Question

A random sample is selected from a population with mean \(\mu=100\) and standard deviation \(\sigma=10 .\) Determine the mean and standard deviation of the \(\bar{x}\) sampling distribution for each of the following sample sizes: a. \(n=9\) b. \(n=15\) c. \(n=36\) d. \(n=50\) c. \(n=100\) f. \(n=400\)

Short answer

For each sample size \(n=9, 15, 36, 50, 100, 400\), the mean of the sampling distribution remains the same at 100, while the standard deviation of the sampling distribution decreases as the sample size increases, and is equal to 3.33, 2.58, 1.67, 1.41, 1.00, 0.50 respectively.

Step-by-step solution

Calculate the mean of the sampling distribution

The mean of the sampling distribution, also known as expected value, which is always equal to the population mean. Therefore, for all sample sizes, the mean \(\mu_{\bar{x}} = 100\).

Calculate the standard deviation of the sampling distribution for \(n=9\)

Using the formula \(\sigma_{\bar{x}} = \sigma / \sqrt{n}\), where \(\sigma = 10\) and \(n = 9\), we find that \(\sigma_{\bar{x}} = 10 / \sqrt{9} = 10 / 3 = 3.33 (to two decimal places)\).

Repeat step 2 for \(n=15, 36, 50, 100, 400\)

Using the same formula \(\sigma_{\bar{x}} = \sigma / \sqrt{n}\) as in step 2, calculate the standard deviation of the sampling distribution for the following sample sizes: \(n = 15, 36, 50, 100, 400\). Rounded to two decimal places, the results are \(\sigma_{\bar{x}} = 2.58, 1.67, 1.41, 1.00, 0.50\), respectively.