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 95% confidence interval is 11.3067.

Answer 1

To calculate the upper limit for a 95% confidence interval for the sample mean, you can use the formula:

\[
\text{Confidence Interval} = \bar{x} \pm z \left( \frac{\sigma}{\sqrt{n}} \right)
\]

where:
- \(\bar{x}\) is the sample mean
- \(z\) is the z-value for the desired confidence level
- \(\sigma\) is the population standard deviation
- \(n\) is the sample size

In this case, we have:
- Sample mean \(\bar{x} = 10\)
- Sample size \(n = 36\)
- Population variance is \(16\), so the population standard deviation \(\sigma = \sqrt{16} = 4\)
- For a 95% confidence level, the z-value is approximately \(1.96\)

Now, we can compute the standard error (SE):

\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{4}{\sqrt{36}} = \frac{4}{6} = \frac{2}{3} \approx 0.6667
\]

Next, we calculate the margin of error (ME):

\[
ME = z \cdot SE = 1.96 \cdot 0.6667 \approx 1.3067
\]

Finally, we find the upper limit of the confidence interval:

\[
\text{Upper Limit} = \bar{x} + ME = 10 + 1.3067 \approx 11.3067
\]

Therefore, the statement "the upper limit for a 95% confidence interval is 11.3067" is **True**.