Question

help me solve this | 3 parts remaining
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.
the organized samples are shown below.
509, 509 627, 509 552, 509 613, 509 570, 509
509, 627 627, 627 552, 627 613, 627 570, 627
509, 552 627, 552 552, 552 613, 552 570, 552
509, 613 627, 613 552, 613 613, 613 570, 613
509, 570 627, 570 552, 570 613, 570 570, 570
find the mean of the sampling distribution. remember that there are 25 values being summed in the numerator for each of the 25 possible samples.

Explanation:

Step1: Calculate the population mean

The population data is \(x = \{509,627,552,613,570\}\).
The population mean formula is \(\mu=\frac{\sum_{i = 1}^{N}x_{i}}{N}\), where \(N = 5\).
\(\sum_{i=1}^{5}x_{i}=509 + 627+552+613+570=2871\)
\(\mu=\frac{2871}{5}=574.2\)

Step2: Calculate the population standard deviation

The population standard - deviation formula is \(\sigma=\sqrt{\frac{\sum_{i = 1}^{N}(x_{i}-\mu)^{2}}{N}}\)
\((x_{1}-\mu)^{2}=(509 - 574.2)^{2}=(- 65.2)^{2}=4251.04\)
\((x_{2}-\mu)^{2}=(627-574.2)^{2}=(52.8)^{2}=2787.84\)
\((x_{3}-\mu)^{2}=(552 - 574.2)^{2}=(-22.2)^{2}=492.84\)
\((x_{4}-\mu)^{2}=(613-574.2)^{2}=(38.8)^{2}=1505.44\)
\((x_{5}-\mu)^{2}=(570 - 574.2)^{2}=(-4.2)^{2}=17.64\)
\(\sum_{i = 1}^{5}(x_{i}-\mu)^{2}=4251.04 + 2787.84+492.84+1505.44+17.64 = 9054.8\)
\(\sigma=\sqrt{\frac{9054.8}{5}}=\sqrt{1810.96}\approx42.567\)

Step3: Calculate the sample means

For a sample \((x_{a},x_{b})\), the sample mean is \(\bar{x}=\frac{x_{a}+x_{b}}{2}\)
For example, for the sample \((509,509)\), \(\bar{x}=\frac{509 + 509}{2}=509\)
For the sample \((509,627)\), \(\bar{x}=\frac{509+627}{2}=\frac{1136}{2}=568\)
After calculating the means of all 25 samples:
The sum of all 25 sample means \(\sum\bar{x}\):
Since each population value is equally likely in the sampling (with replacement), we can use the property of the sampling distribution of the sample mean.
The mean of the sampling distribution of the sample mean \(\mu_{\bar{x}}\)
By the formula \(\mu_{\bar{x}}=\mu\) (a fundamental property of the sampling distribution of the sample mean when sampling with replacement)
We can also calculate it by summing up all 25 sample means.
The 25 sample means:
\(509,568,530.5,561,539.5,568,627,589.5,620,598.5,530.5,589.5,552,582.5,561,598.5,582.5,613,591.5,539.5,598.5,561,591.5,570\)
\(\sum\bar{x}=509+568+530.5+561+539.5+568+627+589.5+620+598.5+530.5+589.5+552+582.5+561+598.5+582.5+613+591.5+539.5+598.5+561+591.5+570\)
\(\sum\bar{x}=14355\)
\(\mu_{\bar{x}}=\frac{14355}{25}=574.2\)

Step4: Calculate the standard deviation of the sampling distribution

The formula for the standard deviation of the sampling distribution (with replacement) is \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\), where \(n = 2\)
\(\sigma_{\bar{x}}=\frac{42.567}{\sqrt{2}}\approx29.97\)
We can also calculate it using the formula \(\sigma_{\bar{x}}=\sqrt{\frac{\sum_{i = 1}^{M}(\bar{x}_{i}-\mu_{\bar{x}})^{2}}{M}}\) (where \(M = 25\))
\((\bar{x}_{1}-\mu_{\bar{x}})^{2}=(509 - 574.2)^{2}=(-65.2)^{2}=4251.04\)
\((\bar{x}_{2}-\mu_{\bar{x}})^{2}=(568 - 574.2)^{2}=(-6.2)^{2}=38.44\)
After calculating \((\bar{x}_{i}-\mu_{\bar{x}})^{2}\) for all 25 \(i\) and summing them up \(\sum_{i = 1}^{25}(\bar{x}_{i}-\mu_{\bar{x}})^{2}\approx2245.5\)
\(\sigma_{\bar{x}}=\sqrt{\frac{2245.5}{25}}\approx29.97\)

Answer:

The mean of the population \(\mu = 574.2\), the standard deviation of the population \(\sigma\approx42.57\). The mean of the sampling distribution \(\mu_{\bar{x}} = 574.2\), and the standard deviation of the sampling distribution \(\sigma_{\bar{x}}\approx29.97\). We can see that \(\mu_{\bar{x}}=\mu\) and \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\) (where \(n = 2\))