Question

find the equation for the least squares regression line of the data described below. tina is a math teacher at a large school. she wonders if her test problems are too wordy. tina is curious whether the wordiness is affecting student performance. for the last several tests, tina computes the average number of words in each question, x, as well as the average percentage scores on the tests, y. average question length average student score 29.90 89.21 33.45 85.36 33.78 71.41 38.92 78.65 42.35 76.78 49.15 51.18 round your answers to the nearest thousandth. y = x +

Explanation:

Explicación:

Paso 1: Calcular sumatorias

Sean $n = 6$ los pares de datos.
Calcular $\sum x$, $\sum y$, $\sum xy$, $\sum x^{2}$.
$\sum x=29.90 + 33.45+33.78+38.92+42.35+49.15 = 229.55$
$\sum y=89.21 + 85.36+71.41+78.65+76.78+51.18 = 452.59$
$\sum xy=(29.90\times89.21)+(33.45\times85.36)+(33.78\times71.41)+(38.92\times78.65)+(42.35\times76.78)+(49.15\times51.18)$
$=2667.379+2855.292+2413.22+3061.058+3243.533+2515.597 = 16755.079$
$\sum x^{2}=29.90^{2}+33.45^{2}+33.78^{2}+38.92^{2}+42.35^{2}+49.15^{2}$
$=894.01+1119.9025+1141.0884+1514.7664+1793.5225+2415.7225 = 8879.0123$

Paso 2: Calcular pendiente $m$

La fórmula para la pendiente $m$ de la recta de regresión lineal es $m=\frac{n\sum xy-\sum x\sum y}{n\sum x^{2}-(\sum x)^{2}}$
$n\sum xy = 6\times16755.079 = 100530.474$
$\sum x\sum y=229.55\times452.59 = 103991.0145$
$n\sum x^{2}=6\times8879.0123 = 53274.0738$
$(\sum x)^{2}=229.55^{2}=52693.2025$
$m=\frac{100530.474 - 103991.0145}{53274.0738 - 52693.2025}=\frac{- 3460.5405}{580.8713}\approx - 5.957$

Paso 3: Calcular intercepto $b$

La fórmula para el intercepto $b$ es $b=\overline{y}-m\overline{x}$, donde $\overline{x}=\frac{\sum x}{n}$ y $\overline{y}=\frac{\sum y}{n}$
$\overline{x}=\frac{229.55}{6}\approx38.258$
$\overline{y}=\frac{452.59}{6}\approx75.432$
$b = 75.432-(-5.957)\times38.258$
$b = 75.432 + 5.957\times38.258$
$b = 75.432+227.803$
$b\approx303.235$

Respuesta:

$y=-5.957x + 303.235$

Answer:

Explicación:

Paso 1: Calcular sumatorias

Sean $n = 6$ los pares de datos.
Calcular $\sum x$, $\sum y$, $\sum xy$, $\sum x^{2}$.
$\sum x=29.90 + 33.45+33.78+38.92+42.35+49.15 = 229.55$
$\sum y=89.21 + 85.36+71.41+78.65+76.78+51.18 = 452.59$
$\sum xy=(29.90\times89.21)+(33.45\times85.36)+(33.78\times71.41)+(38.92\times78.65)+(42.35\times76.78)+(49.15\times51.18)$
$=2667.379+2855.292+2413.22+3061.058+3243.533+2515.597 = 16755.079$
$\sum x^{2}=29.90^{2}+33.45^{2}+33.78^{2}+38.92^{2}+42.35^{2}+49.15^{2}$
$=894.01+1119.9025+1141.0884+1514.7664+1793.5225+2415.7225 = 8879.0123$

Paso 2: Calcular pendiente $m$

La fórmula para la pendiente $m$ de la recta de regresión lineal es $m=\frac{n\sum xy-\sum x\sum y}{n\sum x^{2}-(\sum x)^{2}}$
$n\sum xy = 6\times16755.079 = 100530.474$
$\sum x\sum y=229.55\times452.59 = 103991.0145$
$n\sum x^{2}=6\times8879.0123 = 53274.0738$
$(\sum x)^{2}=229.55^{2}=52693.2025$
$m=\frac{100530.474 - 103991.0145}{53274.0738 - 52693.2025}=\frac{- 3460.5405}{580.8713}\approx - 5.957$

Paso 3: Calcular intercepto $b$

La fórmula para el intercepto $b$ es $b=\overline{y}-m\overline{x}$, donde $\overline{x}=\frac{\sum x}{n}$ y $\overline{y}=\frac{\sum y}{n}$
$\overline{x}=\frac{229.55}{6}\approx38.258$
$\overline{y}=\frac{452.59}{6}\approx75.432$
$b = 75.432-(-5.957)\times38.258$
$b = 75.432 + 5.957\times38.258$
$b = 75.432+227.803$
$b\approx303.235$

Respuesta:

$y=-5.957x + 303.235$