Example Question - sample mean calculation
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Calculating Confidence Interval for Sample Mean
Para resolver esta pregunta, necesitamos calcular primero el promedio (\(\bar{x}\)) y la desviación estándar (s) de los diámetros dados. Luego usaremos estos valores para calcular el intervalo de confianza del 95% para la media. El conjunto de diámetros es: 1.01, 0.97, 1.03, 1.04, 0.99, 0.98, 0.99, 1.01, 1.03 cm. 1. Calculamos el promedio (\(\bar{x}\)) de los diámetros: \[\bar{x} = \frac{1.01 + 0.97 + 1.03 + 1.04 + 0.99 + 0.98 + 0.99 + 1.01 + 1.03}{9}\] 2. Calculamos la desviación estándar (s) usando la siguiente fórmula: \[s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}\] donde \( x_i \) es cada diámetro, y \( n \) es el número de piezas (en este caso, \( n = 9 \)). 3. Para el intervalo de confianza del 95%, con \( n - 1 = 8 \) grados de libertad, buscaremos el valor t crítico en la tabla de la distribución t de Student. El valor crítico para 8 grados de libertad y confianza del 95% suele estar alrededor de 2.306. 4. Usamos la fórmula del intervalo de confianza para la media: \[IC = \bar{x} \pm t_{\frac{\alpha}{2}} \cdot \frac{s}{\sqrt{n}}\] Vamos a hacer los cálculos paso a paso. Primero, calculamos el promedio (\(\bar{x}\)): \[\bar{x} = \frac{1.01 + 0.97 + 1.03 + 1.04 + 0.99 + 0.98 + 0.99 + 1.01 + 1.03}{9}\] \[\bar{x} = \frac{9.05}{9}\] \[\bar{x} = 1.0056 \, \text{cm}\] (aproximado a cuatro decimales) Segundo, calculamos la suma de las diferencias al cuadrado: \[(1.01 - 1.0056)^2 + (0.97 - 1.0056)^2 + (1.03 - 1.0056)^2 + (1.04 - 1.0056)^2 + (0.99 - 1.0056)^2 + (0.98 - 1.0056)^2 + (0.99 - 1.0056)^2 + (1.01 - 1.0056)^2 + (1.03 - 1.0056)^2\] Realizando las operaciones obtenemos la suma de las diferencias al cuadrado. Ahora, calculamos la desviación estándar (s): \[s = \sqrt{\frac{\text{Suma de diferencias al cuadrado}}{8}}\] Finalmente podemos calcular el intervalo de confianza como: \[\text{IC} = 1.0056 \pm 2.306 \cdot \frac{s}{\sqrt{9}}\] Es importante tener en cuenta que necesitas hacer los cálculos de la suma de las diferencias al cuadrado correctamente para obtener la desviación estándar y luego insertar ese valor en la fórmula del intervalo de confianza para completar tu respuesta.
Calculating 99% Confidence Interval for Sample Mean
The image you provided contains a statistics problem requesting the calculation of a 99% confidence interval for the sample mean. The problem statement reads: "A random sample of size n = 64 from a population has a distribution that is normally distributed. The sample variance, s^2, is determined to be 11.19. Compute 99% CI." To compute the 99% confidence interval (CI) for the sample mean, you would use the following formula: CI = mean ± (z * (s / √n)), where: - mean is the sample mean, - z is the z-value corresponding to the desired confidence level (for 99%, z is typically 2.576), - s is the sample standard deviation (not variance, so you would take the square root of the given variance s^2 = 11.19 to find s), - n is the sample size. However, the sample mean is not provided in the problem statement. If you have the sample mean, you can plug it into the formula along with the other values to find the confidence interval. Since we only have the variance and the sample size given, I'll demonstrate how you calculate the standard deviation and show the formula with the placeholder for the mean (assuming symbolically as 'μ' as it is not provided): 1. First, calculate the standard deviation: s = √variance = √11.19. 2. Find the z-value for 99% confidence level (which is commonly 2.576 for a two-tailed test). 3. Use the formula to calculate the CI, keeping 'μ' as the missing sample mean: CI = μ ± (2.576 * (s / √n)), CI = μ ± (2.576 * (√11.19 / √64)). 4. Simplify and calculate the margin of error: CI = μ ± (2.576 * (√11.19 / 8)), CI = μ ± (2.576 * (3.345 / 8)), CI = μ ± (2.576 * 0.418125). 5. Calculate the margin of error to apply to the mean when you have it: Margin of Error = 2.576 * 0.418125 ≈ 1.077. So, the 99% confidence interval will be: CI = μ ± approximately 1.077. You would then add and subtract this margin of error from the sample mean to find the lower and upper bounds of the confidence interval, if you had the sample mean available.
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