Question

according to a recent publication, the mean price of new mobile homes is $62,200. assume a standard deviation of $7300. let \\( \overline { x } \\) denote the mean price of a sample of new mobile homes.
a. for samples of size 100, find the mean and standard deviation of \\( \overline { x } \\). interpret your results in words.
b. repeat part (a) with \\( n = 200 \\).
a. for the 100 mobile homes sampled, the mean and standard deviation of are $ and $, respectively.
(round to the nearest cent as needed )
all possible sample mean prices
the prices of the mobile homes

Explanation:

Step1: Calculate the mean of \(\bar{x}\) for \(n = 100\)

By the property of the sampling distribution of the sample mean \(\mu_{\bar{x}}=\mu\). Given \(\mu = 62200\), so \(\mu_{\bar{x}}=62200\).

Step2: Calculate the standard deviation of \(\bar{x}\) for \(n = 100\)

The formula for the standard deviation of the sample mean is \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\). Given \(\sigma = 7300\) and \(n = 100\), then \(\sigma_{\bar{x}}=\frac{7300}{\sqrt{100}}=\frac{7300}{10}=730\).

Step3: Calculate the mean of \(\bar{x}\) for \(n = 200\)

Again, \(\mu_{\bar{x}}=\mu = 62200\) (since the mean of the sampling distribution of the sample mean is equal to the population mean regardless of sample size).

Step4: Calculate the standard deviation of \(\bar{x}\) for \(n = 200\)

Using the formula \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\), with \(\sigma = 7300\) and \(n = 200\), we have \(\sigma_{\bar{x}}=\frac{7300}{\sqrt{200}}\approx\frac{7300}{14.1421}\approx516.2\)

Answer:

a. For all possible sample mean prices (when \(n = 100\)), the mean is \(\$62200\) and the standard deviation is \(\$730\). This means that if we take many samples of size \(n = 100\) from the population of mobile - home prices, the average of the sample means will be \(\$62200\), and the spread (variability) of these sample means around the population mean is \(\$730\).
b. For all possible sample mean prices (when \(n = 200\)), the mean is \(\$62200\) and the standard deviation is approximately \(\$516.2\). As the sample size increases from \(n = 100\) to \(n = 200\), the spread (variability) of the sample means around the population mean decreases.