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Samples
Método de mínimos cuadrados
Nombre del estudiante
Institución
X 8 2 11 6 5 4 12 9 6 1
y 3 10 3 6 8 12 1 4 9 14
Solución
Media de valores X = (8+2+11+6+5+4+12+9+6+1) /10 =64/10=6.4
Media de los valores de y = (3+10+3+6+8+12+1+4+9+14) /10 =70/10=7
La ecuación de la línea recta es y = bx + a
Las ecuaciones normales son
∑y = b∑x + an
∑xy = b∑x 2 + a∑x
X Y X2 XY
8 3 64 24
2 10 4 20
11 3 121 33
6 6 36 36
5 8 25 40
4 12 16 48
12 1 144 12
9 4 81 36
6 9 36 54
1 14 1 14
∑x = 64 ∑Y=70 ∑x2 _ _ = 528 ∑XY = 317
Sustituyendo estos valores en las ecuaciones normales
64b + 10a =70
64a + 528b =317
Resolviendo la ecuación
b= -1.1064
a= -0.08108
Las ecuaciones se convierten
Y=a + bx; Y= -0.08108 - 1.1064x
Referencias
Fowler, J., Jarvis, P. y Chevannes , M. (2021). Estadísticas prácticas para la enfermería y el cuidado de la salud. John Wiley & Sons.
Grove, SK y Cipher, DJ (2019). Estadísticas para la investigación en enfermería-e-book: un libro de trabajo para la práctica basada en la evidencia. Elsevier Ciencias de la Salud.
Kim, M., Mallory, C. y Valerio, T. (2020). Estadísticas para la práctica basada en la evidencia en enfermería. Editores Jones & Bartlett.