91Ó°ÊÓ

The size of the sample is\(n = 25\).

The sample standard deviation is\(s = 0.24\).

The level of confidence is 95%.

The degrees of freedom (df) is computed as,

\(\begin{array}{c}df = n - 1\\ = 25 - 1\\ = 24\end{array}\)

Using the Chi-square table, the critical values are\(\chi _L^2 = 12.401\)and\(\chi _R^2 = 39.364\).

The 95% confidence interval estimate of\(\sigma \)is computed as,

\(\begin{array}{c}\sqrt {\frac{{\left( {n - 1} \right){s^2}}}{{\chi _R^2}}} < \sigma < \sqrt {\frac{{\left( {n - 1} \right){s^2}}}{{\chi _L^2}}} \\\sqrt {\frac{{\left( {25 - 1} \right){{0.24}^2}}}{{39.364}}} < \sigma < \sqrt {\frac{{\left( {25 - 1} \right){{0.24}^2}}}{{12.401}}} \\0.19 < \sigma < 0.33\end{array}\)

Therefore, the 95% confidence interval estimate of\(\sigma \)is\(0.19\;mg < \sigma < 0.33\;mg\).