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Upper confidence bound for \(\mu \) and lower confidence bound for \(\mu \) are given by,

\(\overline x + {t_{\alpha /2,n - 1}} \cdot \frac{s}{{\sqrt n }}\)-upper bound,

\(\overline x - {t_{\alpha /2,n - 1}} \cdot \frac{s}{{\sqrt n }}\)-lower bound.

With confidence level of \(100(1 - \alpha )\% \). Then distribution the random sample is taken from is normal.

Given that,

\(\begin{array}{l}n = 5\\\overline x = 23.3\\s = 4.1\end{array}\)

And assuming normality, the confidence intervals for \(\mu \) and \(\sigma \), as well as prediction interval can be computed.

Upper confidence bound for \(\mu \) and lower confidence bound for \(\mu \) are given by,

\(\overline x + {t_{\alpha /2,n - 1}} \cdot \frac{s}{{\sqrt n }}\)-upper bound,

\(\overline x - {t_{\alpha /2,n - 1}} \cdot \frac{s}{{\sqrt n }}\)-lower bound.

With confidence level of \(100(1 - \alpha )\% \). Then distribution the random sample is taken from is normal.

To obtain \(95\% \) confidence interval, that \(t\) value is needed for \(\alpha /2 = 0.025\) and \(5 - 1 = 4\) degrees of freedom, where \(\alpha \) was obtained from,

\(\begin{array}{l}100(1 - \alpha ) = 95\\\alpha = 0.05\\\alpha /2 = 0.025\end{array}\)

The \(t\) value can be found at the appendix of the book in the table of values and it is

\(\begin{array}{l}{t_{\alpha /2,n - 1}} = {t_{0.025,4}}\\{t_{\alpha /2,n - 1}} = 2.777\end{array}\)

The \(95\% \) confidence interval now becomes,

\(\begin{array}{l}\left( {\overline x - {t_{\alpha /2,n - 1}} \cdot \frac{s}{{\sqrt n }},\overline x + {t_{\alpha /2,n - 1}} \cdot \frac{s}{{\sqrt n }}} \right)\\ = \left( {24.3 - 2.777\frac{{4.1}}{{\sqrt 5 }},24.3 + 2.777\frac{{4.1}}{{\sqrt 5 }}} \right)\\ = (19.21,29.39)\end{array}\)

Confidence interval for the variance, with confidence level \(100(1 - \alpha )\% \) of a normal population has upper bound,\(\frac{{(n - 1){s^2}}}{{{X^2}_{1 - \alpha /2,n - 1}}}\) and lower bound\(\frac{{(n - 1){s^2}}}{{{X^2}_{\alpha /2,n - 1}}}\).

The confidence intervals for the standard deviation has bounds that can be obtained by taking square root of the corresponding bounds of the confidence interval for the variance. By substituting \(\alpha /2\) with \(\alpha \) the corresponding upper and lower bounds are obtained.

From \(100(1 - \alpha ) = 0.9\),\(\alpha \) is \(0.1\), the square value , where degrees of freedom are \(5 - 1 = 4\) is,

\(\begin{array}{l}{X^2}_{1 - \alpha /2,n - 1} = {X^2}_{0.9,4}\\ = 1.064\end{array}\)

Which can be found at the appendix. The upper bound for \(\sigma \) becomes,

\(\begin{array}{l}\sqrt {\frac{{(n - 1){s^2}}}{{{X^2}_{1 - \alpha /2,n - 1}}}} = \sqrt {\frac{{4{{(4.1)}^2}}}{{1.064}}} \\ = 7.9495\end{array}\)

The upper bound for \(\sigma \) is\(7.9495\).

Two sided prediction interval is

\(\left( {\overline x - {t_{\alpha /2,n - 1}} \cdot s\sqrt {1 + \frac{1}{n}} ,\overline x + {t_{\alpha /2,n - 1}} \cdot s\sqrt {1 + \frac{1}{n}} } \right)\)

To compute a \(95\% \) prediction interval first the \(t\) value is necessary. As in (a), the value is,

\(\begin{array}{l}{t_{\alpha /2,n - 1}} = {t_{0.025,4}}\\{t_{\alpha /2,n - 1}} = 2.777\end{array}\)

A \(95\% \) prediction interval is,

\(\begin{array}{l}\left( {\overline x - {t_{\alpha /2,n - 1}} \cdot s\sqrt {1 + \frac{1}{n}} ,\overline x + {t_{\alpha /2,n - 1}} \cdot s\sqrt {1 + \frac{1}{n}} } \right)\\ = \left( {24.3 - 2.777(4.1)\sqrt {1 + \frac{1}{5}} ,24.3 + 2.777(4.1)\sqrt {1 + \frac{1}{5}} } \right)\\ = (11.8276,36.7724)\end{array}\)

Hence, The \(95\% \) prediction interval is \((11.8276,36.7724)\).