Question

Internal standard calibration curve. Figure 5-10 is a graph of ${\mathbf{A}}_{\mathbf{x}}\mathbf{/}{\mathbf{A}}_{\mathbf{s}}\mathbf{}\mathbf{versus}\mathbf{}\mathbf{\left[}\mathbf{X}\mathbf{\right]}\mathbf{/}\mathbf{\left[}\mathbf{S}\mathbf{\right]}\mathbf{=}\mathbf{(}\mathbf{mol}\mathbf{\%}\mathbf{vinyl}\mathbf{}\mathbf{acetate}\mathbf{}\mathbf{units}\mathbf{}\mathbf{)}\mathbf{}\mathbf{/}\mathbf{(}\mathbf{mol}\mathbf{\%}\mathbf{ethylene}\mathbf{}\mathbf{units}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{q}\mathbf{/}\mathbf{p}$ in Reaction 5-13.

(d) From the uncertainty${\mathbf{u}}_{\mathbf{b}}$of the intercept, find the$\mathbf{95}\mathbf{\%}$confidence interval for the intercept. Does this interval include the theoretical value of zero?

Short answer

The$95\%$confidence interval for the intercept$95\%confidence=\pm 0.160$

Step-by-step solution

Define the equation

We will use this equation to calculate the standard uncertainty in $x$.

${\mathrm{u}}_{\mathrm{x}}=\frac{{\mathrm{s}}_{\mathrm{y}}}{\left|\mathrm{m}\right|}\sqrt{\frac{1}{\mathrm{k}}+\frac{1}{\mathrm{n}}+\frac{{\left(\mathrm{y}-\overline{\mathrm{y}}\right)}^2}{{\mathrm{m}}^2\sum _{}^{}{\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}^2}}$

${\mathrm{s}}_{\mathrm{y}}$is the standard deviation of $\mathrm{y}$,

is the absolute value of the slope,

k is the number of replicate measurements of the unknown,

n is the number of data points for the calibration line,

y is the mean value of y

x is the mean value of x.

Obtain the 95% confidence interval.

Now, let's calculate the $95\%$confidence interval for $\frac{\left[X\right]}{\left[S\right]}$

$$\begin{gathered} 95\%\mathrm{confidence}=\pm {\mathrm{tu}}_{\mathrm{x}} \\ 95\%\mathrm{confidence}=\pm 2.78.019495 \\ \mathrm{confidence}=\pm 0.541 \end{gathered}$$

Now again use the same formula

$$\begin{gathered} 95\%\mathrm{confidence}=\pm {\mathrm{tu}}_{\mathrm{b}}{\mathrm{u}}_{\mathrm{b}} \\ =\mathrm{o}.\mathrm{o}574\mathrm{t}=2.7895\mathrm{confidence} \\ =\pm 0.160 \end{gathered}$$

Thus confidence interval$95\%confidence=\pm 0.160$