To construct a 99% confidence interval for the mean number of years of education, we can use the formula for the confidence interval for the mean when the population standard deviation is unknown (using the t-distribution):
\[ \bar{x} \pm t^* \left( \frac{s}{\sqrt{n}} \right) \]
Where:
- \(\bar{x}\) is the sample mean
- \(t^*\) is the t-score corresponding to the desired confidence level and degrees of freedom
- \(s\) is the sample standard deviation
- \(n\) is the sample size
Given values:
- Sample mean (\(\bar{x}\)) = 9.77 years
- Sample standard deviation (\(s\)) = 9.53 years
- Sample size (\(n\)) = 833
- Confidence level = 99%
First, we find the degrees of freedom (df):
\[ df = n - 1 = 833 - 1 = 832 \]
Next, we look up the t-score for a 99% confidence level and 832 degrees of freedom. The t-score can be found using statistical tables or software (or using a calculator). For a 99% confidence level, the t-score is approximately 2.626.
Now, we can compute the standard error (SE):
\[ SE = \frac{s}{\sqrt{n}} = \frac{9.53}{\sqrt{833}} \approx \frac{9.53}{28.86} \approx 0.3305 \]
Now, apply the t-score to find the margin of error (ME):
\[ ME = t^* \times SE = 2.626 \times 0.3305 \approx 0.8684 \]
Finally, we can construct the confidence interval:
\[ \text{Lower limit} = \bar{x} - ME = 9.77 - 0.8684 \approx 8.90 \]
\[ \text{Upper limit} = \bar{x} + ME = 9.77 + 0.8684 \approx 10.64 \]
Thus, the 99% confidence interval for the mean number of years of education is approximately:
\[ [8.90 < \mu < 10.64] \]
Therefore, the final answer is:
A 99% confidence interval for the mean number of years of education is \([8.90 < \mu < 10.64]\).
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