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MODULO 5
Actividad 2
Keishlany Caraballo
Universidad Ana G. Méndez
STAT 300
05/04/2024
MODULO 5
Actividad 1.
1.Se toma una muestra de 49 observaciones de una población
normal con una desviación estándar de 10. La media de la muestra es de 55. Determine el intervalo de confianza de 99% de la media poblacional. Muestra-49 Desviación estándar-10 Media-55 Z-99% Negativo =55-2.575*6/√49 =55-
2.575* 6/7 =55-2.575*0.86 =55-22.1 =32.9 Positivo =55+2.575* 6/√49 =55+2.575* 6/7 =55+2.575*0.86 =55+22.1 =77.1 El 99% de la media de una población normal
esta entre un 32.9 y 77.1.
2.
Una empresa de investigación llevó a cabo una encuesta para
determinar la cantidad media que los fumadores gastan en cigarrillos durante una semana. La empresa descubrió que la distribución de cantidades que gastan por semana tendía a seguir una distribución normal, con una desviación estándar de $5. Una muestra de 49 fumadores reveló que la media de gastos semanales era de $40. ¿Cuál es el estimador puntual de la media de la población? Explique su respuesta. Con el nivel de confianza de 95%, determine el intervalo de confianza de la media poblacional. Explique su significado.
Limite superior del intervalo: 21,84
Limite inferior del intervalo:18,16
Explicación:
Limite superior del intervalo: 655.93
Limite inferior del intervalo: 601.67
Explicación paso a paso:
MODULO 5
Intervalo de confianza
Datos:
μ = $20
σ = $5
n= 49
Nivel de confianza 99%
Nivel de significancia α= 1-0,99 = 0,01
Zα/2= 0,01/2 =0,005 = -2,58 según la tabla de distribución Normal
Intervalo de confianza:
(μ)99% = μ ± Zα/2 * σ /√n
(μ )99% =20± 2,58*5/√49
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MODULO 5
(μ)99% =20 ±1,84
Limite superior del intervalo: 21,84
Limite inferior del intervalo:18,16
3.
Una empresa que se dedica al cuidado de envejecientes contempla ofrecer de forma conjunta servicio de cuido de niños para sus empleados. Como parte del estudio de viabilidad del proyecto, desean calcular el costo medio semanal por el cuidado de los niños. Una muestra de 10 empleados que recurren este servicio revela las siguientes cantidades gastadas la semana pasada. 100, 107, 90, 97, 92, 105, 101, 91, 99, 104. Construya el intervalo de confianza de 90% de la media poblacional. Interprete el resultado.
Tenemos los gastos de los empleados:107 92 97 95 105 101 91 99 95 104Media = 107+92+97+95+105+101+91+99+95+104 /10Media = 98,6.Desviación estándar = √∑ (X -μ) ²/n = √234,12/10 = 4,8μ = 98,6σ= 4,8n = 10Confianza 90% = 0,90α = 1-0.90 = 0.1Zα/2 = 0.1/2 = 0.05 = 0.51994[μ]1-α = μ +- Zα/2 σ/√n: [μ]90% = 98.6 +- 0.52* 4.8/√10: [μ]90% = 98.6 +- 0.79: [μ]90% =97.81Por lo tanto, el costo medio semanal por el cuidado de niños es de 97.81
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