I have transcribed your data for use in R. (Please proofread.)
town = c(520, 520, 525, 487.5, 480, 475, 480)
vill = c(555, 547.5, 530, 550, 555, 600, 530,
610, 600, 580, 600, 580)
boxplot(town, vill, names=c("Town", "Village"),
horizontal=T, col="skyblue2")
stripchart(town, at=1, add=T, meth="stack", pch=19, col="red")
stripchart(vill, at=2, add=T, meth="stack", pch=19, col="red")

According to the Welch two-sample t test (not assuming equal variances)
there a highly significant difference between Town and Village emissions:
P-value nearly 0.
t.test(town, vill)
Welch Two Sample t-test
data: town and vill
t = -6.057, df = 15.367, p-value = 1.979e-05
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-96.71324 -46.44152
sample estimates:
mean of x mean of y
498.2143 569.7917
This should be no surprise. The boxplot (with individual data values as red dots) shows that there is a complete separation in the scores; the largest Town value is below the smallest Village value.
If we use all 19 values (Town and Village taken together, sample mean 543.42) to test
$H_0: \mu \le 550$ against $H_a: \mu > 550,$ then a t test (obviously) will not
reject $H_0.$
all = c(town, vill)
t.test(all, mu=550, alte="gr")
One Sample t-test
data: all
t = -0.65323, df = 18, p-value = 0.7391
alternative hypothesis: true mean is greater than 550
95 percent confidence interval:
525.9566 Inf
sample estimates:
mean of x
543.4211
However, if we test the same hypothesis, using only the
(higher-emission) Village data, then $H_0$ is rejected at the 5% level, but not at the 1% level---with P-value 0.018.
t.test(vill, mu=550, alte="gr")
One Sample t-test
data: vill
t = 2.3895, df = 11, p-value = 0.01795
alternative hypothesis: true mean is greater than 550
95 percent confidence interval:
554.9168 Inf
sample estimates:
mean of x
569.7917
If the rules of the 'emission violation' game are to use all 19 observations without further investigation, then we can't claim that
the cars are 'dirty.' However, the manufacturer would be well advised
to expect that violations will likely be found in more-localized studies.