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Eva8 [605]
3 years ago
11

Ayudenmee plis

Mathematics
1 answer:
inessss [21]3 years ago
7 0

Answer:

El intervalo, con un nivel de confianza de 99%, para estimar el sueldo medio poblacional de los trabajadores es de ($1807, $2193).

Step-by-step explanation:

El primero paso es el encuentro de lo \alpha, que es la subtracion de 1 por lo nivel de confianza e divididos por 2. Entonces:

\alpha = \frac{1 - 0.99}{2} = 0.005

Ahora, buscamos a z en la tabela Z de manera que z tienga uno p-value de 1 - \alpha.

Entonces, z con uno p-value de 1 - 0.005 = 0.995, o sea, z = 2.575.

La margen de error es:

M = z\frac{\sigma}{\sqrt{n}}

En que \sigma es la desviación estándar y n es el tamaño de la muestra.

σ=$600. Si se tomó una muestra de 64 trabajadores

Entonces \sigma = 600, n = 64

Margen de error:

M = z\frac{\sigma}{\sqrt{n}}

M = 2.575\frac{600}{\sqrt{64}}

M = 193

El limite inferior es la subtración de la media de la muestra por la margen de error. Entonces 2000 - 193 = $1807.

El limite inferior es la suma de la media de la muestra con la margen de error. Entonces 2000 + 193 = $2193.

El intervalo, con un nivel de confianza de 99%, para estimar el sueldo medio poblacional de los trabajadores es de ($1807, $2193).

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