91Ó°ÊÓ

For comparing two different locations of analysis, use both the test and the test.

The F-test is done as shown below.

$$\begin{gathered} F_{calc}=\frac{s_1^2}{s_2^2} \\ =\frac{{\left(30.0\right)}^2}{{\left(29.8\right)}^2} \\ =1.01. \end{gathered}$$

$F_{table}\left(95\%CL\right)≈1.84.$

$F_{table}\left(99\%CL\right)≈2.35.$

Since $F_{calc}<F_{table}$at both 95% and 99% confidence levels, there is no significant difference between the variances and the standard deviations of the analysis done in two different locations at both confidence levels.

The t-test is done as shown below.

The conclusion in our F test result states that $F_{calc}<F_{table}$ at both confidence levels. Now, perform a case 2a of the -test.

$$\begin{gathered} s_{pooled}=\sqrt{\frac{s_1^2\left(n_1-1\right)+s_2^2\left(n_2-1\right)}{n_1+n_2-n_t}} \\ =\sqrt{\frac{{\left(30.0\right)}^2\left(32-1\right)+{\left(29.8\right)}^2\left(32-1\right)}{32+32-2}} \\ =29.90016722 \end{gathered}$$

$$\begin{gathered} t_{calc}=\frac{\left|{\overline{\mathrm{x}}}_1-{\overline{\mathrm{x}}}_2\right|}{{\mathrm{s}}_{\mathrm{pooled}}}\sqrt{\frac{{\mathrm{n}}_1{\mathrm{n}}_2}{{\mathrm{n}}_1+{\mathrm{n}}_2}} \\ =\frac{\left|31.4-52.9\right|}{29.90016722}\sqrt{\frac{\left(32\right)\left(32\right)}{32+32}} \\ =2.88 \end{gathered}$$

$$\begin{gathered} t_{table}\left(95\%CL\right)≈2.042. \\ {t}_{table}\left(99\%CL\right)≈2.750. \end{gathered}$$

Since $t_{calc}>t_{table}$at both 95% and 99% confidence levels, there is a significant difference between the means of the analysis done in two different locations at both confidence levels.