Answer: C. Yes, because -2.77 falls in the critical region .
Step-by-step explanation:
Let [tex]\mu[/tex] be the population mean .
As per given , we have
[tex]H_0:\mu=50\\\\ H_a: \mu<50[/tex]
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 : [tex]t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}[/tex]
Also, it is given that ,
n= 48
[tex]\overline{x}=46[/tex]
s= 10
[tex]t=\dfrac{46-50}{\dfrac{10}{\sqrt{48}}}=\dfrac{-4}{\dfrac{10}{6.93}}\\\\=\dfrac{-4}{1.443}\approx-2.77[/tex]
Degree of freedom = df = n-1= 47
Using t-distribution , we have
Critical value =[tex]t_{\alpha,df}=t_{0.025,47}=2.0117[/tex]
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
