Use STDEV function in excel to find the standard deviation. Then, \(\displaystyle{s}={0.2215}.\)
And the variance is
\(\displaystyle{s}^{{{2}}}={\left({0.2215}\right)}^{{{2}}}\)

\(\displaystyle={0.04906}\) Step 2 At 98% level, \(\displaystyle\alpha={1}-{98}\%\)

\(\displaystyle={1}-{0.98}\)

\(\displaystyle={0.02}\) And The degree of freedom \(\displaystyle={n}-{1}\)

\(\displaystyle={10}-{1}\)

\(\displaystyle={9}\) Step 3 The critical values are \(\displaystyle{X}^{{{2}}}_{\left\lbrace\alpha,{n}-{1}\right\rbrace}={X}^{{{2}}}_{\left\lbrace{0.02},{9}\right\rbrace}\)

\(\displaystyle={19.679}\) And \(\displaystyle{X}^{{{2}}}_{\left\lbrace{1}-\alpha,{n}-{1}\right\rbrace}={X}^{{{2}}}_{\left\lbrace{0.98},{9}\right\rbrace}\)

\(\displaystyle={2.5324}\) Step 4 The 98% confidence interval for variance is \(\displaystyle{\left[{\frac{{{\left({n}-{1}\right)}{s}^{{{2}}}}}{{{x}^{{{2}}}_{\left\lbrace\alpha,{n}-{1}\right\rbrace}}}}\leq\sigma^{{{2}}}\leq{\frac{{{\left({n}-{1}\right)}{s}^{{{2}}}}}{{{x}^{{{2}}}_{\left\lbrace{1}-\alpha,{n}-{1}\right\rbrace}}}}\right]}={\frac{{{9}{\left({0.04906}\right)}}}{{{19.679}}}}\leq\sigma^{{{2}}}\leq{\frac{{{9}{\left({0.04906}\right)}}}{{{2.5324}}}}\)

\(\displaystyle={\frac{{{0.44154}}}{{{19.679}}}}\leq\sigma^{{{2}}}\leq{\frac{{{0.44154}}}{{{2.5324}}}}\)

\(\displaystyle={0.022}\leq\sigma^{{{2}}}\leq{0.174}\) The 98% confidence interval for standard deviation is \(\displaystyle{\left[\sqrt{{{\frac{{{\left({n}-{1}\right)}{s}^{{{2}}}}}{{{X}^{{{2}}}_{\left\lbrace\alpha,{n}-{1}\right\rbrace}}}}}}\leq\sigma\leq\sqrt{{{\frac{{{\left({n}-{1}\right)}{s}^{{{2}}}}}{{'{X}^{{{2}}}_{\left\lbrace{1}-\alpha,{n}-{1}\right\rbrace}}}}}}\right]}=\sqrt{{{0.022}}}\leq\sigma\leq\sqrt{{{0.174}}}\)

\(\displaystyle={0.150}\leq\sigma\leq{0.417}\)

\(\displaystyle={0.04906}\) Step 2 At 98% level, \(\displaystyle\alpha={1}-{98}\%\)

\(\displaystyle={1}-{0.98}\)

\(\displaystyle={0.02}\) And The degree of freedom \(\displaystyle={n}-{1}\)

\(\displaystyle={10}-{1}\)

\(\displaystyle={9}\) Step 3 The critical values are \(\displaystyle{X}^{{{2}}}_{\left\lbrace\alpha,{n}-{1}\right\rbrace}={X}^{{{2}}}_{\left\lbrace{0.02},{9}\right\rbrace}\)

\(\displaystyle={19.679}\) And \(\displaystyle{X}^{{{2}}}_{\left\lbrace{1}-\alpha,{n}-{1}\right\rbrace}={X}^{{{2}}}_{\left\lbrace{0.98},{9}\right\rbrace}\)

\(\displaystyle={2.5324}\) Step 4 The 98% confidence interval for variance is \(\displaystyle{\left[{\frac{{{\left({n}-{1}\right)}{s}^{{{2}}}}}{{{x}^{{{2}}}_{\left\lbrace\alpha,{n}-{1}\right\rbrace}}}}\leq\sigma^{{{2}}}\leq{\frac{{{\left({n}-{1}\right)}{s}^{{{2}}}}}{{{x}^{{{2}}}_{\left\lbrace{1}-\alpha,{n}-{1}\right\rbrace}}}}\right]}={\frac{{{9}{\left({0.04906}\right)}}}{{{19.679}}}}\leq\sigma^{{{2}}}\leq{\frac{{{9}{\left({0.04906}\right)}}}{{{2.5324}}}}\)

\(\displaystyle={\frac{{{0.44154}}}{{{19.679}}}}\leq\sigma^{{{2}}}\leq{\frac{{{0.44154}}}{{{2.5324}}}}\)

\(\displaystyle={0.022}\leq\sigma^{{{2}}}\leq{0.174}\) The 98% confidence interval for standard deviation is \(\displaystyle{\left[\sqrt{{{\frac{{{\left({n}-{1}\right)}{s}^{{{2}}}}}{{{X}^{{{2}}}_{\left\lbrace\alpha,{n}-{1}\right\rbrace}}}}}}\leq\sigma\leq\sqrt{{{\frac{{{\left({n}-{1}\right)}{s}^{{{2}}}}}{{'{X}^{{{2}}}_{\left\lbrace{1}-\alpha,{n}-{1}\right\rbrace}}}}}}\right]}=\sqrt{{{0.022}}}\leq\sigma\leq\sqrt{{{0.174}}}\)

\(\displaystyle={0.150}\leq\sigma\leq{0.417}\)