Answer
Yes, see explanations.
Work Step by Step
From the data set (1-Birmingham, 2-Chicago), we have
$n1=25, \bar X1=72.92, s_1^2=30.24$ and $n2=25, \bar X2=70.80, s_2^2=33.83$
a. State the hypotheses and identify the claim.
$H_o: \mu1-\mu2=0$
$H_a: \mu1-\mu2\gt0$ (claim, right tail test)
b. Find the critical value(s).
$\alpha=0.10, df=24, t_c=1.318$
c. Compute the test value.
$t=\frac{72.92-70.80-0}{\sqrt {\frac{30.24}{25}+\frac{33.83}{25}}}=1.324$
d. Make the decision.
Because $t\gt t_c$, we reject the null hypothesis.
e. Summarize the results.
At α= 0.10, it can be concluded that it is warmer in Birmingham.