Answer
$T\lt w_{1-α}$: null hypothesis is rejected.
There is enough evidence to conclude that the median pH of rain in Houston is greater than the median pH of rain near Chicago.
Work Step by Step
$H_0:M_{Houston}=M_{Chicago}$ versus $H_1:M_{Houston}\gt M_{Chicago}$
$pH~of~rain~~~~~~Sample~~~~~~~~Rank$
$~~~~~~4.11~~~~~~~~~~~Houston~~~~~~~~~~~1$
$~~~~~~4.22~~~~~~~~~~~Chicago~~~~~~~~~~~~2$
$~~~~~~4.25~~~~~~~~~~~Houston~~~~~~~~~~~3$
$~~~~~~4.35~~~~~~~~~~~Chicago~~~~~~~~~~~~4$
$~~~~~~4.36~~~~~~~~~~~Chicago~~~~~~~~~~~~5$
$~~~~~~4.40~~~~~~~~~~~Chicago~~~~~~~~~~6.5$
$~~~~~~4.40~~~~~~~~~~~Chicago~~~~~~~~~~6.5$
$~~~~~~4.45~~~~~~~~~~~Chicago~~~~~~~~~~~~8$
$~~~~~~4.46~~~~~~~~~~~Houston~~~~~~~~~~~9$
$~~~~~~4.49~~~~~~~~~~~Chicago~~~~~~~~~~~10$
$~~~~~~4.52~~~~~~~~~~~Chicago~~~~~~~~~~~11$
$~~~~~~4.54~~~~~~~~~~~Chicago~~~~~~~~~~~12$
$~~~~~~4.63~~~~~~~~~~~Chicago~~~~~~~~~~~13$
$~~~~~~4.64~~~~~~~~~~~Chicago~~~~~~~~~~~14$
$~~~~~~4.65~~~~~~~~~~~Houston~~~~~~~~~~15$
$~~~~~~4.69~~~~~~~~~~~Chicago~~~~~~~~~~~17$
$~~~~~~4.69~~~~~~~~~~~Chicago~~~~~~~~~~~17$
$~~~~~~4.69~~~~~~~~~~~Houston~~~~~~~~~~17$
$~~~~~~4.71~~~~~~~~~~~Houston~~~~~~~~~~19$
$~~~~~~4.75~~~~~~~~~~~Chicago~~~~~~~~~~~20$
$~~~~~~4.76~~~~~~~~~~~Houston~~~~~~~~~~21$
$~~~~~~4.93~~~~~~~~~~~Houston~~~~~~~~~~22$
$~~~~~~5.10~~~~~~~~~~~Houston~~~~~~~~~~23$
$~~~~~~5.14~~~~~~~~~~~Houston~~~~~~~~~~24$
$~~~~~~5.22~~~~~~~~~~~Houston~~~~~~~~~~25$
$~~~~~~5.22~~~~~~~~~~~Houston~~~~~~~~~~26$
Small-sample case:
$S=1+3+9+15+17+19+21+22+23+24+25+26=205$
$T=S-\frac{n_1(n_1+1)}{2}=207-\frac{12(12+1)}{2}=205-78=127$
$w_α=w_{0.05}=52$
(According to table XIII, for $n_1=12$, $n_2=14$ and $α=0.05$)
Critical value:
$w_{1-α}=n_1n_2-w_α$
$w_{0.95}=12\times14-52=116$
Since $T\gt w_{1-α}$, we reject the null hypothesis.
