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shutvik [7]
3 years ago
9

En una muestra de 50 restaurantes de comida rápida, la venta media fue de $3.000, y la desviación estándar, $200. El intervalo d

e confianza del 95% es
Mathematics
1 answer:
Step2247 [10]3 years ago
8 0

Respuesta:

(2945.411; 3054.589)

Explicación paso a paso:

Dado ;

Tamaño de la muestra, n = 50

Media, xbar = 3000

Desviación estándar, s = 200

Nivel de confianza, Zcrítico al 95% = 1,96

El intervalo de confianza se define como:

Xbar ± margen de error

Margen de error = Zcrítico * s / sqrt (n)

Margen de error = 1,96 * 200 / sqrt (50)

Margen de error = 54.589

Límite inferior = (3000 - 54.589) = 2945.411

Límite superior = (3000 + 54.589) = 3054.589

(2945.411; 3054.589)

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