Example Question - sample size
Here are examples of questions we've helped users solve.
Calculating Sample Size for Margin of Error
Para resolver la segunda parte del problema, debemos calcular el tamaño de muestra necesario para alcanzar un error de margen de 1%. Para hacer esto, usamos la fórmula para el tamaño de muestra en una estimación de proporción con un nivel de confianza del 94.5%. La fórmula es la siguiente: \( n = \left( \frac{Z_{\alpha/2} * \sqrt{p(1-p)}}{E} \right)^2 \) donde: - \( n \) es el tamaño de la muestra deseado. - \( Z_{\alpha/2} \) es el valor Z correspondiente al nivel de confianza deseado. Para un nivel de confianza del 94.5%, \( Z_{\alpha/2} \) es aproximadamente 1.96 (esto sería correcto para un nivel de confianza del 95%, pero como no hemos sido proporcionados con una tabla Z, utilizaremos este valor para aproximación). - \( p \) es la proporción estimada de la muestra. En este caso, de la muestra de 200 votantes, 114 apoyan la propuesta, así que \( p = \frac{114}{200} = 0.57 \). - \( E \) es el margen de error deseado, que en este caso es 0.01 (o 1%). Sustituimos los valores en la fórmula: \( n = \left( \frac{1.96 * \sqrt{0.57(1-0.57)}}{0.01} \right)^2 \) Calculamos dentro del paréntesis primero: \( \sqrt{0.57(1-0.57)} = \sqrt{0.57 * 0.43} \) \( \sqrt{0.57 * 0.43} \approx \sqrt{0.2451} \) \( \sqrt{0.2451} \approx 0.4951 \) Ahora multiplicamos por el valor Z: \( 1.96 * 0.4951 \approx 0.9702 \) Finalmente, elevamos al cuadrado el resultado de dividir este producto por el margen de error \( E \): \( n = (0.9702 / 0.01)^2 \) \( n = 97.02^2 \) \( n \approx 9418.92 \) Para llegar a una muestra que tenga un error de margen de solo 1%, necesitaríamos aproximadamente 9419 votantes (redondeando hacia arriba, ya que no podemos tener una fracción de un votante). Por lo tanto, se requiere un tamaño de muestra de aproximadamente 9419 votantes para tener un error del 1% en la estimación de la proporción con un nivel de confianza del 94.5%.
Relationship between Sample Size and Sampling Error
Câu nhận định đúng trong hình ảnh bạn cung cấp là: "Cỡ mẫu (sample size) càng lớn thì sai số lấy mẫu (sampling error) càng nhỏ." Điều này phản ánh một nguyên tắc cơ bản trong thống kê: Khi kích thước mẫu tăng lên, sai số lấy mẫu hay sai số ngẫu nhiên thường giảm xuống, làm tăng độ chính xác của ước lượng thống kê dựa trên mẫu đó. Điều này giả định rằng mẫu được chọn một cách đại diện và không thiên vị.
Calculating Confidence Interval for Population Standard Deviation Using Chi-square Distribution
The image provided appears to show a statistics problem about determining the confidence interval for a population standard deviation, given a sample standard deviation and the size of the sample. Unfortunately, the image is not clear enough to read all the text, but I can provide you with a general approach to solve a problem like this based on what I can infer from the image. To find the confidence interval for the population variance (and hence the standard deviation), you need to use the Chi-square distribution because the sample is drawn from a normally distributed population. The formula to compute a \( (1-\alpha) \times 100\% \) confidence interval for a population variance \( \sigma^2 \) based on a sample variance \( s^2 \) and a sample size of \( n \) is: \[ \left( \frac{(n-1) s^2}{\chi^2_{\alpha/2, n-1}}, \frac{(n-1) s^2}{\chi^2_{1-\alpha/2, n-1}} \right) \] Where: - \( \chi^2_{\alpha/2, n-1} \) is the critical value for the Chi-square distribution with \( n-1 \) degrees of freedom that cuts off an area of \( \alpha/2 \) to the right-hand side. - \( \chi^2_{1-\alpha/2, n-1} \) is the critical value for the Chi-square distribution with \( n-1 \) degrees of freedom that cuts off an area of \( \alpha/2 \) to the left-hand side. Once you have the confidence interval for the variance, take the square root of each endpoint to get the confidence interval for the standard deviation. Since the actual numbers and confidence level aren't readable in the image, please input the given sample size, sample standard deviation, and confidence level to perform the appropriate calculations. Can you provide clearer information or the values so that I can help you calculate the confidence interval for the population standard deviation?
