = 44.1$

Paso 2: Aplicar la fórmula del coeficiente de correlación

La fórmula para el coeficiente de correlación $r$ es:
\[r=\frac{n\sum xy-\sum x\sum y}{\sqrt{n\sum x^2 - (\sum x)^2}\sqrt{n\sum y^2-(\sum y)^2}}\]
Sustituimos los valores:
\[n\sum xy=6\times44.1 = 264.6\]
\[\sum x\sum y=16\times18.4 = 294.4\]
\[n\sum x^2=6\times66 = 396\]
\[(\sum x)^2=16^2 = 256\]
\[n\sum y^2=6\times58.38 = 350.28\]
\[(\sum y)^2=18.4^2 = 338.56\]

\[r=\frac{264.6 - 294.4}{\sqrt{396 - 256}\sqrt{350.28-338.56}}\]
\[r=\frac{-29.8}{\sqrt{140}\sqrt{11.72}}\]
\[r=\frac{-29.8}{\sqrt{140\times11.72}}\]
\[r=\frac{-29.8}{\sqrt{1640.8}}\]
\[r=\frac{-29.8}{40.5068}\]
\[r\approx - 0.736\]

Respuesta:

$-0.736$