Question

find the correlation coefficient, r, of the data described below. the owner of a bicycle store wonders if the employees who take the longest lunch breaks also make the fewest bicycle sales. one day, he notes the length of each salespersons lunch break (in minutes), x, as well as the number of bicycles he or she sold that day, y. length of lunch break (in minutes) bicycles 25 8 28 6 31 7 31 4 34 2 40 2 round your answer to the nearest thousandth. r =

Explanation:

Explicación:

Paso 1: Calcular las sumas básicas

Sean $x_i$ la duración de la comida y $y_i$ el número de bicicletas vendidas. Tenemos $n = 6$ datos.
$\sum_{i = 1}^{n}x_i=25 + 28+31+31+34+40=189$
$\sum_{i = 1}^{n}y_i=8 + 6+7+4+2+2=29$
$\sum_{i = 1}^{n}x_i^2=25^2+28^2+31^2+31^2+34^2+40^2=625+784+961+961+1156+1600 = 6087$
$\sum_{i = 1}^{n}y_i^2=8^2+6^2+7^2+4^2+2^2+2^2=64+36+49+16+4+4 = 173$
$\sum_{i = 1}^{n}x_iy_i=25\times8+28\times6+31\times7+31\times4+34\times2+40\times2=200 + 168+217+124+68+80 = 857$

Paso 2: Calcular la media de $x$ y $y$

$\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}=\frac{189}{6}=31.5$
$\bar{y}=\frac{\sum_{i = 1}^{n}y_i}{n}=\frac{29}{6}\approx4.833$

Paso 3: Calcular la covarianza y las desviaciones estándar

$S_{xy}=\sum_{i = 1}^{n}x_iy_i - n\bar{x}\bar{y}=857-6\times31.5\times4.833=857 - 900.006=- 43.006$
$S_{xx}=\sum_{i = 1}^{n}x_i^2 - n\bar{x}^2=6087-6\times31.5^2=6087 - 5953.5 = 133.5$
$S_{yy}=\sum_{i = 1}^{n}y_i^2 - n\bar{y}^2=173-6\times(4.833)^2=173 - 139.997\approx33.003$
$S_x=\sqrt{S_{xx}}=\sqrt{133.5}\approx11.554$
$S_y=\sqrt{S_{yy}}=\sqrt{33.003}\approx5.745$

Paso 4: Calcular el coeficiente de correlación

$r=\frac{S_{xy}}{S_xS_y}=\frac{-43.006}{11.554\times5.745}=\frac{-43.006}{66.397}\approx - 0.648$

Respuesta:

$-0.648$

Answer:

Explicación:

Paso 1: Calcular las sumas básicas

Sean $x_i$ la duración de la comida y $y_i$ el número de bicicletas vendidas. Tenemos $n = 6$ datos.
$\sum_{i = 1}^{n}x_i=25 + 28+31+31+34+40=189$
$\sum_{i = 1}^{n}y_i=8 + 6+7+4+2+2=29$
$\sum_{i = 1}^{n}x_i^2=25^2+28^2+31^2+31^2+34^2+40^2=625+784+961+961+1156+1600 = 6087$
$\sum_{i = 1}^{n}y_i^2=8^2+6^2+7^2+4^2+2^2+2^2=64+36+49+16+4+4 = 173$
$\sum_{i = 1}^{n}x_iy_i=25\times8+28\times6+31\times7+31\times4+34\times2+40\times2=200 + 168+217+124+68+80 = 857$

Paso 2: Calcular la media de $x$ y $y$

$\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}=\frac{189}{6}=31.5$
$\bar{y}=\frac{\sum_{i = 1}^{n}y_i}{n}=\frac{29}{6}\approx4.833$

Paso 3: Calcular la covarianza y las desviaciones estándar

$S_{xy}=\sum_{i = 1}^{n}x_iy_i - n\bar{x}\bar{y}=857-6\times31.5\times4.833=857 - 900.006=- 43.006$
$S_{xx}=\sum_{i = 1}^{n}x_i^2 - n\bar{x}^2=6087-6\times31.5^2=6087 - 5953.5 = 133.5$
$S_{yy}=\sum_{i = 1}^{n}y_i^2 - n\bar{y}^2=173-6\times(4.833)^2=173 - 139.997\approx33.003$
$S_x=\sqrt{S_{xx}}=\sqrt{133.5}\approx11.554$
$S_y=\sqrt{S_{yy}}=\sqrt{33.003}\approx5.745$

Paso 4: Calcular el coeficiente de correlación

$r=\frac{S_{xy}}{S_xS_y}=\frac{-43.006}{11.554\times5.745}=\frac{-43.006}{66.397}\approx - 0.648$

Respuesta:

$-0.648$