Question

In Figure 5-6, the x-intercept is $\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{89}\mathit{m}\mathit{M}$and its standard uncertainty is$\mathbf{0}\mathbf{.}{\mathbf{09}}_{\mathbf{8}}\mathit{m}\mathit{M}$. Find the$\mathbf{90}\mathbf{\%}$and$\mathbf{99}\mathbf{\%}$confidence intervals for the intercept.

Short answer

The$90\%$and$90\%$confidence interval for the intercept has to be calculated as$\pm 0.19\mathrm{Mm}$and$\pm 034\mathrm{Mm}$.

Step-by-step solution

Concept used

Confidence Intervals:

The confidence interval is given by the equation:

Confidence interval $=\overline{X}\pm \frac{ts}{\sqrt{n}}$

$=\overline{x}\pm t{u}_x($since standard uncertainty $\left(u_x\right)=s/\sqrt{n})$

Where,

$\overline{x}$is mean

$s$is standard deviation

$t$is Student's t

$n$is number of measurements

$u_x$is standard uncertainty

Calculation of confidence interval

Given

$x$-intercept $=-2.89\mathrm{mM}$

Standard uncertainty $=0.098\mathrm{mM}$

There are points

So, the degree of freedom is

$9-2=7$

For $90\%$confidence,

$\mathrm{t}=1.895$

Confidence interval is $\pm \left(1.895\right)\left(0.98\mathrm{mM}\right)=\pm 0.19\mathrm{mM}$

For $90\%$confidence,

$t=3.500$

Confidence interval is$\pm \left(3.500\right)\left(0.98\mathrm{mM}\right)=\pm 0.19\mathrm{mM}$