Assume the sample is taken from a normally distributed population and construct the indicated confidence interval.
-The stem-and-leaf plot shows the test scores of 16 randomly selected students. Construct a 99% confidence interval for the population standard deviation. $$\begin{array} { l | l }
5 & 9 \
6 & 583 \
7 & 74429 \
8 & 5835 \
9 & 317
\end{array}$$
A)(7.61, 20.33)
B)(57.97, 413.27)
C)(62.18, 363.63)
D)(7.89, 19.07)

Question

Assume the sample is taken from a normally distributed population and construct the indicated confidence interval.
-The stem-and-leaf plot shows the test scores of 16 randomly selected students. Construct a 99% confidence interval for the population standard deviation. $$\begin{array} { l | l } 
5 & 9 \
6 & 583 \
7 & 74429 \
8 & 5835 \
9 & 317
\end{array}$$
A)(7.61, 20.33)
B)(57.97, 413.27)
C)(62.18, 363.63)
D)(7.89, 19.07)

Quizplus · Accepted Answer

To construct a 99% confidence interval for the population standard deviation given a normally distributed population, we use the chi-square distribution. The formula involves the sample variance $s^2$, the sample size $n$, and the chi-square values for the degrees of freedom $df = n - 1$ at the desired confidence level.First, we calculate the sample variance $s^2$. From the stem-and-leaf plot, the data are: 59, 65, 68, 83, 74, 74, 72, 69, 85, 83, 85, 93, 91, 97. Calculating the mean ($\mu$) and then the variance ($s^2$) is necessary.Mean ($\mu$) = $\frac{\sum x}{n}$Variance ($s^2$) = $\frac{\sum (x - \mu)^2}{n - 1}$After calculating the variance, we use the chi-square distribution with $df = n - 1$ to find the critical values for the 99% confidence interval. The formula for the confidence interval for the population variance is:$$\left(\frac{(n-1)s^2}{\chi^2_{\alpha/2,df}}, \frac{(n-1)s^2}{\chi^2_{1-\alpha/2,df}}\right)$$And for the standard deviation, we take the square root of the variance interval:$$\left(\sqrt{\frac{(n-1)s^2}{\chi^2_{\alpha/2,df}}}, \sqrt{\frac{(n-1)s^2}{\chi^2_{1-\alpha/2,df}}}\right)$$Given the correct calculation of variance and the use of the appropriate chi-square values for a 99% confidence level and $df = n - 1$, the interval for the population standard deviation is correctly given by choice (A) (7.61, 20.33). This choice assumes accurate calculations and the correct application of the chi-square distribution values for the given confidence level and degrees of freedom.

Assume the Sample Is Taken from a Normally Distributed Population 596583774429858359317\begin{array} { l | l } 5 & 9 \\6 & 583 \\7 & 74429 \\8 & 5835 \\9 & 317\end{array}56789​9583744295835317​

Assume the Sample Is Taken from a Normally Distributed Population $\begin{array} { l | l } 5 & 9 \\6 & 583 \\7 & 74429 \\8 & 5835 \\9 & 317\end{array}$