91Ó°ÊÓ

Comparing the two samples in this problem, undergo the Fand t tests for both the subparts.

Indicator 1 vs. Indicator 2:

TheF-test is done as follows:

$$\begin{gathered} {\mathrm{F}}_{\mathrm{calc}}=\frac{{\mathrm{s}}_1^2}{{\mathrm{s}}_2^2} \\ =\frac{{\left(0.00225\right)}^2}{{\left(0.00098\right)}^2} \\ =5.27. \end{gathered}$$

$F_{table}≈2.2$(based on Table 4-3; degrees of freedom for $s_1=27$and $s_2=17$).

Since $F_{calc}>F_{table}$,then there is a significant difference between the variances and the standard deviations of the two indicators.

TheT-test is done as follows:

Based on the F test result,$F_{calc}>F_{table}$,so we will perform a case 2b of t test,

$$\begin{gathered} {\mathrm{t}}_{\mathrm{calc}}=\frac{\left|{\overline{\mathrm{x}}}_1-{\overline{\mathrm{x}}}_2\right|}{\sqrt{\left({\mathrm{s}}_1^2/{\mathrm{n}}_1\right)+\left({\mathrm{s}}_2^2/{\mathrm{n}}_2\right)}} \\ =\frac{\left|0.9565-0.08686\right|}{\sqrt{{\displaystyle \frac{{\left(0.00225\right)}^2}{28}}+{\displaystyle \frac{{\left(0.00098\right)}^2}{18}}}} \\ =18.16 \end{gathered}$$

$t_{table}=2.021$(based on Table 4-4 ; degrees of freedom =40;95% confidence level).

Since $t_{calc}>t_{table}$,there is a significant difference between the means of the two indicators.

Indicator 2 vs. Indicator 3:

F-test(Note: we let ${\overline{\mathbf{x}}}_{\mathbf{1}}\mathbf{,}{\mathbf{s}}_{\mathbf{1}}$, and ${\mathbf{n}}_{\mathbf{1}}$ be equal to indicator 3’s statistical values).

$$\begin{gathered} F_{calc}=\frac{s_1^2}{s_2^2} \\ =\frac{{\left(0.00113\right)}^2}{{\left(0.00098\right)}^2} \\ =1.33. \end{gathered}$$

$F_{table}≈2.2$(Based on Table 4-3 ; degrees of freedom for $s_1=28$and $s_2=17$).

Since $F_{calc}<F_{table},$there is no significant difference between the variances and the standard deviations of the two indicators.

A t-testis done as follows:

Since the conclusion in our test result was $F_{calc}<F_{table},$perform 2a of t test.

$$\begin{gathered} s_{pooled}=\sqrt{\frac{s_1^2\left(n_1-1\right)+s_2^2\left(n_2-1\right)}{n_1+n_2-n_t}} \\ =\sqrt{\frac{{\left(0.00113\right)}^2\left(29-1\right)+{\left(0.00098\right)}^2\left(18-1\right)}{29+18-2}} \\ =1.07579428×{10}^{-3} \end{gathered}$$

$$\begin{gathered} t_{calc}=\frac{\left|{\overline{x}}_1-{\overline{x}}_2\right|}{s_{pooled}}\sqrt{\frac{n_1{n}_2}{n_1+n_2}} \\ =\frac{\left|0.08641-0.08686\right|}{1.07579428×{10}^{-3}}\sqrt{\frac{\left(29\right)\left(18\right)}{29+18}} \\ =1.39≈1.4. \end{gathered}$$

$t_{table}≈2.021$(based on Table 4-4 ; degrees of freedom).

$$\begin{gathered} =n_1+n_2-n_t \\ =29+18-2, \\ =45. \end{gathered}$$,

(confidence level)

Since $t_{calc}<t_{table},$there is no significant difference between the means of two indicators.