Question

find the correlation coefficient, r, of the data described below. a sleep specialist wants to know if meditating before bedtime can help people fall asleep. in a study, he asked the subjects to meditate for various lengths of time just before going to bed. the specialist tracked the subjects’ meditation times, x, and how long it took them to fall asleep, y. both times were recorded in minutes.

meditation time	time needed to fall asleep
10	21
22	5
36	20
45	10
48	12

round your answer to the nearest thousandth. r =

Explanation:

Explicación:

Paso 1: Calcular las sumas necesarias

Sean $x$ los tiempos de meditación y $y$ los tiempos para dormirse. Tenemos $n = 6$ datos.
Calculamos $\sum x=6 + 10+22 + 36+45+48=167$, $\sum y=12 + 21+5+20+10+12=80$, $\sum xy=6\times12 + 10\times21+22\times5+36\times20+45\times10+48\times12=72+210 + 110+720+450+576=2138$, $\sum x^{2}=6^{2}+10^{2}+22^{2}+36^{2}+45^{2}+48^{2}=36+100+484+1296+2025+2304=6245$, $\sum y^{2}=12^{2}+21^{2}+5^{2}+20^{2}+10^{2}+12^{2}=144+441+25+400+100+144=1254$.

Paso 2: Aplicar la fórmula del coeficiente de correlación de Pearson

La fórmula 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:
$n\sum xy=6\times2138 = 12828$, $\sum x\sum y=167\times80 = 13360$, $n\sum x^{2}=6\times6245=37470$, $(\sum x)^{2}=167^{2}=27889$, $n\sum y^{2}=6\times1254 = 7524$, $(\sum y)^{2}=80^{2}=6400$.
$n\sum x^{2}-(\sum x)^{2}=37470 - 27889=9581$, $n\sum y^{2}-(\sum y)^{2}=7524 - 6400 = 1124$.
$\sqrt{n\sum x^{2}-(\sum x)^{2}}\sqrt{n\sum y^{2}-(\sum y)^{2}}=\sqrt{9581}\sqrt{1124}\approx97.88\times33.53\approx3270.91$.
$n\sum xy-\sum x\sum y=12828 - 13360=-532$.
$r=\frac{- 532}{3270.91}\approx - 0.163$.

Respuesta:

$-0.163$

Answer:

Explicación:

Paso 1: Calcular las sumas necesarias

Sean $x$ los tiempos de meditación y $y$ los tiempos para dormirse. Tenemos $n = 6$ datos.
Calculamos $\sum x=6 + 10+22 + 36+45+48=167$, $\sum y=12 + 21+5+20+10+12=80$, $\sum xy=6\times12 + 10\times21+22\times5+36\times20+45\times10+48\times12=72+210 + 110+720+450+576=2138$, $\sum x^{2}=6^{2}+10^{2}+22^{2}+36^{2}+45^{2}+48^{2}=36+100+484+1296+2025+2304=6245$, $\sum y^{2}=12^{2}+21^{2}+5^{2}+20^{2}+10^{2}+12^{2}=144+441+25+400+100+144=1254$.

Paso 2: Aplicar la fórmula del coeficiente de correlación de Pearson

La fórmula 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:
$n\sum xy=6\times2138 = 12828$, $\sum x\sum y=167\times80 = 13360$, $n\sum x^{2}=6\times6245=37470$, $(\sum x)^{2}=167^{2}=27889$, $n\sum y^{2}=6\times1254 = 7524$, $(\sum y)^{2}=80^{2}=6400$.
$n\sum x^{2}-(\sum x)^{2}=37470 - 27889=9581$, $n\sum y^{2}-(\sum y)^{2}=7524 - 6400 = 1124$.
$\sqrt{n\sum x^{2}-(\sum x)^{2}}\sqrt{n\sum y^{2}-(\sum y)^{2}}=\sqrt{9581}\sqrt{1124}\approx97.88\times33.53\approx3270.91$.
$n\sum xy-\sum x\sum y=12828 - 13360=-532$.
$r=\frac{- 532}{3270.91}\approx - 0.163$.

Respuesta:

$-0.163$