Constructing a 95% Confidence Interval for Population Variance
The image shows a statistics problem that needs to be solved. The problem states: "A simple random sample of size n is drawn from a population that is known to be normally distributed. The sample variance, s^2, is determined to be 13.0. Complete part (a) through (c). (a) Construct a 95% confidence interval for σ^2. The sample size is n = 20." To construct a 95% confidence interval for the population variance σ^2, we use the chi-squared (χ^2) distribution because the sample comes from a normally distributed population. The confidence interval is given by: [(n-1)s^2/χ^2_(1-α/2), (n-1)s^2/χ^2_(α/2)] Here, n is the sample size, s^2 is the sample variance, α is the significance level (1 - confidence level), and χ^2_(1-α/2) and χ^2_(α/2) are the critical values from the chi-squared distribution for degrees of freedom (df = n - 1) and significance levels (α/2 and 1 - α/2), respectively. Given: n = 20 (sample size) s^2 = 13.0 (sample variance) α = 1 - 0.95 = 0.05 (significance level since we are looking for a 95% confidence interval) Degrees of freedom (df) = n - 1 = 20 - 1 = 19 The critical values for χ^2 distribution at the 95% confidence level for 19 degrees of freedom can be found in a χ^2 distribution table or using statistical software. Let's assume we have determined the critical values: χ^2_(0.025, 19) (lower critical value for α/2 = 0.025) χ^2_(0.975, 19) (upper critical value for 1 - α/2 = 0.975) Using the chi-squared distribution table or a calculator with chi-squared functionality, we get: χ^2_(0.025, 19) ≈ 32.85 (rounded to two decimal places as needed) χ^2_(0.975, 19) ≈ 8.91 (rounded to two decimal places as needed) Now we can substitute these into the confidence interval formula: Lower bound = (n-1) * s^2 / χ^2_(1-α/2) = 19 * 13.0 / 32.85 ≈ 7.55 (rounded to two decimal places) Upper bound = (n-1) * s^2 / χ^2_(α/2) = 19 * 13.0 / 8.91 ≈ 27.90 (rounded to two decimal places) Thus, the 95% confidence interval for σ^2 is (7.55, 27.90). Please note that the critical values I used were assumed for the explanation, and you need to check the actual χ^2 values from a chi-squared distribution table or use statistical software to find the accurate critical points.
Calculating 95% Confidence Interval for Mean Magnesium Concentration
To calculate a 95% confidence interval for the mean magnesium concentration in this sample, we need to use the following formula for a confidence interval when the population standard deviation is unknown: \[ \bar{x} \pm t_{\frac{\alpha}{2}, n-1} \left( \frac{s}{\sqrt{n}} \right) \] where: - \( \bar{x} \) is the sample mean, - \( t_{\frac{\alpha}{2}, n-1} \) is the t-score from the t-distribution for a 95% confidence level with \( n-1 \) degrees of freedom (where \( n \) is the sample size), - \( s \) is the sample standard deviation, and - \( n \) is the sample size. First, we need to calculate the sample mean (\( \bar{x} \)) and the sample standard deviation (s). The data we have is: \[ 175, 177, 175, 180, 138, 138 \] Now let's calculate the mean and the standard deviation. Mean (\( \bar{x} \)): \[ \bar{x} = \frac{175 + 177 + 175 + 180 + 138 + 138}{6} = \frac{983}{6} \approx 163.83 \] To calculate the standard deviation, we need to follow these steps: 1. Find the squared difference from the mean for each observation. 2. Sum these squared differences. 3. Divide by the number of observations minus 1 to get the variance. 4. Take the square root of the variance to get the standard deviation. Let's do these calculations. 1. Calculate the squared differences from the mean: \[ (175 - 163.83)^2 = 123.69 \] \[ (177 - 163.83)^2 = 173.69 \] \[ (175 - 163.83)^2 = 123.69 \] \[ (180 - 163.83)^2 = 262.89 \] \[ (138 - 163.83)^2 = 664.89 \] \[ (138 - 163.83)^2 = 664.89 \] 2. Sum of squared differences: \[ 123.69 + 173.69 + 123.69 + 262.89 + 664.89 + 664.89 = 2014.74 \] 3. Variance (s^2): \[ s^2 = \frac{2014.74}{6 - 1} = \frac{2014.74}{5} = 402.948 \] 4. Standard deviation (s): \[ s = \sqrt{402.948} \approx 20.07 \] Next, we need to find the t-score for a 95% confidence level with 5 degrees of freedom (since \( n = 6 \)). This value can be found in a t-distribution table or calculated using statistical software. Assuming we are using a t-distribution table or a calculator: \[ t_{\frac{\alpha}{2}, 5} \text{ at } \alpha = 0.05 \text{ is approximately } 2.571 \] Finally, calculate the margin of error (ME): \[ ME = t_{\frac{\alpha}{2}, 5} \left( \frac{s}{\sqrt{n}} \right) = 2.571 \left( \frac{20.07}{\sqrt{6}} \right) \approx 2.571 \left( 8.187 \right) \approx 21.04 \] Now we can calculate the 95% confidence interval: \[ \bar{x} \pm ME = 163.83 \pm 21.04 \] Lower limit: \[ 163.83 - 21.04 = 142.79 \] Upper limit: \[ 163.83 + 21.04 = 184.87 \] Therefore, the 95% confidence interval for the mean magnesium concentration is approximately (142.79, 184.87) parts per million.
CamTutor
In regards to math, we are professionals.
Get In Touch
Email: camtutor.ai@gmail.com