To determine if the statement is true or false, we need to calculate the 90% confidence interval for the sample mean.
Given:
- Sample mean (\( \bar{x} \)) = 10
- Population variance (\( \sigma^2 \)) = 16
- Population standard deviation (\( \sigma \)) = \( \sqrt{16} = 4 \)
- Sample size (\( n \)) = 36
- Confidence level = 90%
The formula for the confidence interval for the mean is given by:
\[ \bar{x} \pm z \frac{\sigma}{\sqrt{n}} \]
Where \( z \) is the z-score corresponding to the desired confidence level. For a 90% confidence interval, the critical z-value (from z-tables) is approximately 1.645.
Now, we calculate the margin of error:
\[ \text{Margin of Error} = z \cdot \frac{\sigma}{\sqrt{n}} = 1.645 \cdot \frac{4}{\sqrt{36}} = 1.645 \cdot \frac{4}{6} = 1.645 \cdot \frac{2}{3} = 1.0967 \]
Now we calculate the lower limit of the confidence interval:
\[ \text{Lower Limit} = \bar{x} - \text{Margin of Error} = 10 - 1.0967 = 8.9033 \]
Thus, the lower limit for the 90% confidence interval is indeed 8.9033.
Therefore, the statement is True.