91Ó°ÊÓ

The mean (\(\mu\)) and standard deviation (\(\sigma\)) for the large-sample approximation can be determined using the following formulas: $$\mu = \frac{n_1(n_1 + n_2 + 1)}{2}$$ $$\sigma = \sqrt{\frac{n_1n_2(n_1 + n_2 + 1)}{12}}$$ Plug in the given values of \(n_1 = 20\) and \(n_2 = 25\): $$\mu = \frac{20(20 + 25 + 1)}{2} = 460$$ $$\sigma = \sqrt{\frac{20 \times 25(20 + 25 + 1)}{12}} = \sqrt{2200} = 46.9$$

The z-score can be calculated using the following formula: $$z = \frac{T_1 - \mu}{\sigma}$$ Plug in the given values of \(T_1 = 252\), \(\mu = 460\), and \(\sigma = 46.9\): $$z = \frac{252 - 460}{46.9} = -4.44$$

With the z-score, we can now calculate the two-tailed p-value, as we do not know if the alternative hypothesis is an increase or decrease in the distribution. Look up the p-value for the left tail using -4.44 in a standard normal table or use a calculator that can compute the p-value directly from the z-score. The p-value for -4.44 from a standard normal table is approximately equal to 0.00001. Then multiply by 2 for the two-tailed test: $$p \text{-value} = 2 \times 0.00001 = 0.00002$$ The p-value is 0.00002, which means there is a very small probability of observing this difference in ranks by random chance, assuming the null hypothesis is true. Since the p-value is small, we conclude that there is a significant difference between the two population distributions.