91Ó°ÊÓ

The problem is based on the Least square method.

The formula of standard uncertainty is as follows:

${\mathit{u}}_{\mathbf{x}}\mathbf{=}\frac{{\mathbf{S}}_{\mathbf{y}}}{\left|m\right|}\sqrt{\frac{\mathbf{1}}{\mathbf{k}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{n}}\mathbf{+}\frac{{\mathbf{(}\mathbf{y}\mathbf{-}\overline{\mathbf{y}}\mathbf{)}}^{\mathbf{2}}}{{\mathbf{m}}^{\mathbf{2}}\underset{\mathbf{}}{\mathbf{∑}}{\mathbf{(}{\mathbf{x}}_{\mathbf{i}}\mathbf{-}\overline{\mathbf{x}}\mathbf{)}}^{\mathbf{2}}}}\mathbf{.}$

Here, the standard uncertainty is $u_x$, the standard deviation of y is $s_y$ , the total number of measurement is n, the value of the measurement of the unknown is k, the slope is m, the mean of y measurements is $\overline{y}$, the mean of x measurements is $\overline{x}$.

(a)

The formula of absorbance is$\mathbf{absorbance}\mathbf{}\mathbf{=}\mathbf{m}\mathbf{×}\mathbf{(}\mathbf{μ²\mu }\mathbf{}\mathbf{}\mathbf{of}\mathbf{}\mathbf{protein}\mathbf{}\mathbf{)}\mathbf{+}\mathbf{b}\mathbf{.}$

Rearrange the formula for $\mathrm{μ}$g of protein as follows:

$\mathrm{μ}g$of protein $=\frac{absorbance-b}{m}.$

Check page number 85 (figure 4.13 ) in the textbook for the values of the slope and the intercept.

The value of the slope is 0.0163 and the intercept is 0.0047.

Calculate the corrected absorbance as follows:

Subtract the absorbance of the blank from the absorbance of the measured to calculate the corrected absorbance:

Corrected absorbance$=0.264-0.095$

$=0.169.$

$\mathrm{μ}g$of protein $=\frac{0.169-0.0047}{0.0163}$

$\mathrm{μ}g$of protein $=\frac{0.1643}{0.0163}$

$\mathrm{μ}g$of protein $=10.079\mathrm{μ}g$

So, the amount of protein is$10.1\mathrm{μ}g.$

(b)

Use standard uncertainty formula to calculate$u_x$

Refer to figure 4-13 in the textbook (page number 85) for the value of $s_y$.

The value of $s_y$is 0.0046.

Refer to figure 4-13 in the textbook (page number 87) for the values of $x_i, y, \overline{x}, and\overline{y}.$The values of localid="1663566988931" $\overline{x}$are $0,0,0,5.0,5.0,5.0,10.0,10.0,10.0,15.0,15.0,20.0,20.0,20.0$and $\overline{x}$is 9.643$\mathrm{μ}g$. The values of y is 0.302 and $\overline{y}$is 0.1618.

Calculate$\underset{}{∑}{\left(x_i-\overline{x}\right)}^2$value as follows:

$$\begin{gathered} \underset{}{∑}(x_i-\overline{x}{)}^2=(0-9.643{)}^2+(0-9.643{)}^2+(0-9.64.643{)}^2+ \\ +(10-9.643{)}^2+(10-9.643{)}^2+(10-9.643{)}^2 \\ +(20-9.643{)}^2+(20-9.643{)}^2+(20-9.643{)}^2 \\ \underset{}{∑}(x_i-\overline{x}{)}^2=92.99+92.99+92.99+21.56+ \\ +28.7+28.7+107.27+107.27+107.27 \\ \underset{}{∑}(x_i-\overline{x}{)}^2=723.25. \end{gathered}$$

Replace 0.0163 for m,0.0059 for $s_y$, 4 for k, 14 for n, 724 for $\underset{}{∑}(x_i-\overline{x}{)}^2, 0.302$ for y, and 0.1618 for $\overline{y}$and calculate$u_x$.

$$\begin{gathered} u_x=\frac{0.005}{\left|0.0163\right|\sqrt{{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{14}}+{\displaystyle \frac{{\left(0.302-0.161\right)}^1}{{\left(0.0163\right)}^2\left(723.25\right)}}}} \\ =0.31\sqrt{\frac{1}{4}+\frac{1}{14}+\frac{0.0197}{0.1922}} \\ =0.31×0.65. \\ n{u}_x=0.31×0.65 \\ {u}_x=0.2015. \end{gathered}$$

So, the standard uncertainty $u_x$is$Â\pm 0.2015.$

The formula to find the confidence interval for an unknown measurement x is as follows:

Confidence intervals for $x=Â\pm t{u}_x.$

Here, the value of Student's is tand the standard uncertainty is$u_x$.

For the degree of freedom 12, the value of t for the 95% confidence interval is $2.179.

From the solution of part (b),$u_x$is 0.2015.

Replace 0.2015 for $u_x$and 2.179 for tand calculate the confidence interval.

Confidence intervals for$x=Â\pm 2.179×0.2015.$

Confidence intervals for $x=Â\pm 0.439.$

Thus, the standard uncertainty $u_x$is 0.2015 $\mathrm{μ}g$and the 95% confidence interval is$10.08\mathrm{μ}gÂ\pm 0.44.$