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.

1 answer

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.