Chapter 4, Problem 4.33P

(a)

Interpretation Introduction

Interpretation:

For measurement giving y value of $2.58$, the corresponding x value and its standard uncertainty has to be found out.

Concept Introduction:

Uncertainty with a Calibration Curve:

Propagation for the equation $y=mx+b$ (but not $y=mx)$ gives,

The standard uncertainty in x is given as:

$u_x=\frac{s_y}{\left|m\right|}\sqrt{\frac{1}{k}+\frac{1}{n}+\frac{{(y-\overline{y})}^2}{m^2{\displaystyle \sum {(x_i-\overline{x})}^2}}}$

Where,

$u_x$ is standard uncertainty for $x$

$y$ is the corrected absorbance of unknown

$x_i$ is the mass of sample in standards

$\overline{y}$ is the average of $y$ values

$\overline{x}$ is the average of $x$ values

k is the number of replicate measurements

Answer to Problem 4.33P

The corresponding x and its standard uncertainty value for y value of $2.58$ is $2.0\pm 0.3$

Explanation of Solution

Given data:

Consider least-squares in Figure 4-11

A new single measurement gives a y value of $2.58$

Finding of x value and its standard uncertainty:

The equation of straight line from figure 4-11 is $y=mx+b\Rightarrow y=0.61538x+1.34615$

On rearranging, we get,

$\begin{array}{l}x=\frac{y-b}{m}\\ \text{ =}\frac{2.58-1.35}{0.615}\\ \text{ =2}\text{.00}\end{array}$

The value of $\overline{y}$ and $\overline{x}$ is calculated as follows,

$\begin{array}{l}\overline{y}=(2+3+4+5)/4\\ \text{ =3}\text{.5}\\ \overline{x}=(1+3+4+6)/4\\ \text{ =3}\text{.5}\end{array}$

The ${\displaystyle \sum {(x_i-\overline{x})}^2}$ is found as follows,

$\begin{array}{l}{\displaystyle \sum {(x_i-\overline{x})}^2}={(}^1+{(}^3+{(}^4+{(}^6\\ \text{ =13}\text{.0}\end{array}$

The standard uncertainty in x is given as,

$\begin{array}{l}{u}_x=\frac{s_y}{\left|m\right|}\sqrt{\frac{1}{k}+\frac{1}{n}+\frac{{(y-\overline{y})}^2}{m^2{\displaystyle \sum {(x_i-\overline{x})}^2}}}\\ \text{ =}\frac{0.19612}{\left|0.61538\right|}\sqrt{\frac{1}{1}+\frac{1}{4}+\frac{{(2.58-3.5)}^2}{{(0.61538)}^2(13.0)}}\\ \text{ =0}\text{.38}\\ \text{ =0}\text{.3}\text{(rounded off)}\end{array}$

Therefore, the value for x is $2.0\pm 0.3$

Conclusion

The corresponding x and its standard uncertainty value for y value of $2.58$ is found out as $2.0\pm 0.3$

(b)

Interpretation Introduction

Interpretation:

The four times measurement of y gives an average value of $2.58$, the corresponding x value and its standard uncertainty based on four measurements has to be found out.

Concept Introduction:

Uncertainty with a Calibration Curve:

Propagation for the equation $y=mx+b$ (but not $y=mx)$ gives,

The standard uncertainty in x is given as:

$u_x=\frac{s_y}{\left|m\right|}\sqrt{\frac{1}{k}+\frac{1}{n}+\frac{{(y-\overline{y})}^2}{m^2{\displaystyle \sum {(x_i-\overline{x})}^2}}}$

Where,

$u_x$ is standard uncertainty for $x$

$y$ is the corrected absorbance of unknown

$x_i$ is the mass of sample in standards

$\overline{y}$ is the average of $y$ values

$\overline{x}$ is the average of $x$ values

k is the number of replicate measurements

Answer to Problem 4.33P

The corresponding x and its standard uncertainty value based on four measurements is   $2.0\pm 0.2$

Explanation of Solution

Given data:

Consider least-squares in Figure 4-11

The y is measured four times and its average value is $2.58$

The numbers in subscript denotes insignificant figures.

