Question

suppose x has a normal distribution with mean μ = 23 and standard deviation σ = 9. describe the distribution of x values for sample size n = 4. describe the distribution of x values for sample size n = 16. describe the distribution of x values for sample size n = 100. how do the x distributions compare for the various samples sizes? the means are the same, but the standard deviations are decreasing with increasing sample size. the standard deviations are the same, but the means are increasing with increasing sample size. the means are the same, but the standard deviations are increasing with increasing sample size. the means and standard deviations are the same regardless of sample size. the standard deviations are the same, but the means are decreasing with increasing sample size.

Explanation:

Step1: Mean of sample mean

The mean of the sampling distribution of the sample mean \(\bar{x}\) is equal to the population mean \(\mu\). So, for all sample sizes \(n = 4\), \(n=16\), and \(n = 100\), \(\mu_{\bar{x}}=\mu = 23\)

Step2: Standard deviation of sample mean

The formula for the standard deviation of the sampling distribution of the sample mean (also known as the standard error) is \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\)

• For \(n = 4\):

\(\sigma_{\bar{x}}=\frac{9}{\sqrt{4}}=\frac{9}{2}=4.5\)

• For \(n = 16\):

\(\sigma_{\bar{x}}=\frac{9}{\sqrt{16}}=\frac{9}{4} = 2.25\)

• For \(n = 100\):

\(\sigma_{\bar{x}}=\frac{9}{\sqrt{100}}=\frac{9}{10}=0.9\)

Step3: Comparing distributions

As \(n\) increases (\(4<16<100\)), \(\mu_{\bar{x}}\) remains constant (\(\mu_{\bar{x}} = 23\) for all \(n\)) and \(\sigma_{\bar{x}}\) decreases (\(4.5>2.25>0.9\))

Answer:

• For \(n = 4\): \(\mu_{\bar{x}}=23\), \(\sigma_{\bar{x}} = 4.5\)
• For \(n = 16\): \(\mu_{\bar{x}}=23\), \(\sigma_{\bar{x}}=2.25\)
• For \(n = 100\): \(\mu_{\bar{x}}=23\), \(\sigma_{\bar{x}}=0.9\)
• The means are the same, but the standard deviations are decreasing with increasing sample size.