To determine whether the statement is true or false, we need to calculate the 99% confidence interval for the sample mean using the given information.
- Sample Size (n): 36
- Sample Mean (\( \bar{x} \)): 10
- Population Variance (\( \sigma^2 \)): 16
- Population Standard Deviation (\( \sigma \)): \( \sqrt{16} = 4 \)
- Confidence Level: 99%
First, we need to calculate the critical z-value for a 99% confidence level. The z-value for a two-tailed confidence level of 99% is approximately 2.576.
Next, we calculate the standard error (SE) of the mean:
\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{4}{\sqrt{36}} = \frac{4}{6} = \frac{2}{3} \approx 0.6667 \]
Now we can calculate the margin of error (ME):
\[ ME = z \cdot SE = 2.576 \cdot 0.6667 \approx 1.718 \]
Now we can determine the confidence interval. The sample mean is 10, so we will calculate the lower and upper limits:
- Lower Limit:
\[ \text{Lower Limit} = \bar{x} - ME = 10 - 1.718 \approx 8.282 \]
- Upper Limit:
\[ \text{Upper Limit} = \bar{x} + ME = 10 + 1.718 \approx 11.718 \]
Thus, the 99% confidence interval for the sample mean is approximately (8.282, 11.718).
The statement claims that the lower limit of the 99% confidence interval is 11.7173, which is incorrect. The lower limit we calculated is approximately 8.282, not 11.7173.
Therefore, the statement is False.