To calculate the upper limit of a 99% confidence interval for the sample mean, you'll use the formula:
\[ \text{Confidence Interval} = \bar{x} \pm z \left(\frac{\sigma}{\sqrt{n}}\right) \]
Where:
- \(\bar{x}\) = sample mean
- \(z\) = z-score corresponding to the desired confidence level
- \(\sigma\) = population standard deviation (the square root of the variance)
- \(n\) = sample size
Given the information:
- Sample mean, \(\bar{x} = 10\)
- Population variance = 16, thus population standard deviation, \(\sigma = \sqrt{16} = 4\)
- Sample size, \(n = 36\)
First, we need to find the z-score for a 99% confidence interval. For a two-tailed test at 99% confidence, the significance level (\(\alpha\)) is 0.01, meaning that each tail has 0.005. Using the z-table, the z-score for 0.005 in the upper tail (or 99% confidence) is approximately \(z \approx 2.576\).
Now we can calculate the standard error (SE):
\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{4}{\sqrt{36}} = \frac{4}{6} = \frac{2}{3} \approx 0.6667 \]
Now calculate the margin of error (ME):
\[ ME = z \cdot SE = 2.576 \cdot 0.6667 \approx 1.7187 \]
Finally, we find the upper limit of the confidence interval:
\[ \text{Upper Limit} = \bar{x} + ME = 10 + 1.7187 \approx 11.7187 \]
Thus, the upper limit for the 99% confidence interval is approximately \(11.7187\), not \(8.2827\).
Therefore, the statement is False.
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