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\begin{tabular}{|c|c|c|c|c|}
\hline Index & Sample & Scores & \multicolumn{2}{|r|}{Sample mean, [tex]$\bar{x}$[/tex]} \\
\hline 1 & Jenny, Jenny & 2,2 & \multicolumn{2}{|r|}{2.00} \\
\hline 2 & Jenny, Jose & 2,3 & \multicolumn{2}{|r|}{2.50} \\
\hline 3 & Jenny, Lisa & 2,4 & \multicolumn{2}{|r|}{3.00} \\
\hline 4 & Jose, Jenny & 3,2 & \multicolumn{2}{|r|}{2.50} \\
\hline 5 & Jose, Jose & 3,3 & \multicolumn{2}{|r|}{3.00} \\
\hline 6 & Jose, Lisa & 3,4 & \multicolumn{2}{|r|}{3.50} \\
\hline 7 & Lisa, Jenny & 4,2 & \multicolumn{2}{|r|}{3.00} \\
\hline 8 & Lisa, Jose & 4,3 & \multicolumn{2}{|r|}{3.50} \\
\hline 9 & Lisa, Lisa & 4,4 & \multicolumn{2}{|r|}{4.00} \\
\hline \multicolumn{3}{|c|}{Compute} & Mean: 3.00 & Standard deviation: 0.58 \\
\hline
\end{tabular}

Use the table 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). Write your answers to two decimal places.

Sagot :

GinnyAnsweridn
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].
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