Assume the sample is taken from a normally distributed population and construct the indicated confidence interval.
-A random sample of 20 women have a mean height of 62.5 inches and a standard deviation of 1.8 inches. Construct a 98% confidence interval for the population variance, $\sigma ^ { 2 }.$
A)(1.7, 8.1)
B)(1.3, 2.8)
C)(0.9, 4.5)
D)(1.8, 8.5)

Question

Quizplus · Accepted Answer

The correct answer is A because the confidence interval for the population variance is given by:$$\frac{(n-1)s^2}{\chi^2_{\alpha/2, n-1}} < \sigma^2 < \frac{(n-1)s^2}{\chi^2_{1-\alpha/2, n-1}}$$where $\chi^2_{\alpha/2, n-1}$ and $\chi^2_{1-\alpha/2, n-1}$ are the chi-square critical values with $\alpha/2$ and $1-\alpha/2$ levels of significance and $n-1$ degrees of freedom.For a 98% confidence level, $\alpha = 0.02$ and $\alpha/2 = 0.01$. Using a chi-square distribution table, we find that $\chi^2_{0.01, 19} = 30.144$ and $\chi^2_{0.99, 19} = 8.907$.Therefore, the 98% confidence interval for the population variance is:$$\frac{(20-1)(1.8)^2}{30.144} < \sigma^2 < \frac{(20-1)(1.8)^2}{8.907}$$$$1.7 < \sigma^2 < 8.1$$

Assume the Sample Is Taken from a Normally Distributed Population σ2.\sigma ^ { 2 }.σ2.

Assume the Sample Is Taken from a Normally Distributed Population $\sigma ^ { 2 }.$