Welcome to Westonci.ca, your ultimate destination for finding answers to a wide range of questions from experts. Our platform connects you with professionals ready to provide precise answers to all your questions in various areas of expertise. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
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].
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.
