Answer: C. Yes, because -2.77 falls in the critical region .
Step-by-step explanation:
Let
be the population mean .
As per given , we have

Since the alternative hypothesis is left-tailed and population standard deviation is not given , so we need to perform a left-tailed t-test.
Test statistic : 
Also, it is given that ,
n= 48

s= 10

Degree of freedom = df = n-1= 47
Using t-distribution , we have
Critical value =
Since, the absolute t-value (|-2.77|=2.77) is greater than the critical value.
So , we reject the null hypothesis.
i.e. -2.77 falls in the critical region.
[Critical region is the region of values that associates with the rejection of the null hypothesis at a given probability level.]
Conclusion : We have sufficient evidence to support the claim that these inspectors are slower than average.
Hence, the correct answer is C. Yes, because -2.77 falls in the critical region