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 to test hypothesis :
[tex]H_0:\mu=50\\\\ H_a:\mu<50[/tex]
Since the alternative hypothesis is left-tailed , so test is a left-tailed test.
Also, population standard deviation is unknown , so we will perform a left tailed t-test.
Sample size : n= 48
Sample mean : [tex]\overline{x}=46[/tex]
sample standard deviation : s= 10
Test statistics : [tex]t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}[/tex]
[tex]t=\dfrac{46-50}{\dfrac{10}{\sqrt{48}}}[/tex]
[tex]t=\dfrac{-4}{\dfrac{10}{6.92820323028}}\approx-2.77[/tex]
Degree of freedom : df = n-1 = 47
For significance level 0.025 and degree of freedom 47 , we have
Critical t-value for left-tailed test =[tex]t^*=t_{0.01, 65}=-2.012[/tex] [Using student's t-distribution table]
Rejection region : we reject null hypothesis for t<-2.012
Decision : Since -2.77(calculated t- value)< -2.012(Critical value) , it means it falls under rejection region.
i.e. We reject the null hypothesis.
We have sufficient evidence to support the alternative hypothesis μ < 5.7 ounces .
i.e. The company have sufficient evidence at α = 0.025 to believe that these inspectors are slower than average.
Hence , the correct answer = c. Yes, because –2.77 falls in the critical region
