Question

High concentration of the toxic element arsenic is all too common in groundwater. The article “Evaluation of Treatment Systems for the Removal of Arsenic from Groundwater” (Practice Periodical of Hazardous, Toxic, and Radioactive Waste Mgmt., 2005: 152–157) reported that for a sample of n = 5 water specimens selected for treatment by coagulation, the sample mean arsenic concentration was 24.3 µg/L, and the sample standard deviation was 4.1. The authors of the cited article used t-based methods to analyze their data, so hopefully had reason to believe that the distribution of arsenic concentration was normal.

a.Calculate and interpret a 95% CI for true average arsenic concentration in all such water specimens.

b.Calculate a 90% upper confidence bound for the standard deviation of the arsenic concentration distribution.

c.Predict the arsenic concentration for a single water specimen in a way that conveys information about precision and reliability.

Short answer

1. The \(95\% \) confidence interval is\((19.21,29.39)\).
2. The upper bound for\(\sigma \) is\(7.9495\).
3. The \(95\% \) prediction interval is \((11.8276,36.7724)\).

Step-by-step solution

Step 1:Confidence bound.

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.

Step 2:Solution for part a).

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,

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

Step 3:95% confidence interval.

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,

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

Step 4:Solution for part b).

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\(7.9495\).

Step 5:Solution for part c).

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 \((11.8276,36.7724)\).