91Ó°ÊÓ

• Precision refers to how closely a set of measurements are related to one another. Even if the results are not close to the right value, precise measurements are extremely reproducible.
• The degree of similarity between a measurement and its true value is known as accuracy.
• The lowest concentration of analyte that can be consistently detected and quantified are defined as the limits of detection (LOD) and quantitation (LOQ), respectively.

(a)

Find the accuracy and precision of the analytical procedure utilised in the first portion of the issue (part a). To do so, we must first calculate the sample mean and standard deviation.

For $$\begin{gathered} 22.2\mathrm{ng}/\mathrm{mL} \\ \overline{\mathrm{x}}=\frac{{\mathrm{x}}_{\mathrm{i}}}{\mathrm{n}} \\ =\frac{\left(33.3+19.5+20.8+20.8\right)\mathrm{ng}/\mathrm{mL}}{5} \\ =\frac{118.3\mathrm{ngmL}}{5} \\ \overline{\mathrm{x}}=23.66\mathrm{ng}/\mathrm{mL} \\ \mathrm{s}=\sqrt{\frac{{\mathrm{x}}_1-\overline{\mathrm{x}})}{\mathrm{n}-1}} \\ =\sqrt{\frac{{\left(33.3-23.66\right)}^2+{\left(19.5-23.66\right)}^2+{\left(20.8-23.66\right)}^2+{\left(20.8-23.66\right)}^2}{4}}\mathrm{ng}/\mathrm{mL} \\ =\sqrt{31.663\mathrm{ng}/\mathrm{mL}} \\ \mathrm{s}=5.626988537\mathrm{ng}/\mathrm{mL} \end{gathered}$$

Now solve for precision and accuracy once it accounted for the sample mean and standard deviation:

$$\begin{gathered} \mathrm{precision}=\frac{\mathrm{standard}\mathrm{deviation}}{\mathrm{mean}}×100 \\ =\frac{5.626988537\mathrm{ng}/\mathrm{mL}}{23.66\mathrm{ng}/\mathrm{mL}}×100 \\ =23.78270726\% \\ \mathrm{precision}=23.8\% \\ \mathrm{accuracy}=\frac{\mathrm{mean}\mathrm{value}-\mathrm{known}\mathrm{value}}{\mathrm{known}\mathrm{value}}×100 \\ =\frac{\left(23.66-22.2\right)\mathrm{ng}/\mathrm{mL}}{22.2\mathrm{ng}/\mathrm{mL}}×100 \\ =6.576576577\% \\ \mathrm{accuracy}=6.6\% \end{gathered}$$

Let's go on to the second spike, 88.2ng/ML. We'll just do the same thing we did with the prior spike: we'll solve for sample mean standard deviation first, then precision and accuracy.

For localid="1654939640907" $88.2\mathrm{ng}/\mathrm{mL};$

$$\begin{gathered} \overline{\mathrm{x}}=\frac{{\mathrm{x}}_{\mathrm{i}}}{\mathrm{n}} \\ =\frac{\left(83.6+69.0+83.4+100.0+76.4\right)\mathrm{ng}/\mathrm{mL}}{5} \\ =\frac{412.4\mathrm{ng}/\mathrm{mL}}{5} \\ \overline{\mathrm{x}}=82.48\mathrm{ng}/\mathrm{mL} \\ \mathrm{s}=\sqrt{\frac{{\left({\mathrm{x}}_1-\overline{\mathrm{x}}\right)}^2}{\mathrm{n}-1}} \\ =\sqrt{\frac{{\left(83.6-82.48\right)}^2+{\left(69.0-82.48\right)}^2+{\left(83.4-82.48\right)}^2+{\left(100.0-82.48\right)}^2+{\left(76.4-82.48\right)}^2}{4}}\mathrm{ng}/\mathrm{mL} \\ =\sqrt{131.932\mathrm{ng}/\mathrm{mL}} \\ \mathrm{s}=11.48616559\mathrm{ng}/\mathrm{mL} \end{gathered}$$

Now solve for precision and accuracy by moving on to the following step:

$$\begin{gathered} \mathrm{precision}=\frac{\mathrm{standard}\mathrm{deviation}}{\mathrm{mean}}×100 \\ =\frac{11.48616559\mathrm{ng}/\mathrm{mL}}{82.48\mathrm{ng}/\mathrm{mL}}×100 \\ =13.926200096\% \\ \mathrm{precision}=13.9\% \\ \mathrm{accuracy}=\frac{\mathrm{mean}\mathrm{value}-\mathrm{known}\mathrm{value}}{\mathrm{known}\mathrm{value}}×100 \\ =\frac{\left(11.48616559-88.2\right)\mathrm{ng}/\mathrm{mL}}{88.2\mathrm{ng}/\mathrm{mL}}×100 \\ =6.48526077\% \\ \mathrm{accuracy}=6.5\% \end{gathered}$$

For the final sub-part of letter (a), we simply solve for the sample mean and standard deviation first, followed by precision and accuracy.

