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 99% confidence interval is 8.2827.

1 answer

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.