91Ó°ÊÓ

Comparing a measured result with an accepted result that is known, use Case 1 of the test. Note that the given value of standard reference material is the accepted value.

From the formula of the confidence interval (CI), we isolate tand solve for$t_{calc}$:

$$\begin{gathered} \overline{x}=\frac{{∑}_i{x}_i}{n} \\  =\frac{98.6+98.4+97.2+94.6+96.2}{5} \\ =97. \end{gathered}$$

role="math" localid="1663318053416" $$\begin{gathered} s=\sqrt{\frac{{\displaystyle \underset{i}{∑{\left(x_i-\overline{x}\right)}^2}}}{n-1}} \\ =\sqrt{\frac{{\left(98.6-97\right)}^2+{\left(98.4-97\right)}^2+{\left(97.2-97\right)}^2+{\left(94.6-97\right)}^2+{\left(96.2-97\right)}^2}{5-1}} \\ =1.655294536 \end{gathered}$$

$$\begin{gathered} t_{calc}=\left(\overline{x}-\mathrm{μ}\right)\frac{\sqrt{n}}{s} \\ =\left(97-94.6\right)\frac{\sqrt{5}}{1.655294536} \\ =3.24 \end{gathered}$$

$t_{table}=2.776$(based on Table 4-4 degrees of freedom =4;95%CL).

Since $t_{calc}>t_{table}$there is a significant difference between the measured value and the accepted value at 95% confidence level.

Adding one measurement (=94.5) , we have:

$$\begin{gathered} \overline{x}=\frac{{∑}_i{x}_i}{n} \\  =\frac{98.6+98.4+97.2+94.6+96.2+94.5}{6} \\ =96.583 \end{gathered}$$

$$\begin{gathered} s=\sqrt{\frac{{\displaystyle \underset{i}{∑{\left(x_i-\overline{x}\right)}^2}}}{n-1}} \\ =\sqrt{\frac{\begin{array}{c}{\left(98.6-96.583\right)}^2+{\left(98.4-96.583\right)}^2+{\left(97.2-96.583\right)}^2\\ +{\left(94.6-96.583\right)}^2+{\left(96.2-96.583\right)}^2+{\left(94.5-96.583\right)}^2\end{array}}{6-1}} \\ =1.79823988. \end{gathered}$$

$$\begin{gathered} t_{calc}=\left(\overline{x}-\mathrm{μ}\right)\frac{\sqrt{n}}{s} \\ =\left(96.583-94.6\right)\frac{\sqrt{5}}{1.79823988} \\ =2.70. \end{gathered}$$

localid="1663318719798" $t_{table}=2.571$(based on Table 4-4; degrees of freedom=5;95%CL).

Since $t_{calc}>t_{table}$, there is still a significant difference between the measured value and the accepted value at 95% confidence level despite adding one measurement that is equal to 94.5.