Certainly! Let's find the 95% confidence interval for the population mean [tex]\(\mu\)[/tex] using the given sample data. The sample consists of 11 values: [tex]\(50.9, 26.2, 79.8, 32.7, 62.7, 45.7, 51.5, 95.5, 50.2, 22.8, 64.1\)[/tex]. Here is the detailed step-by-step solution:
1. Calculate the sample mean [tex]\(\bar{x}\)[/tex]:
[tex]\[ \bar{x} \approx 52.92 \][/tex]
2. Calculate the sample standard deviation [tex]\(s\)[/tex]. Given:
[tex]\[ s \approx 22.03 \][/tex]
3. Determine the sample size [tex]\(n\)[/tex]:
[tex]\[ n = 11 \][/tex]
4. Calculate the standard error of the mean (SEM):
[tex]\[ \text{SEM} = \frac{s}{\sqrt{n}} \approx \frac{22.03}{\sqrt{11}} \approx 6.64 \][/tex]
5. For a 95% confidence interval, the z-score (critical value for standard normal distribution) is approximately 1.96.
6. Calculate the margin of error (MOE):
[tex]\[ \text{MOE} = z \cdot \text{SEM} \approx 1.96 \cdot 6.64 \approx 13.02 \][/tex]
7. Determine the confidence interval:
- Lower bound of the confidence interval:
[tex]\[ \bar{x} - \text{MOE} \approx 52.92 - 13.02 \approx 39.90 \][/tex]
- Upper bound of the confidence interval:
[tex]\[ \bar{x} + \text{MOE} \approx 52.92 + 13.02 \approx 65.94 \][/tex]
Thus, the 95% confidence interval for the population mean [tex]\(\mu\)[/tex] is approximately [tex]\((39.90, 65.94)\)[/tex].
So, we can be 95% confident that the true population mean [tex]\(\mu\)[/tex] lies within the interval [tex]\(39.90\)[/tex] to [tex]\(65.94\)[/tex].
