As per given , we have to test hypothesis :
[tex]H_0:\mu=5.7\\\\ H_a:\mu<0.57[/tex] , where [tex]\mu[/tex] = Population mean.
Since the alternative hypothesis is left--tailed , so test is a left-tailed test.
Sample size : n=66
Sample mean : [tex]\overline{x}=5.43[/tex]
sample standard deviation : s= 0.25
Also, population standard deviation is not given , so we will perform a left tailed t-test.
Test statistics : [tex]t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}[/tex]
[tex]t=\dfrac{5.43-5.7}{\dfrac{0.25}{\sqrt{66}}}[/tex]
[tex]t=\dfrac{-0.27}{\dfrac{0.25}{8.12403840464}}\approx-8.77[/tex]
For significance level 0.05 and degree of freedom 65 (df=n-1), we have
Critical t-value =[tex]t^*=t_{0.01, 65}=-1.669[/tex] [Using student's t-distribution table]
Decision : Since -8.77(calculated t- value)< -1.669(Critical value) , it means it falls under rejection region.
i.e. We reject the null hypothesis.
Conclusion : We have sufficient evidence to support the alternative hypothesis μ < 5.7 ounces .
