Elementary Statistics (12th Edition)

Published by Pearson
ISBN 10: 0321836960
ISBN 13: 978-0-32183-696-0

Chapter 7 - Estimates and Sample Sizes - 7-4 Estimating a Population Standard Deviation or Variance - Basic Skills and Concepts: 5

Answer

df=24, $X_{L}^2=9.886$, $ X_{R}^2=45.559$, $\mu$ is between 0.174 and 0.374.

Work Step by Step

$\alpha=1-0.99=0.01.$ By using the table we can find the critical chi-square values with with $df=sample \ size-1=25-1=24$. $X_{L}^2= X_{0.995}^2=9.886$ $ X_{R}^2= X_{0.005}^2=45.559$ Hence the confidence interval:$\mu$ is between $\sqrt{\frac{(n-1)\cdot s^2}{ X_{R}^2}}=\sqrt{\frac{(24)\cdot 0.24^2}{45.559}}=0.174$ and $\sqrt{\frac{(n-1)\cdot s^2}{ X_{L}^2}}=\sqrt{\frac{(24)\cdot 0.24^2}{9.886}}=0.374$
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