# Chapter 7 - Estimates and Sample Sizes - 7-4 Estimating a Population Standard Deviation or Variance - Basic Skills and Concepts - Page 369: 6

df=19, $X_{L}^2=6.844$, $X_{R}^2=38.582$, $\mu$ is between 0.0288 and 0.0685.

#### Work Step by Step

$\alpha=1-0.99=0.01.$ By using the table we can find the critical chi-square values with with $df=sample \ size-1=20-1=19$. $X_{L}^2= X_{0.995}^2=6.844$ $X_{R}^2= X_{0.005}^2=38.582$ Hence the confidence interval:$\mu$ is between $\sqrt{\frac{(n-1)\cdot s^2}{ X_{R}^2}}=\sqrt{\frac{(19)\cdot 0.04111^2}{38.582}}=0.0288$ and $\sqrt{\frac{(n-1)\cdot s^2}{ X_{L}^2}}=\sqrt{\frac{(19)\cdot 0.04111^2}{6.844}}=0.0685.$

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