Question

a certain population has mean \\( \mu = 100 \\), and standard deviation \\( \sigma = 17 \\).
if you choose srss of size \\( n \\), what will be the mean and standard deviation of the sampling distribution for the sample mean, \\( \overline { x } \\) ?
a) \\( n = 4 \\),
\\( \mu _ { \overline { x } } = \\)
\\( \sigma _ { \overline { x } } = \\)
b) \\( n = 16 \\),
\\( \mu _ { \overline { x } } = \\)
\\( \sigma _ { \overline { x } } = \\)
c) \\( n = 100 \\),
\\( \mu _ { \overline { x } } = 17 \\)
\\( \sigma _ { \overline { x } } = \\)
d) if the population distribution from which the samples is chosen is skewed, which sampling distribution will have approximately a normal distribution?

Explanation:

Step1: Recall Sampling Distribution Properties

For the sampling distribution of the sample mean \(\bar{x}\), the mean \(\mu_{\bar{x}}\) is equal to the population mean \(\mu\), and the standard deviation (standard error) \(\sigma_{\bar{x}}\) is given by \(\frac{\sigma}{\sqrt{n}}\), where \(\sigma\) is the population standard deviation and \(n\) is the sample size.

Step2: Solve for Part A (\(n = 4\))

• Mean: \(\mu_{\bar{x}}=\mu = 100\)
• Standard Deviation: \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{17}{\sqrt{4}}=\frac{17}{2} = 8.5\)

Step3: Solve for Part B (\(n = 16\))

• Mean: \(\mu_{\bar{x}}=\mu = 100\)
• Standard Deviation: \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{17}{\sqrt{16}}=\frac{17}{4}=4.25\)

Step4: Solve for Part C (\(n = 100\))

• Mean: \(\mu_{\bar{x}}=\mu = 100\) (the previous answer of 17 was incorrect)
• Standard Deviation: \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{17}{\sqrt{100}}=\frac{17}{10} = 1.7\)

Part A:

\(\mu_{\bar{x}}=\boldsymbol{100}\), \(\sigma_{\bar{x}}=\boldsymbol{8.5}\)

Part B:

\(\mu_{\bar{x}}=\boldsymbol{100}\), \(\sigma_{\bar{x}}=\boldsymbol{4.25}\)

Part C:

\(\mu_{\bar{x}}=\boldsymbol{100}\), \(\sigma_{\bar{x}}=\boldsymbol{1.7}\)

Part D:

By the Central Limit Theorem, if the population is not normal, the sampling distribution of \(\bar{x}\) will be approximately normal when the sample size \(n\) is large (usually \(n\geq30\)). Among the given sample sizes (\(n = 4\), \(n = 16\), \(n = 100\)), \(n = 100\) is the largest and satisfies \(n\geq30\), so the sampling distribution for \(n = 100\) will be approximately normal.

Answer:

Step1: Recall Sampling Distribution Properties

For the sampling distribution of the sample mean \(\bar{x}\), the mean \(\mu_{\bar{x}}\) is equal to the population mean \(\mu\), and the standard deviation (standard error) \(\sigma_{\bar{x}}\) is given by \(\frac{\sigma}{\sqrt{n}}\), where \(\sigma\) is the population standard deviation and \(n\) is the sample size.

Step2: Solve for Part A (\(n = 4\))

• Mean: \(\mu_{\bar{x}}=\mu = 100\)
• Standard Deviation: \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{17}{\sqrt{4}}=\frac{17}{2} = 8.5\)

Step3: Solve for Part B (\(n = 16\))

• Mean: \(\mu_{\bar{x}}=\mu = 100\)
• Standard Deviation: \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{17}{\sqrt{16}}=\frac{17}{4}=4.25\)

Step4: Solve for Part C (\(n = 100\))

• Mean: \(\mu_{\bar{x}}=\mu = 100\) (the previous answer of 17 was incorrect)
• Standard Deviation: \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{17}{\sqrt{100}}=\frac{17}{10} = 1.7\)

Part A:

\(\mu_{\bar{x}}=\boldsymbol{100}\), \(\sigma_{\bar{x}}=\boldsymbol{8.5}\)

Part B:

\(\mu_{\bar{x}}=\boldsymbol{100}\), \(\sigma_{\bar{x}}=\boldsymbol{4.25}\)

Part C:

\(\mu_{\bar{x}}=\boldsymbol{100}\), \(\sigma_{\bar{x}}=\boldsymbol{1.7}\)

Part D:

By the Central Limit Theorem, if the population is not normal, the sampling distribution of \(\bar{x}\) will be approximately normal when the sample size \(n\) is large (usually \(n\geq30\)). Among the given sample sizes (\(n = 4\), \(n = 16\), \(n = 100\)), \(n = 100\) is the largest and satisfies \(n\geq30\), so the sampling distribution for \(n = 100\) will be approximately normal.