Question

for the following situation, find the mean and standard deviation of the population. list all samples (with replacement) of the given size from that population and find the mean of each. find the mean and standard deviation of the sampling distribution and compare them with the mean and standard deviation of the population. the word counts of five essays are 509, 627, 552, 613, and 570. use a sample size of 2. find the standard deviation of the sampling distribution. since there are nine different samples, n is equal to 25. refer to table above for the sample means.

σ = \sqrt{\frac{\sum(x - \mu)^{2}}{n}} = \sqrt{\frac{(509 - 574.2)^{2}+(568 - 574.2)^{2}+...+(570 - 574.2)^{2}}{25}} = \square (round to two decimal places as needed.)

Explanation:

Step1: Calculate each term in the numerator

First, calculate \((509 - 574.2)^{2}=(- 65.2)^{2}=4251.04\)
\((568 - 574.2)^{2}=(-6.2)^{2}=38.44\)
Since there are 25 terms (samples), but we can also use the formula for the standard deviation of the sampling distribution of the sample mean \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\)

First, find the population mean \(\mu=\frac{509 + 627+552+613+570}{5}=\frac{2871}{5}=574.2\)

Population standard deviation \(\sigma=\sqrt{\frac{(509 - 574.2)^{2}+(627 - 574.2)^{2}+(552 - 574.2)^{2}+(613 - 574.2)^{2}+(570 - 574.2)^{2}}{5}}\)

\((627 - 574.2)^{2}=(52.8)^{2}=2787.84\), \((552 - 574.2)^{2}=(-22.2)^{2}=492.84\), \((613 - 574.2)^{2}=(38.8)^{2}=1505.44\), \((570 - 574.2)^{2}=(-4.2)^{2}=17.64\)

\(\sigma=\sqrt{\frac{4251.04 + 2787.84+492.84+1505.44+17.64}{5}}=\sqrt{\frac{9054.8}{5}}=\sqrt{1810.96}\approx42.56\)

For sampling distribution of sample mean with \(n = 2\), \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{42.56}{\sqrt{2}}\approx29.96\)

If we calculate directly from the formula \(\sigma=\sqrt{\frac{\sum(x-\mu)^{2}}{N}}\)

Let's assume we calculate each \((x-\mu)^{2}\) for 25 samples:

The number of times each value is repeated:
For example, when we take samples with replacement of size \(n = 2\) from a population of size \(N = 5\), the total number of samples is \(N^{n}=25\)

The mean of the sampling distribution of the sample mean \(\mu_{\bar{x}}=\mu = 574.2\)

If we calculate \(\sum(x-\mu)^{2}\) for all 25 sample - means:

Another way:
The formula for the standard deviation of the sampling distribution of the sample mean (when sampling with replacement) is \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\)

We know \(\sigma\approx42.56\) and \(n = 2\)

Answer:

\(29.96\)