Solution Found!
Nitrite ( ) was measured by two methods in rainwater and unchlorinated drinking water
Chapter 4, Problem 4-22(choose chapter or problem)
Nitrite \(\left(\mathrm{NO}_{2}^{-}\right)\) was measured by two methods in rainwater and unchlorinated drinking water. The results \(\pm\) standard deviation (number of samples) are
\(\begin{array}{lll}
\text { Sample source } & \text { Gas chromatography } & \text { Spectrophotometry } \\
\hline \text { Rainwater } & 0.069 \pm 0.005 \ \mathrm{mg} / \mathrm{L}(n=7) & 0.063 \pm 0.008 \ \mathrm{mg} / \mathrm{L}(n=5) \\
\text { Drinking water } & 0.078 \pm 0.007 \ \mathrm{mg} / \mathrm{L}(n=5) & 0.087 \pm 0.008 \ \mathrm{mg} / \mathrm{L}(n=5)
\end{array}\)
SOURCE: I. Sarudi and I. Nagy, Talanta 1995, 42, 1099.
(a) Do the two methods agree with each other at the 95% confidence level for both rainwater and drinking water?
(b) For each method, does the drinking water contain significantly more nitrite than the rainwater (at the 95% confidence level)?
Questions & Answers
QUESTION:
Nitrite \(\left(\mathrm{NO}_{2}^{-}\right)\) was measured by two methods in rainwater and unchlorinated drinking water. The results \(\pm\) standard deviation (number of samples) are
\(\begin{array}{lll}
\text { Sample source } & \text { Gas chromatography } & \text { Spectrophotometry } \\
\hline \text { Rainwater } & 0.069 \pm 0.005 \ \mathrm{mg} / \mathrm{L}(n=7) & 0.063 \pm 0.008 \ \mathrm{mg} / \mathrm{L}(n=5) \\
\text { Drinking water } & 0.078 \pm 0.007 \ \mathrm{mg} / \mathrm{L}(n=5) & 0.087 \pm 0.008 \ \mathrm{mg} / \mathrm{L}(n=5)
\end{array}\)
SOURCE: I. Sarudi and I. Nagy, Talanta 1995, 42, 1099.
(a) Do the two methods agree with each other at the 95% confidence level for both rainwater and drinking water?
(b) For each method, does the drinking water contain significantly more nitrite than the rainwater (at the 95% confidence level)?
ANSWER:Step 1 of 5
a) Use a paired t-test to determine if the two methods agree with each other at the 95 % confidence level.
The results \(\pm\) standard deviation for two methods in rainwater and drinking water are as follows:
\(\begin{array}{|l|l|l|}
\hline \text { Sample source } & \text { Gas chromatography } & \text { Spectrophotometry } \\
\hline \text { Rainwater } & 0.069 \pm 0.005 \ \mathrm{mg} / \mathrm{L}(\mathrm{n}=7) & 0.063 \pm 0.008 \ \mathrm{mg} / \mathrm{L}(\mathrm{n}=5) \\
\hline \text { Drinking water } & 0.078 \pm 0.007 \ \mathrm{mg} / \mathrm{L}(\mathrm{n}=5) & 0.087 \pm 0.008 \ \mathrm{mg} / \mathrm{L}(\mathrm{n}=5) \\
\hline
\end{array}\)
Since the standard deviation is not the same for both sets of measurements, use the following equations for calculate the t value.
\(t_{\text {calculated }}=\frac{\left|\overline{x_{1}}-\overline{x_{2}}\right|}{\sqrt{s_{1}^{2} / n_{1}+s_{2}^{2} / n_{2}}} \dots \dots(1)\)
Here, \(\overline{x_{1}}, s_{1}^{2}\) and \(n_{1}\) are the mean value, standard deviation, and the number of measurements for method $1. \(\bar{x}_{2}, s_{2}^{2}\) and \(n_{2}\) are the mean value, standard deviation, and the number of measurements for method 2 .
\(\text { Degrees of freedom }=\frac{\left(s_{1}^{2} / n_{1}+s_{2}^{2} n_{2}\right)^{2}}{\frac{\left(s_{1}^{2} / n_{1}\right)^{2}}{n_{1}-1}+\frac{\left(s_{2}^{2} / n_{2}\right)^{2}}{n_{2}-1}} \dots \dots (2)\)
If value of \(t_{\text {calculated }}\) is greater than \(t_{\text {table }}\) at the 95 % confidence level, the two results are different.
For rainwater:
Substitute 0.069 for \(\bar{x}_{1}\), 0.005 for \(s_{1}\), 7 for \(n_{1}\), 0.063 for \(\bar{x}_{2}\), 0.008 for \(s_{2}\) and 5 for \(n_{2}\) in equation ( 1 ).
\(\begin{array}{l}
\begin{array}{l}
=\frac{|(0.069)-(0.063)|}{\sqrt{\left(\frac{(0.005)^{2}}{7}\right)+\left(\frac{(0.008)^{2}}{5}\right)}} \\
=\frac{0.0060}{0.004}
\end{array}\\
=1.5
\end{array}\)
Therefore, the value of \(t_{\text {calculated }}\) is 1.5 .