To find the 90% confidence interval for the population mean based on the given sample, we will follow these steps:
- Calculate the sample mean (\( \bar{x} \)).
- Calculate the sample standard deviation (\( s \)).
- Determine the critical value (\( t^* \)) for a 90% confidence level.
- Calculate the margin of error.
- Construct the confidence interval.
Step 1: Calculate the sample mean (\( \bar{x} \))
The sample data is: \[ 77, 44.4, 33.6, 63.1, 68.1, 70.6, 52, 43.5, 26.4, 72.6, 30.9, 29.4 \]
First, calculate the sum of the sample values:
\[ \text{Sum} = 77 + 44.4 + 33.6 + 63.1 + 68.1 + 70.6 + 52 + 43.5 + 26.4 + 72.6 + 30.9 + 29.4 = 683.8 \]
Then, the sample mean (\( \bar{x} \)) is:
\[ \bar{x} = \frac{\text{Sum}}{n} = \frac{683.8}{12} \approx 56.9833 \]
Step 2: Calculate the sample standard deviation (\( s \))
We will first calculate the sum of squared deviations from the mean:
\[ \text{Sum of Squared Deviations} = \sum (x_i - \bar{x})^2 \]
Each term \( (x_i - \bar{x})^2 \) is calculated as follows:
- For 77: \( (77 - 56.9833)^2 \approx 400.6338 \)
- For 44.4: \( (44.4 - 56.9833)^2 \approx 158.4526 \)
- For 33.6: \( (33.6 - 56.9833)^2 \approx 568.6851 \)
- For 63.1: \( (63.1 - 56.9833)^2 \approx 37.8409 \)
- For 68.1: \( (68.1 - 56.9833)^2 \approx 125.8235 \)
- For 70.6: \( (70.6 - 56.9833)^2 \approx 178.1312 \)
- For 52: \( (52 - 56.9833)^2 \approx 4.8311 \)
- For 43.5: \( (43.5 - 56.9833)^2 \approx 179.8831 \)
- For 26.4: \( (26.4 - 56.9833)^2 \approx 926.0673 \)
- For 72.6: \( (72.6 - 56.9833)^2 \approx 236.0750 \)
- For 30.9: \( (30.9 - 56.9833)^2 \approx 679.5150 \)
- For 29.4: \( (29.4 - 56.9833)^2 \approx 743.0652 \)
Now summing these squared deviations:
\[ \text{Total} = 400.6338 + 158.4526 + 568.6851 + 37.8409 + 125.8235 + 178.1312 + 4.8311 + 179.8831 + 926.0673 + 236.0750 + 679.5150 + 743.0652 \approx 3140.0968 \]
Now, calculate the sample variance (\( s^2 \)):
\[ s^2 = \frac{\text{Sum of Squared Deviations}}{n - 1} = \frac{3140.0968}{12 - 1} \approx \frac{3140.0968}{11} \approx 285.463 \]
Then take the square root to find the sample standard deviation (\( s \)):
\[ s \approx \sqrt{285.463} \approx 16.91 \]
Step 3: Determine the critical value (\( t^* \))
For a 90% confidence level with \( n - 1 = 11 \) degrees of freedom, we look up \( t^* \) in the t-distribution table.
\( t_{0.05, 11} \approx 1.796 \) (rounded to 3 decimal places)
Step 4: Calculate the margin of error (ME)
The margin of error is calculated as:
\[ \text{ME} = t^* \times \frac{s}{\sqrt{n}} = 1.796 \times \frac{16.91}{\sqrt{12}} \approx 1.796 \times 4.886 = 8.796 \]
Step 5: Construct the confidence interval
Finally, the confidence interval is given by:
\[ \left( \bar{x} - \text{ME}, \bar{x} + \text{ME} \right) = \left( 56.9833 - 8.796, 56.9833 + 8.796 \right) \]
Calculating the endpoints:
\[ \text{Lower Limit} = 56.9833 - 8.796 \approx 48.1873 \] \[ \text{Upper Limit} = 56.9833 + 8.796 \approx 65.7793 \]
Thus, the 90% confidence interval is:
\[ \text{90% C.I.} = (48.19, 65.78) \]
Final result with two decimal places as requested:
\[ \text{90% C.I.} = (48.19, 65.78) \]