Question

Consider the least-squares problem in Figure 4-11.

(a) Suppose that a single new measurement produces a yvalue of 2.58. Find the corresponding xvalue and its standard uncertainty, ${\mathbf{u}}_{\mathbf{x}}$.

(b) Suppose you measure yfour times and the average is 2.58. Calculate ${\mathbf{u}}_{\mathbf{x}}$based on four measurements, not one.

(c) Find the 95%confidence intervals for (a) and (b).

Short answer

(a) The corresponding x value and its standard uncertainty value ${\mathit{u}}_{\mathbf{x}}$ for a y value of 2.58 is $2.00\pm 0.3$.

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

(c) 95% confidence interval for (a) is $2.0\pm 0.8.$

95% confidence interval for (b) is$2.0\pm 0.6.$

Step-by-step solution

Uncertainty with a Calibration Curve

The propagation for the equation $\mathit{y}\mathbf{=}\mathit{m}\mathit{x}\mathbf{+}\mathit{b}$(but noty=mx) gives,

The standard uncertainty inxis mathematically presented as:

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

Here,

$u_x$, is the standard uncertainty for x,

y is the corrected absorbance of the unknown,

$x_i$is the mass of the sample in standards,

$\overline{y}$is the average of the value of y,

$\overline{x}$is the average of the value of x, and

k is the number of replicate measurements.

Standard uncertainty

(a)

Consider the least-squares in Figure 4-11. A new single measurement gives a y value of 2.58.

Find the value of x and its standard uncertainty as shown below.

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

On rearranging, we get,

$$\begin{gathered} x=\frac{y-b}{m} \\ =\frac{2.58-1.35}{0.615} \\ =2.00 \end{gathered}$$

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

$$\begin{gathered} \overline{y}=\left(2+3+4+5\right)/4 \\ 3.5\overline{x}=\left(1+3+4+6\right)/4 \\ =3.5. \end{gathered}$$

The $\sum _{}{\left(x_1-\overline{x}\right)}^2$is found as follows:

$$\begin{gathered} \sum _{}{\left(x_1-\overline{x}\right)}^2 \\ =13.0x. \end{gathered}$$

Standard Uncertainty in x

The standard uncertainty inis mathematically presented as:

$$\begin{gathered} u_{\mathrm{x}}=\frac{{\mathrm{S}}_{\mathrm{y}}}{\left|\mathrm{m}\right|}\sqrt{\frac{1}{\mathrm{k}}+\frac{1}{\mathrm{n}}+\frac{{(\mathrm{y}-\overline{\mathrm{y}})}^2}{{\mathrm{m}}^2\sum _{}{({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}})}^2}}. \\ =\frac{0.19612}{\left|0.61538\right|}\sqrt{\frac{1}{1}+\frac{1}{4}+\frac{{\left(2.58-3.5\right)}^2}{{\left(0.61538\right)}^2\left(13.0\right)}} \\ =0.38 \end{gathered}$$

Therefore, the value for xis $2.0\pm 0.38.$

Conclusion:

The corresponding value and its standard uncertainty value ${\mathit{u}}_{\mathbf{x}}$ for a y value of 2.58 is $2.00\pm 0.3.$

Value of y and x 

(b)

Consider the least-squares in Figure 4-11. The y is measured four times and its average value is 2.58 The number in the subscript denotes the insignificant figures.

Find the value of x and its standard uncertainty as shown below.

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

On rearranging, we get,

$$\begin{gathered} x=\frac{y-b}{m} \\ =\frac{2.58-1.35}{0.615} \\ x=2.00. \end{gathered}$$

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

$$\begin{gathered} \overline{y}=\left(2+3+4+5\right)/4 \\ 3.5\overline{x}=\left(1+3+4+6\right)/4 \\ =3.5. \end{gathered}$$

The $\sum _{}{\left(x_i-\overline{x}\right)}^2$is found as follows:

$$\begin{gathered} \sum _{}{\left(x_i-\overline{x}\right)}^2 \\ =13.0. \end{gathered}$$

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

The standard uncertainty in x is calculated as:

$$\begin{gathered} u_x=\frac{s}{\left|m\right|}\sqrt{\frac{1}{k}+\frac{1}{n}+\frac{{\left(y-\overline{y}\right)}^2}{m^2\sum _{}{\left(x_i-\overline{x}\right)}^2}} \\ =0.26 \end{gathered}$$

Therefore, the value for x is$2.00\pm 0.26.$

Confidence Intervals

Confidence interval$\mathbf{=}\overline{\mathbf{x}}\mathbf{\pm }\frac{\mathbf{t}\mathbf{s}}{\sqrt{\mathbf{n}}}\mathbf{=}\overline{\mathbf{x}}\mathbf{\pm }\mathit{t}{\mathit{u}}_{\mathbf{x}}$ since the standard uncertainty is$\mathbf{\left(}{\mathit{u}}_{\mathbf{x}}\mathbf{\right)}\mathbf{=}\mathit{s}\mathbf{/}\sqrt{\mathbf{n}}\mathbf{)}\mathbf{.}$

Here,

$\overline{x}$is the mean,

s is the standard deviation,

n is the number of measurements, and

$u_x$is the standard uncertainty.

Result of the Measurement

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

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

The number in the subscript denotes the insignificant figures.

Calculation of confidence intervals:

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

Step 7:  Confidence Interval

(c)

confidence interval for part (a) is calculated as follows:

Confidence interval$=\overline{x}\pm t{u}_x.$

confidence interval$=2.00\pm (2.353\times 0.38)$

$$\begin{gathered} =2.00\pm (0.89) \\ =2.0\pm (0.8)(roundedoff). \end{gathered}$$

90% confidence interval for part (b) is calculated as follows:

Confidence interval $=\overline{x}\pm t{u}_x.$

confidence interval$=2.00\pm (2.353\times 0.26)$

$$\begin{gathered} =2.00\pm (0.61) \\ =2.0\pm (0.6)(\mathrm{rounded}\mathrm{off}). \end{gathered}$$

Conclusion:

90% confidence interval for (a) is calculated as $2.00\pm 0.8.$

90% confidence interval for (b) is calculated as $2.0\pm 0.6.$