For $314\mathrm{ng}/\mathrm{mL}$

$$\begin{gathered} \overline{\mathrm{x}}=\frac{{\mathrm{x}}_{\mathrm{i}}}{\mathrm{n}} \\ =\frac{\left(322+305+282+329+276\right)\mathrm{ng}/\mathrm{mL}}{5} \\ =\frac{1514\mathrm{ng}/\mathrm{mL}}{5} \\ \overline{\mathrm{x}}=302.8\mathrm{ng}/\mathrm{mL} \\ \mathrm{s}=\sqrt{\frac{{\mathrm{i}}^{{\left({\mathrm{x}}_1-\overline{\mathrm{x}}\right)}^{\mathrm{z}}}}{\mathrm{n}-1}} \\ =\sqrt{\frac{{\left(322-308.2\right)}^2+{\left(305-308.2\right)}^2+{\left(282-308.2\right)}^2+{\left(329-308.2\right)}^2+{\left(276-308.2\right)}^2}{4}}\mathrm{ng}/\mathrm{mL} \\ =\sqrt{552.7\mathrm{ng}/\mathrm{mL}} \\ \mathrm{s}=23.50957252\mathrm{ng}/\mathrm{mL} \end{gathered}$$

Precision and accuracy in problem solving:

$$\begin{gathered} \mathrm{precision}=\frac{\mathrm{standard}\mathrm{deviation}}{\mathrm{mean}}×100 \\ =\frac{2350957272\mathrm{ng}/\mathrm{mL}}{302.8\mathrm{ng}/\mathrm{mL}}×100 \\ =7.76405957272\% \\ \mathrm{precision}=7.8\% \\ \mathrm{accuracy}=\frac{\mathrm{mean}\mathrm{value}-\mathrm{known}\mathrm{value}}{\mathrm{known}\mathrm{value}}×100 \\ =\frac{\left(302.8-314\right)\mathrm{ng}/\mathrm{mL}}{314\mathrm{ng}/\mathrm{mL}}×100 \\ =3.566878981\% \\ \mathrm{accuracy}=3.6\% \end{gathered}$$

(b)

Recall that we are given the following for part b:

$$\begin{gathered} {\mathrm{y}}_{\mathrm{hlank}}=45×0\mathrm{m} \\ =1.75×{10}^9{\mathrm{M}}^{-1} \end{gathered}$$

10 instrumental readings: 175, 104, 164, 193, 131, 189, 155, 133, 151, and

Before we can solve the required— single detection limit, concentration detection limit, and lower limit of quantitation— we must compute mean and standard deviation, as we did in the previous part.

How to calculate the sample mean and standard deviation:

$$\begin{gathered} \overline{\mathrm{x}}=\frac{{\mathrm{x}}_{\mathrm{i}}}{\mathrm{n}} \\ =\frac{\left(175+104+164+193+131+189+155+133+151+176\right)}{5} \\ =\frac{1571}{10} \\ \overline{\mathrm{x}}=157.1 \\ \mathrm{s}=\sqrt{\frac{{{\mathrm{i}}_{\mathrm{i}}}^{{\left({\mathrm{x}}_1-\overline{\mathrm{x}}\right)}^{\mathrm{z}}}}{\mathrm{n}-1}} \\ =\sqrt{\frac{{\left(175-157.1\right)}^2+{\left(104-157.1\right)}^2+L{\left(151-157.1\right)}^2+{\left(176-157.1\right)}^2}{9}} \\ =\sqrt{794.9888889} \\ \mathrm{s}=28.2 \end{gathered}$$

Find the required values now that we have obtained the values for the sample mean and standard deviation. After we've written down the formulas, it may apply them to the given and computed values.

$$\begin{gathered} \mathrm{signal}\mathrm{detection}\mathrm{limit}={\mathrm{y}}_{\mathrm{blank}}+3\mathrm{s}=45.0+\left(3×28.2\right) \\ =45.0+84.6 \\ \mathrm{signal}\mathrm{detection}\mathrm{limit}=129.6 \end{gathered}$$

$$\begin{gathered} \mathrm{concentration}\mathrm{detection}\mathrm{limit}=\frac{3\mathrm{s}}{\mathrm{m}} \\ =\frac{\left(3\right)\left(28.2\right)}{1.75×{10}^9{\mathrm{M}}^{-1}} \\ =\frac{84.6}{1.75×{10}^9{\mathrm{M}}^{-1}} \\ =4.834285714×{10}^{-8}\mathrm{M} \\ \mathrm{concentration}\mathrm{detection}\mathrm{limit}=4.8×{10}^{-8}\mathrm{M} \end{gathered}$$

$$\begin{gathered} \mathrm{lower}\mathrm{limit}\mathrm{of}\mathrm{quantitation}=\frac{10\mathrm{s}}{\mathrm{m}} \\ =\frac{\left(10\right)\left(28.2\right)}{1.75×{10}^9\mathrm{M}} \\ =\frac{282.0}{1.75×{10}^9{\mathrm{M}}^{-1}} \\ =1.611428571×{10}^{-7}\mathrm{M} \end{gathered}$$