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
![H_0:\mu=50\\\\ H_a: \mu](https://tex.z-dn.net/?f=H_0%3A%5Cmu%3D50%5C%5C%5C%5C%20H_a%3A%20%5Cmu%3C50)
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 : ![t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7B%5Coverline%7Bx%7D-%5Cmu%7D%7B%5Cdfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%7D)
Also, it is given that ,
n= 48
![\overline{x}=46](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%3D46)
s= 10
![t=\dfrac{46-50}{\dfrac{10}{\sqrt{48}}}=\dfrac{-4}{\dfrac{10}{6.93}}\\\\=\dfrac{-4}{1.443}\approx-2.77](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7B46-50%7D%7B%5Cdfrac%7B10%7D%7B%5Csqrt%7B48%7D%7D%7D%3D%5Cdfrac%7B-4%7D%7B%5Cdfrac%7B10%7D%7B6.93%7D%7D%5C%5C%5C%5C%3D%5Cdfrac%7B-4%7D%7B1.443%7D%5Capprox-2.77)
Degree of freedom = df = n-1= 47
Using t-distribution , we have
Critical value =![t_{\alpha,df}=t_{0.025,47}=2.0117](https://tex.z-dn.net/?f=t_%7B%5Calpha%2Cdf%7D%3Dt_%7B0.025%2C47%7D%3D2.0117)
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