Answer :
To find [tex]\(\mu_x\)[/tex] (the mean of the sampling distribution of the sample mean) and [tex]\(\sigma_x\)[/tex] (the standard deviation of the sampling distribution of the sample mean), follow these steps:
1. List the sample means:
We have the sample means given in the table:
- Sample 1: 2.00
- Sample 2: 2.50
- Sample 3: 3.00
- Sample 4: 2.50
- Sample 5: 3.00
- Sample 6: 3.50
- Sample 7: 3.00
- Sample 8: 3.50
- Sample 9: 4.00
2. Calculate the mean of the sample means ([tex]\(\mu_x\)[/tex]):
[tex]\[ \mu_x = \frac{\sum \bar{x}}{n} \][/tex]
[tex]\(\sum \bar{x}\)[/tex] is the sum of all sample means.
Calculate [tex]\(\sum \bar{x}\)[/tex]:
[tex]\[ 2.00 + 2.50 + 3.00 + 2.50 + 3.00 + 3.50 + 3.00 + 3.50 + 4.00 = 27.00 \][/tex]
Now, divide by the number of samples [tex]\(n = 9\)[/tex]:
[tex]\[ \mu_x = \frac{27.00}{9} = 3.00 \][/tex]
3. Calculate the standard deviation of the sample means ([tex]\(\sigma_x\)[/tex]):
[tex]\[ \sigma_x = \sqrt{\frac{\sum (\bar{x} - \mu_x)^2}{n}} \][/tex]
Where [tex]\(\sum (\bar{x} - \mu_x)^2\)[/tex] is the sum of the squared differences between each sample mean and [tex]\(\mu_x\)[/tex].
First, find each [tex]\(\bar{x} - \mu_x\)[/tex]:
[tex]\[ (2.00 - 3.00), (2.50 - 3.00), (3.00 - 3.00), (2.50 - 3.00), (3.00 - 3.00), (3.50 - 3.00), (3.00 - 3.00), (3.50 - 3.00), (4.00 - 3.00) \][/tex]
Which simplifies to:
[tex]\[ -1.00, -0.50, 0.00, -0.50, 0.00, 0.50, 0.00, 0.50, 1.00 \][/tex]
Now, square each difference:
[tex]\[ (-1.00)^2, (-0.50)^2, (0.00)^2, (-0.50)^2, (0.00)^2, (0.50)^2, (0.00)^2, (0.50)^2, (1.00)^2 \][/tex]
Which gives us:
[tex]\[ 1.00, 0.25, 0.00, 0.25, 0.00, 0.25, 0.00, 0.25, 1.00 \][/tex]
Now, sum these squared differences:
[tex]\[ 1.00 + 0.25 + 0.00 + 0.25 + 0.00 + 0.25 + 0.00 + 0.25 + 1.00 = 3.00 \][/tex]
Now, divide by [tex]\(n\)[/tex]:
[tex]\[ \frac{3.00}{9} = 0.3333 \][/tex]
Finally, take the square root:
[tex]\[ \sigma_x = \sqrt{0.3333} \approx 0.577 \][/tex]
Rounded to two decimal places:
[tex]\[ \sigma_x \approx 0.58 \][/tex]
So, the mean of the sampling distribution of the sample mean [tex]\(\mu_x\)[/tex] is [tex]\(3.00\)[/tex] and the standard deviation of the sampling distribution of the sample mean [tex]\(\sigma_x\)[/tex] is [tex]\(0.58\)[/tex].
1. List the sample means:
We have the sample means given in the table:
- Sample 1: 2.00
- Sample 2: 2.50
- Sample 3: 3.00
- Sample 4: 2.50
- Sample 5: 3.00
- Sample 6: 3.50
- Sample 7: 3.00
- Sample 8: 3.50
- Sample 9: 4.00
2. Calculate the mean of the sample means ([tex]\(\mu_x\)[/tex]):
[tex]\[ \mu_x = \frac{\sum \bar{x}}{n} \][/tex]
[tex]\(\sum \bar{x}\)[/tex] is the sum of all sample means.
Calculate [tex]\(\sum \bar{x}\)[/tex]:
[tex]\[ 2.00 + 2.50 + 3.00 + 2.50 + 3.00 + 3.50 + 3.00 + 3.50 + 4.00 = 27.00 \][/tex]
Now, divide by the number of samples [tex]\(n = 9\)[/tex]:
[tex]\[ \mu_x = \frac{27.00}{9} = 3.00 \][/tex]
3. Calculate the standard deviation of the sample means ([tex]\(\sigma_x\)[/tex]):
[tex]\[ \sigma_x = \sqrt{\frac{\sum (\bar{x} - \mu_x)^2}{n}} \][/tex]
Where [tex]\(\sum (\bar{x} - \mu_x)^2\)[/tex] is the sum of the squared differences between each sample mean and [tex]\(\mu_x\)[/tex].
First, find each [tex]\(\bar{x} - \mu_x\)[/tex]:
[tex]\[ (2.00 - 3.00), (2.50 - 3.00), (3.00 - 3.00), (2.50 - 3.00), (3.00 - 3.00), (3.50 - 3.00), (3.00 - 3.00), (3.50 - 3.00), (4.00 - 3.00) \][/tex]
Which simplifies to:
[tex]\[ -1.00, -0.50, 0.00, -0.50, 0.00, 0.50, 0.00, 0.50, 1.00 \][/tex]
Now, square each difference:
[tex]\[ (-1.00)^2, (-0.50)^2, (0.00)^2, (-0.50)^2, (0.00)^2, (0.50)^2, (0.00)^2, (0.50)^2, (1.00)^2 \][/tex]
Which gives us:
[tex]\[ 1.00, 0.25, 0.00, 0.25, 0.00, 0.25, 0.00, 0.25, 1.00 \][/tex]
Now, sum these squared differences:
[tex]\[ 1.00 + 0.25 + 0.00 + 0.25 + 0.00 + 0.25 + 0.00 + 0.25 + 1.00 = 3.00 \][/tex]
Now, divide by [tex]\(n\)[/tex]:
[tex]\[ \frac{3.00}{9} = 0.3333 \][/tex]
Finally, take the square root:
[tex]\[ \sigma_x = \sqrt{0.3333} \approx 0.577 \][/tex]
Rounded to two decimal places:
[tex]\[ \sigma_x \approx 0.58 \][/tex]
So, the mean of the sampling distribution of the sample mean [tex]\(\mu_x\)[/tex] is [tex]\(3.00\)[/tex] and the standard deviation of the sampling distribution of the sample mean [tex]\(\sigma_x\)[/tex] is [tex]\(0.58\)[/tex].
