Question

A Standard Reference Material is certified to contain 94.6 ppm of an organic contaminant in soil. Your analysis gives values of 98.6,98.4,97.2,94.6, and 96.2ppm. Do your results differ from the expected result at the 95% confidence level? If you made one more measurement and found 94.5, would your conclusion change?

Short answer

The measured value significantly differs from the accepted value at 95% confidence level for both the original data and the one that is added with another measurement.

Step-by-step solution

t-test for Case 1

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$t_{calc}$:

$$\begin{gathered} \overline{x}=\frac{{\sum }_i{x}_i}{n} \\ \hspace{0.33em}=\frac{98.6+98.4+97.2+94.6+96.2}{5} \\ =97. \end{gathered}$$

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

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

$t_{table}=2.776$(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.

t-test for Case 2

Adding one measurement (=94.5) , we have:

$$\begin{gathered} \overline{x}=\frac{{\sum }_i{x}_i}{n} \\ \hspace{0.33em}=\frac{98.6+98.4+97.2+94.6+96.2+94.5}{6} \\ =96.583 \end{gathered}$$

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

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

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.