Sure, let's solve this step-by-step.
### (a) Find the standard error of the mean for each sampling situation:
The standard error of the mean (SEM) is calculated using the formula:
[tex]\[ \text{SEM} = \frac{\sigma}{\sqrt{n}} \][/tex]
where [tex]\(\sigma\)[/tex] is the population standard deviation and [tex]\(n\)[/tex] is the sample size.
Let's calculate the SEM for each given situation:
1. Situation (a): [tex]\(\sigma = 26\)[/tex], [tex]\(n = 4\)[/tex]
[tex]\[
\text{SEM}_{\text{a}} = \frac{26}{\sqrt{4}} = \frac{26}{2} = 13.00
\][/tex]
2. Situation (b): [tex]\(\sigma = 26\)[/tex], [tex]\(n = 16\)[/tex]
[tex]\[
\text{SEM}_{\text{b}} = \frac{26}{\sqrt{16}} = \frac{26}{4} = 6.50
\][/tex]
3. Situation (c): [tex]\(\sigma = 26\)[/tex], [tex]\(n = 64\)[/tex]
[tex]\[
\text{SEM}_{\text{c}} = \frac{26}{\sqrt{64}} = \frac{26}{8} = 3.25
\][/tex]
So, the standard errors are:
- For (a): 13.00
- For (b): 6.50
- For (c): 3.25
### (b) What happens to the standard error each time you quadruple the sample size?
Let's observe the trend when the sample size is quadrupled:
- From [tex]\(n = 4\)[/tex] to [tex]\(n = 16\)[/tex]: [tex]\( \text{SEM}_{\text{a}} \to \text{SEM}_{\text{b}} \)[/tex]
- From [tex]\(n = 16\)[/tex] to [tex]\(n = 64\)[/tex]: [tex]\( \text{SEM}_{\text{b}} \to \text{SEM}_{\text{c}} \)[/tex]
When the sample size is quadrupled:
[tex]\[
\text{Ratio of SEMs} = \frac{\text{SEM}_{n}}{\text{SEM}_{4n}}
\][/tex]
For [tex]\(n = 4\)[/tex] to [tex]\(n = 16\)[/tex]:
[tex]\[
\text{Ratio} = \frac{13.00}{6.50} = 2.0
\][/tex]
For [tex]\(n = 16\)[/tex] to [tex]\(n = 64\)[/tex]:
[tex]\[
\text{Ratio} = \frac{6.50}{3.25} = 2.0
\][/tex]
Therefore, each time you quadruple the sample size, the standard error is reduced by half.
