Question

Use 6.84 days as the planned value for the standard deviation ASSUME 95% confidence in what size the sample must be to have a margin of error of 1.5 days? If the precision statement was made with 90% confidence, what size sample should be made to have a margin of error of two days?

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Answer to a math question Use 6.84 days as the planned value for the standard deviation ASSUME 95% confidence in what size the sample must be to have a margin of error of 1.5 days? If the precision statement was made with 90% confidence, what size sample should be made to have a margin of error of two days?

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Eliseo
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Para determinar el tamaño de muestra requerido para un margen de error (E), un nivel de confianza y una desviación estándar (σ) determinados, utilizamos la fórmula para el tamaño de muestra n: n = \left(\frac{Z \cdot \sigma}{E}\right)^2 dónde: - Z es el valor Z correspondiente al nivel de confianza deseado, - \sigma es la desviación estándar, - E es el margen de error. ### Nivel de confianza del 95 % con un margen de error de 1,5 días Para un nivel de confianza del 95%, el valor Z es aproximadamente 1,96. Dado: - \sigma = 6,84 días, - E = 1,5 días. n = \left(\frac{1.96 \cdot 6.84}{1.5}\right)^2 n = \izquierda(\frac{13.4064}{1.5}\derecha)^2 n = \izquierda(8.9376\derecha)^2 n \aproximadamente 79,88 Como el tamaño de la muestra debe ser un número entero, redondeamos hacia arriba: n \aproximadamente 80 Por lo tanto, se necesita un tamaño de muestra de 80 para tener un margen de error de 1,5 días con un 95% de confianza. ### Nivel de confianza del 90 % con un margen de error de 2 días Para un nivel de confianza del 90%, el valor Z es aproximadamente 1,645. Dado: - \sigma = 6,84 días, - E = 2 días. n = \left(\frac{1.645 \cdot 6.84}{2}\right)^2 n = \izquierda(\frac{11.2518}{2}\derecha)^2 n = \izquierda(5.6259\derecha)^2 n \aproximadamente 31,65 Como el tamaño de la muestra debe ser un número entero, redondeamos hacia arriba: n \aproximadamente 32 Entonces, se necesita un tamaño de muestra de 32 para tener un margen de error de 2 días con un 90% de confianza.

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