Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation σ.
Assume that the population has a normal distribution. Round the confidence interval limits to one more decimal place
than is used for the original set of data.
-The amounts (in ounces) of juice in eight randomly selected juice bottles are: $\begin{array} { l l l l } 15.2 & 15.1 & 15.9 & 15.5 \end{array}$
$\begin{array} { l l l l l } 15.6 & 15.1 & 15.8 & 15.0 \end{array}$
Find a $98 \%$ confidence interval for the population standard deviation $\sigma$.
A) $0.21 \mathrm { oz }

Question

Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation σ.
Assume that the population has a normal distribution. Round the confidence interval limits to one more decimal place
than is used for the original set of data.
-The amounts (in ounces) of juice in eight randomly selected juice bottles are: $\begin{array} { l l l l } 15.2 & 15.1 & 15.9 & 15.5 \end{array}$
$\begin{array} { l l l l l } 15.6 & 15.1 & 15.8 & 15.0 \end{array}$
Find a $98 \%$ confidence interval for the population standard deviation $\sigma$.
A) $0.21 \mathrm { oz } < \sigma < 0.71 \mathrm { oz }$
B) $0.20 \mathrm { oz } < \sigma < 0.71 \mathrm { oz }$
C) $0.21 \mathrm { oz } < \sigma < 0.82 \mathrm { oz }$
D) $0.22 \mathrm { oz } < \sigma < 0.86 \mathrm { oz }$

Quizplus · Accepted Answer

First, we need to find the sample standard deviation $s$:
$$s=\sqrt{\frac{\sum_{i=1}^n(x_i-\bar{x})^2}{n-1}}=\sqrt{\frac{(15.2-15.5)^2+(15.1-15.5)^2+(15.9-15.5)^2+(15.5-15.5)^2+(15.6-15.5)^2+(15.1-15.5)^2+(15.8-15.5)^2+(15-15.5)^2}{8-1}}\approx0.39$$
Next, we use the chi-squared distribution to find the critical values $c_1$ and $c_2$ such that the area between $c_1$ and $c_2$ is the desired confidence level of $98\%$, with $n-1=7$ degrees of freedom:
$$\chi_{0.01/2}^2(7)<c_1<\chi_{0.99/2}^2(7)$$
Using a chi-squared table or calculator, we find:
$$2.17<c_1<18.48$$
Then, we use the formula for the confidence interval of the population standard deviation:
$$\sqrt{\frac{(n-1)s^2}{c_2}}<\sigma<\sqrt{\frac{(n-1)s^2}{c_1}}$$
Plugging in the values, we get:
$$\sqrt{\frac{7(0.39)^2}{18.48}}<\sigma<\sqrt{\frac{7(0.39)^2}{2.17}}$$
Simplifying, we get:
$$0.21<\sigma<0.82$$
Rounding to one more decimal place than the original data, we get:
$$0.21 \mathrm { oz } < \sigma < 0.82 \mathrm { oz }$$

Use the Given Degree of Confidence and Sample Data to Find