Question

True or False: A sample size of 36 provides a sample mean of 10. If the population variance is known to be 16 and the population distribution is assumed to be normal, the upper limit for a 90% confidence interval is 11.0967.

Answer 1

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:

1. Find \( 1.645 \cdot 0.6667 \):

\[ 1.645 \cdot 0.6667 \approx 1.0967 \]

2. 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**.