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 lower limit for a 50% confidence interval is 9.5507.

Answer 1

To determine if the statement is true or false, we need to calculate the 50% confidence interval for the sample mean based on the given data.

1. **Given information**:
- Sample mean (\( \bar{x} \)) = 10
- Sample size (\( n \)) = 36
- Population variance (\( \sigma^2 \)) = 16, so the population standard deviation (\( \sigma \)) = \( \sqrt{16} = 4 \)

2. **Finding z-value for 50% confidence interval**:
For a 50% confidence interval, we have 25% in each tail of the normal distribution. The z-score that corresponds to 25% in the left tail is approximately -0.674 (you can look this up in a z-table or use statistical software).

3. **Calculating the standard error (SE)**:
\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{4}{\sqrt{36}} = \frac{4}{6} = \frac{2}{3} \approx 0.6667
\]

4. **Calculating the margin of error (ME)**:
\[
ME = z \times SE = -0.674 \times 0.6667 \approx -0.4495
\]

5. **Finding the lower limit of the confidence interval**:
\[
\text{Lower limit} = \bar{x} + ME = 10 - 0.4495 \approx 9.5505
\]

From this calculation, the lower limit for the 50% confidence interval is approximately \( 9.5505 \), which is very close to \( 9.5507 \).

Considering rounding and the approximation used for the z-score, we can say that the provided lower limit in the statement (9.5507) is very close to our calculation. Therefore, depending on rounding conventions, we can conclude that the statement is essentially true within reasonable precision.

**Final Answer: True**