Answer
$t_0\lt -t_α$: null hypothesis is rejected.
There is statistical evidence that mean wait-time has decreased. Now, notice that the difference is $43-42.3=0.7~minutes=42~seconds$. But, no customer would be more satisfied because he/she waited 42 seconds less. That is, the results have no practical significance.
Work Step by Step
$H_0:~µ=43$ versus $H_1:~µ\lt43$
Requirement:
$n\geq30$
$n=250$, so:
$d.f.=n-1=250$
$t_0=\frac{x ̅-µ_0}{\frac{s}{\sqrt n}}=\frac{42.3-43}{\frac{4.2}{\sqrt {250}}}=-2.64$
Left-tailed test:
$t_α=t_{0.05}=1.660$
(According to Table VI, for d.f. = 100, the closest value to 249, and area in right tail = 0.05)
So, $-t_α=-1.660$
Since $t_0\lt -t_α$, we reject the null hypothesis.
