## 4 Examples of the Procedure  
Three examples are now shown to demonstrate the simple technique. The first two examples used simulated data, and the third used real-life datasets. First, let a random sample 1 of size 40 for a variable \(X\) from \(\Gamma(k_{1},\theta_{1})=\Gamma(2,4)\) be 7.4323, 8.7579, 3.8367, 24.3859, 4.1931, 1.4071, 2.1765, 13.8280, 2.1577, 8.9442, 3.6841, 3.6455, 5.8649, 2.2743, 3.5843, 4.0561, 8.9005, 10.6677, 12.9202, 21.4182, 4.9585, 7.7933, 1.4325, 2.4112, 11.1183, 11.2643, 7.2044, 6.7496, 7.0007, 2.2562, 2.7139, 9.4639, 4.3929, 9.1965, 16.9821, 4.4714, 12.9340, 7.5178, 13.2414, 12.3011. Let a random sample 2 of size 40 for a variable \(Y\) from \(\Gamma(k_{2},\theta_{2})=\Gamma(4,2)\) be 3.5289, 7.5509, 15.9686, 2.9356, 4.2433, 22.0422, 12.0441, 3.2122, 6.0766, 3.7120, 17.4808, 14.7796, 17.8730, 3.8967, 7.1522, 10.7450, 7.7302, 9.9103, 6.2068, 3.4237, 12.2881, 3.8782, 5.0942, 13.1173, 4.6774, 7.0516, 11.4397, 4.5981, 6.1976, 12.9560, 15.7683, 9.1313, 6.0904, 3.1014, 4.2042, 13.4389, 4.5231, 4.5774, 10.8752, 5.0082. The original values were rounded off to four decimal places. Their visualization is shown in Figure 2. \begin{table}
\begin{tabular}{c c c c c c} \hline \(k_{1}\) & \(k_{2}\) & \(\theta_{1}\) & \(\theta_{2}\) & \(\Delta\) in Mean & Power [12] & Power \\ \hline
1 & 2 & 1 & 0.75 & 0.50 & 0.679 & 0.024 \\
1 & 2 & 1 & 1.00 & 1.00 & 0.410 & 0.190 \\
1 & 2 & 1 & 1.25 & 1.50 & 0.225 & 0.490 \\
1 & 5 & 1 & 0.75 & 2.75 & 0.999 & 0.999 \\
1 & 5 & 1 & 1.00 & 4.00 & 0.994 & 1.000 \\
1 & 5 & 1 & 1.25 & 5.25 & 0.965 & 1.000 \\
2.5 & 5 & 1 & 0.75 & 1.25 & 0.967 & 0.150 \\
2.5 & 5 & 1 & 1.00 & 2.50 & 0.768 & 0.800 \\
2.5 & 5 & 1 & 1.25 & 3.75 & 0.450 & 0.999 \\
5 & 5 & 1 & 0.75 & 1.25 & 0.278 & 0.050 \\
5 & 5 & 1 & 1.00 & 0.00 & 0.170 & 0.000 \\
5 & 5 & 1 & 1.25 & 1.25 & 0.900 & 0.015 \\
5 & 10 & 1 & 0.75 & 2.50 & 0.998 & 0.530 \\
5 & 10 & 11 & 1.00 & 5.00 & 0.967 & 0.990 \\
5 & 10 & 1 & 1.25 & 7.50 & 0.574 & 1.000 \\ \hline \end{tabular}
\end{table} 
Table 14: Comparison of test powers for the method given in [12] and the method proposed in this text (the last column). For the method in [12], the sample sizes are both equal to 10, and the powers are taken from [12]. For the method proposed here, the sample sizes are both equal to 30. The two distributions have the same means of 8; hence, the null hypothesis holds, and the variances are 32 and 16, respectively. The test criterion in abs. value is \(|\overline{x}-\overline{y}|=|7.7385-8.4634|=0.725\), the percentile-based critical value equals \(1.96\widehat{\sigma a}^{1/2}=1.96\sqrt{40^{-1}[\widehat{\sigma a}(X)+\widehat {\sigma a}(Y)]}=1.96\sqrt{40^{-1}[27.206+23.587]}=2.208\), and the hypothesis is correctly accepted. The variances \(\widehat{\sigma a}(X)\), \(\widehat{\sigma a}(Y)\) were calculated with the \((n+1)^{-1}\) divisor. The \(d\) multiples were taken as \(\pm 1.96\) since scenario 1 is observed here. In the second example, let a sample 1 of 40 for a variable \(X\) from \(\Gamma(k_{1},\theta_{1})=\Gamma(3,4)\) be 20.2621, 5.0067, 6.8178, 17.9724, 5.7980, 47.9481, 24.0818, 16.9974, 15.6469, 2.2784, 7.0815, 16.5170, 12.8472, 20.7937, 11.4743, 10.6894, 10.7756, 15.0273, 6.6600, 21.2111, 10.5329, 18.7154, 12.5374, 8.01664, 5.4640, 26.8381, 16.0145, 20.7234, 5.3591, 11.9667, 14.6437, 8.9770, 9.0279, 10.5961, 17.2995, 9.2122, 27.5701, 9.9853, 15.5422, 2.3258. Let a sample 2 of size 40 for a variable \(Y\) from \(\Gamma(k_{2},\theta_{2})=\Gamma(4,2.5)\) be 16.7113, 9.3561, 19.1426, 6.7874, 8.5002, 7.2641, 29.7096, 12.4152, 20.7611, 4.3977, 6.2623, 9.2883, 7.7831, 16.7462, 3.8046, 7.8470, 4.5086, 11.9305, 18.3177, 12.5083, 4.6306, 5.6554, 17.0151, 6.7073, 6.0264, 10.4029, 15.1457, 14.8109, 7.0398, 3.3200, 10.5470, 5.5386, 10.2651, 6.7952, 17.0514, 3.7160, 4.1363, 13.9024, 8.0307, 12.7817. The data visualization is in Figure 3.  
The population means are 12 and 10 this time, respectively, so \(H_{1}\) holds. The variances are 48 and 25, respectively.