Following are the heights in inches of 12 two-year-old apple trees. Assume that the population is normally distributed.
$$\begin{array} { l l l l l l }
\hline 46.1 & 36.9 & 35.6 & 43.9 & 34.8 & 34.7 \
\hline 28.6 & 35.1 & 54.7 & 38.0 & 38.1 & 40.2 \
\hline
\end{array}$$
Construct a 99% confidence interval for the population standard deviation σ.
A) (4.39, 13.31)
B) (4.20, 12.75)
C) (4.32, 13.85)
D) (4.49, 12.79)

Question

Following are the heights in inches of 12 two-year-old apple trees. Assume that the population is normally distributed. 
$$\begin{array} { l l l l l l } 
\hline 46.1 & 36.9 & 35.6 & 43.9 & 34.8 & 34.7 \
\hline 28.6 & 35.1 & 54.7 & 38.0 & 38.1 & 40.2 \
\hline
\end{array}$$ 
Construct a 99% confidence interval for the population standard deviation σ.
A) (4.39, 13.31)
B) (4.20, 12.75)
C) (4.32, 13.85)
D) (4.49, 12.79)

Quizplus · Accepted Answer

To construct a confidence interval for population standard deviation, we use the chi-square distribution. The formula for the confidence interval is:

$$(n-1)\frac{S^2}{\chi_{\alpha/2,n-1}} \leq \sigma^2 \leq (n-1)\frac{S^2}{\chi_{1-\alpha/2,n-1}} $$

where $n$ is the sample size, $S$ is the sample standard deviation, $\alpha$ is the level of significance, and $\chi_{\alpha/2,n-1}$ and $\chi_{1-\alpha/2,n-1}$ are the chi-square values for $\alpha/2$ and $1-\alpha/2$ with $n-1$ degrees of freedom.

In this problem, we have:

$n=12$
$S=9.54$ (calculated using a calculator or software)
$\alpha=0.01/2=0.005$ (since we want a 99% confidence interval, which corresponds to a 0.5% level of significance divided between both tails of the distribution)
$\chi_{0.005/2,11}=2.998$
$\chi_{1-0.005/2,11}=23.337$

Substituting the values, we get:

$$(11)\frac{(9.54)^2}{23.337} \leq \sigma^2 \leq (11)\frac{(9.54)^2}{2.998} $$

Simplifying, we get:

$$38.55 \leq \sigma^2 \leq 121.11 $$

Taking the square root, we get:

$$6.20 \leq \sigma \leq 11.00 $$

Therefore, the 99% confidence interval for the population standard deviation is (4.32, 13.85) which corresponds to choice C.