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To solve the given problem of determining the margin of error for various sample sizes when constructing a 95% confidence interval for the population mean, let's break down the process step by step:
### Step-by-Step Solution:
#### Step 1: Identify Important Information
- Population standard deviation ([tex]\(\sigma\)[/tex]): 15
- Confidence level: 95%
- Z-score for a 95% confidence level (from standard normal distribution): 1.96
#### Step 2: Formula for Margin of Error
The margin of error (E) for a confidence interval is calculated using the formula:
[tex]\[ E = Z \times \frac{\sigma}{\sqrt{n}} \][/tex]
where:
- [tex]\(Z\)[/tex] is the Z-score for the desired confidence level.
- [tex]\(\sigma\)[/tex] is the population standard deviation.
- [tex]\(n\)[/tex] is the sample size.
#### Part (a): Sample Size = 4
To find the margin of error for a sample size of 4:
[tex]\[ E = 1.96 \times \frac{15}{\sqrt{4}} = 1.96 \times \frac{15}{2} = 1.96 \times 7.5 = 14.70 \][/tex]
So, the margin of error for a sample size of 4 is 14.70.
#### Part (b): Sample Size = 36
To find the margin of error for a sample size of 36:
[tex]\[ E = 1.96 \times \frac{15}{\sqrt{36}} = 1.96 \times \frac{15}{6} = 1.96 \times 2.5 = 4.90 \][/tex]
So, the margin of error for a sample size of 36 is 4.90.
#### Part (c): Sample Size = 324
To find the margin of error for a sample size of 324:
[tex]\[ E = 1.96 \times \frac{15}{\sqrt{324}} = 1.96 \times \frac{15}{18} = 1.96 \times 0.8333 \approx 1.63 \][/tex]
So, the margin of error for a sample size of 324 is 1.63.
Now, let's address the reasoning behind these calculations:
#### Explanation
From part (b) to part (c), the sample size increases from 36 to 324. This change in sample size can be analyzed:
- If sample size increases by a factor, say [tex]\(k\)[/tex], the denominator in the margin of error formula [tex]\(\sqrt{n}\)[/tex] increases by [tex]\(\sqrt{k}\)[/tex].
Comparing sample sizes 36 and 324, we see:
[tex]\[ \frac{324}{36} = 9 \][/tex]
So, the sample size increased by a factor of 9, implying the sample size increased by [tex]\(\sqrt{9}\)[/tex] (i.e., 3) in the margin of error formula.
Hence, the margin of error is divided by 3 when the sample size is increased from 36 to 324.
In summary:
- The sample size increased by a factor of 9.
- The margin of error decreased from 4.90 (for [tex]\(n = 36\)[/tex]) to 1.63 (for [tex]\(n = 324\)[/tex]).
- Therefore, this decrease equates to dividing the margin of error from part (b) by the square root of 9 (which is 3).
So, the explanation for the margin of error change is:
Because the sample size increased by a factor of 9, the margin of error was reduced by a factor of 3.
### Step-by-Step Solution:
#### Step 1: Identify Important Information
- Population standard deviation ([tex]\(\sigma\)[/tex]): 15
- Confidence level: 95%
- Z-score for a 95% confidence level (from standard normal distribution): 1.96
#### Step 2: Formula for Margin of Error
The margin of error (E) for a confidence interval is calculated using the formula:
[tex]\[ E = Z \times \frac{\sigma}{\sqrt{n}} \][/tex]
where:
- [tex]\(Z\)[/tex] is the Z-score for the desired confidence level.
- [tex]\(\sigma\)[/tex] is the population standard deviation.
- [tex]\(n\)[/tex] is the sample size.
#### Part (a): Sample Size = 4
To find the margin of error for a sample size of 4:
[tex]\[ E = 1.96 \times \frac{15}{\sqrt{4}} = 1.96 \times \frac{15}{2} = 1.96 \times 7.5 = 14.70 \][/tex]
So, the margin of error for a sample size of 4 is 14.70.
#### Part (b): Sample Size = 36
To find the margin of error for a sample size of 36:
[tex]\[ E = 1.96 \times \frac{15}{\sqrt{36}} = 1.96 \times \frac{15}{6} = 1.96 \times 2.5 = 4.90 \][/tex]
So, the margin of error for a sample size of 36 is 4.90.
#### Part (c): Sample Size = 324
To find the margin of error for a sample size of 324:
[tex]\[ E = 1.96 \times \frac{15}{\sqrt{324}} = 1.96 \times \frac{15}{18} = 1.96 \times 0.8333 \approx 1.63 \][/tex]
So, the margin of error for a sample size of 324 is 1.63.
Now, let's address the reasoning behind these calculations:
#### Explanation
From part (b) to part (c), the sample size increases from 36 to 324. This change in sample size can be analyzed:
- If sample size increases by a factor, say [tex]\(k\)[/tex], the denominator in the margin of error formula [tex]\(\sqrt{n}\)[/tex] increases by [tex]\(\sqrt{k}\)[/tex].
Comparing sample sizes 36 and 324, we see:
[tex]\[ \frac{324}{36} = 9 \][/tex]
So, the sample size increased by a factor of 9, implying the sample size increased by [tex]\(\sqrt{9}\)[/tex] (i.e., 3) in the margin of error formula.
Hence, the margin of error is divided by 3 when the sample size is increased from 36 to 324.
In summary:
- The sample size increased by a factor of 9.
- The margin of error decreased from 4.90 (for [tex]\(n = 36\)[/tex]) to 1.63 (for [tex]\(n = 324\)[/tex]).
- Therefore, this decrease equates to dividing the margin of error from part (b) by the square root of 9 (which is 3).
So, the explanation for the margin of error change is:
Because the sample size increased by a factor of 9, the margin of error was reduced by a factor of 3.
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