To determine if the statement is true or false, we need to calculate the upper limit for the 90% confidence interval for the sample mean given the following:
- Sample size (\( n \)) = 36
- Sample mean (\( \bar{x} \)) = 10
- Population variance (\( \sigma^2 \)) = 16 (therefore, the population standard deviation \( \sigma = \sqrt{16} = 4 \))
- Confidence level = 90%
First, we find the critical value (z-score) for a 90% confidence interval. For a 90% confidence level, the z-score (two-tailed, which means we split the remaining 10% into two tails) corresponds to the 95th percentile (since we take half of 10%):
\[ z = 1.645 \]
Next, we calculate the standard error (SE) of the sample mean:
\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{4}{\sqrt{36}} = \frac{4}{6} = \frac{2}{3} \approx 0.6667 \]
Now, we can calculate the upper limit of the 90% confidence interval using the formula:
\[ \text{Upper Limit} = \bar{x} + z \cdot SE \]
Substituting our values:
\[ \text{Upper Limit} = 10 + 1.645 \cdot 0.6667 \]
Calculating:
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Find \( 1.645 \cdot 0.6667 \):
\[ 1.645 \cdot 0.6667 \approx 1.0967 \]
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Then,
\[ \text{Upper Limit} = 10 + 1.0967 \approx 11.0967 \]
Thus, the upper limit for the 90% confidence interval is indeed 11.0967.
Therefore, the statement is True.