91Ó°ÊÓ

Confidence interval $\mathbf{=}\stackrel{\mathbf{¯}}{\mathbf{x}}\mathbf{Â\pm }\frac{\mathbf{ts}}{\sqrt{\mathbf{n}}}\mathbf{=}\stackrel{\mathbf{¯}}{\mathbf{x}}\mathbf{Â\pm }{\mathbf{tu}}_{\mathbf{x}}$since the standard uncertainty

$\left(u_x\right)\mathbf{=}\mathbf{s}\mathbf{/}\sqrt{\mathbf{n}}$

Here,

$\stackrel{\mathbf{¯}}{\mathbf{x}}$is the mean,

s is the standard deviation,

t is the Student's t,

uis the number of measurements, and

u_xis the standard uncertainty.

The number of measurements of the known point is $\mathrm{n}=10$

The result of the protein is $15.2_2\left(Â\pm 0.4_6\right)\mathrm{μ²\mu }$

The standard uncertainty is ${\mathrm{u}}_{\mathrm{x}}=0.4_6\mathrm{μ²\mu }$

The numbers in the subscript denotes the insignificant figures.

Calculation of Confidence intervals:

The t value corresponding to 90% confidence level for eight degrees of freedom is ${\mathrm{t}}_{90\%}=1.860$

The t value corresponding to 99% confidence level for eight degrees of freedom is ${\mathrm{t}}_{99\%}=3.355$

90% confidence interval is calculated as follows:

Confidence interval $\stackrel{¯}{\mathrm{x}}Â\pm {\mathrm{tu}}_{\mathrm{x}}$

90% confidence interval role="math" localid="1663318037622" $=15.22\left(Â\pm 1.860×0.46\right)$

$$\begin{gathered} =15.2_2\left(Â\pm 0.8_6\right)\mathrm{μ²\mu } \\ =15.2\left(Â\pm 0.8\right)\mathrm{μ²\mu } \end{gathered}$$

99% confidence interval is calculated as follows:

Confidence interval$=\stackrel{¯}{\mathrm{x}}Â\pm {\mathrm{tu}}_{\mathrm{x}}$

90% confidence interval $=15.22\left(Â\pm 3.355×0.46\right)$

$$\begin{gathered} =15.2_2\left(Â\pm 1.5\right)\mathrm{μ²\mu } \\ =15\left(Â\pm 1\right)\mathrm{μ²\mu } \end{gathered}$$