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fredd [130]
3 years ago
10

Una muestra aleatoria si 136 personas de 400 a quien se le aplica un nuevo medicamento experimentaron mejoría Desarrolla un inte

rvalo de confianza del 95% para la verdadera proporción que experimentara alguna mejoría con la medicina
Mathematics
1 answer:
Rudiy273 years ago
3 0

Answer:

Intervalo de confianza

= (0.294, 0.386)

Step-by-step explanation:

La fórmula para el intervalo de confianza para la proporción se da como

p ± z × √p (1 - p) / n

Donde p = x / n

De donde de la pregunta anterior

x = 136 personas

n = 400 personas

p = 136/400

p = 0.34

z = puntuación z del intervalo de confianza del 95% = 1.96

Por lo tanto,

Intervalo de confianza =

0.34 ± 1.96 × √0.34 (1 - 0.34) / 400

= 0.34 ± 1.96 × √0.34 × 0.66/400

= 0.34 ± 1.96 × √0.000561

= 0.34 ± 1.96 × 0.0236854386

= 0.34 ± 0.0464234596

Intervalo de confianza

= 0.34 - 0.0464234596

= 0.2935765404

= 0.34 + 0.0464234596

= 0.3864234596

Hence:Intervalo de confianza

= (0.294, 0.386)

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