Finding of x value and its standard uncertainty:

The equation of straight line from figure 4-11 is $y=mx+b\Rightarrow y=0.61538x+1.34615$

On rearranging, we get,

$\begin{array}{l}x=\frac{y-b}{m}\\ \text{ =}\frac{2.58-1.35}{0.615}\\ \text{ =2}\text{.00}\end{array}$

The value of $\overline{y}$ and $\overline{x}$ is calculated as follows,

$\begin{array}{l}\overline{y}=(2+3+4+5)/4\\ \text{ =3}\text{.5}\\ \overline{x}=(1+3+4+6)/4\\ \text{ =3}\text{.5}\end{array}$

The ${\displaystyle \sum {(x_i-\overline{x})}^2}$ is found as follows,

$\begin{array}{l}{\displaystyle \sum {(x_i-\overline{x})}^2}={(}^1+{(}^3+{(}^4+{(}^6\\ \text{ =13}\text{.0}\end{array}$

Here, the number of replicate measurement is four. Hence, $k=4$

The standard uncertainty in x is given as,

$\begin{array}{l}{u}_x=\frac{s_y}{\left|m\right|}\sqrt{\frac{1}{k}+\frac{1}{n}+\frac{{(y-\overline{y})}^2}{m^2{\displaystyle \sum {(x_i-\overline{x})}^2}}}\\ \text{ =}\frac{0.19612}{\left|0.61538\right|}\sqrt{\frac{1}{4}+\frac{1}{4}+\frac{{(2.58-3.5)}^2}{{(0.61538)}^2(13.0)}}\\ \text{ =0}\text{.26}\\ \text{ =0}\text{.2}\text{(rounded off)}\end{array}$

Therefore, the value for x is $2.0\pm 0.2$

Conclusion

The corresponding x and its standard uncertainty value based on four measurements is found out as $2.0\pm 0.2$

(c)

Interpretation Introduction

Interpretation:

The $95\%$ confidence intervals for (a) and (b) has to be found out.

Concept Introduction:

Confidence Intervals:

The confidence interval is given by the equation:

$\begin{array}{l}\text{Confidence interval = }\overline{x}\pm \frac{ts}{\sqrt{n}}\\ =\overline{x}\pm t{u}_x(\text{since standard uncertainty(}{u}_x)=s/\sqrt{n}\text{) }\end{array}$

Where,

$\overline{x}$ is mean

$s$ is standard deviation

$t$ is Student’s t

n is number of measurements

$u_x$ is standard uncertainty

To Find: The $95\%$ confidence intervals for (a) and (b)

Answer to Problem 4.33P

The $95\%$ confidence intervals for (a) is $2.0\pm 0.8$

The $95\%$ confidence intervals for (b) is $2.0\pm 0.6$

Explanation of Solution

Given data:

The results of the measurement in part (a) is $2.00\pm 0.38$

The results of the measurement in part (b) is $2.00\pm 0.26$

The numbers in the subscript denotes insignificant figures.

Calculation of Confidence intervals:

The t value corresponding to $90\%$ confidence level for three degrees of freedom is $t_{90\%}=2.353$

The $90\%$ confidence interval for part (a) is calculated as follows,

$\begin{array}{l}\text{ confidence interval = }\overline{x}\pm t{u}_x\\ \text{90\% confidence interval =2}\text{.00(}\pm 2.353\times 0.38)\\ \text{ =2}\text{.00(}\pm 0.89)\\ \text{ =2}\text{.0(}\pm 0.8)(\text{rounded off)}\end{array}$

The $90\%$ confidence interval for part (b) is calculated as follows,

$\begin{array}{l}\text{ confidence interval = }\overline{x}\pm t{u}_x\\ \text{90\% confidence interval =2}\text{.00(}\pm 2.353\times 0.26)\\ \text{ =2}\text{.00(}\pm 0.61)\\ \text{ =2}\text{.0(}\pm 0.6)\text{(rounded off)}\end{array}$

Conclusion

The $95\%$ confidence intervals for (a) is calculated as $2.0\pm 0.8$

The $95\%$ confidence intervals for (b) is calculated as $2.0\pm 0.6$