To solve this question, we need to determine several key statistical measures based on the given data: the population mean, the margin of error for a 95% confidence interval, and the resulting confidence interval bounds.
Given:
- Population mean, [tex]\( \mu = 22 \)[/tex]
- Population standard deviation, [tex]\( \sigma = 13 \)[/tex]
- Sample size, [tex]\( N = 85 \)[/tex]
- Confidence level = 95%
- [tex]\( z \)[/tex]-score for a 95% confidence level ≈ 1.96
### Step-by-Step Solution:
1. Determine the population mean:
The population mean [tex]\( \mu \)[/tex] is provided as 22.
2. Calculate the margin of error (E):
The margin of error is calculated using the formula:
[tex]\[
E = z \cdot \frac{\sigma}{\sqrt{N}}
\][/tex]
Plugging in the values:
[tex]\[
E = 1.96 \cdot \frac{13}{\sqrt{85}}
\][/tex]
This evaluates to approximately 2.7637 (to four decimal places).
3. Determine the confidence interval bounds:
The lower bound of the 95% confidence interval is calculated as:
[tex]\[
\text{Lower bound} = \mu - E = 22 - 2.7637 \approx 19.2363
\][/tex]
The upper bound of the 95% confidence interval is calculated as:
[tex]\[
\text{Upper bound} = \mu + E = 22 + 2.7637 \approx 24.7637
\][/tex]
Summarizing the results:
- The population mean is [tex]\( \mu = 22 \)[/tex]
- The margin of error for a 95% confidence interval is approximately 2.7637
- Therefore, the 95% confidence interval of the population mean is from approximately 19.2363 to 24.7637
### Final Answer:
- The population mean is 22.
- For a 95% confidence interval, the approximate margin of error is 2.7637.
- The 95% confidence interval of the population mean is from 19.2363 to 24.7637.
