A machine is used to fill containers with a liquid product. Fill volume can be assumed to be normally distributed. A random sample of ten containers is selected, and the net contents (oz) are as follows: 12.03, 12.01, 12.04, 12.02, 12.05, 11.98, 11.96, 12.02, 12.05, and 11.99.(a) Suppose that the manufacturer wants to be sure that the mean net contents exceeds 12 oz. What conclusions can be drawn from the data (use α= 0.01).(b) Construct a 95% two-sided confidence interval on the mean fill volume.

Answer:There is no statistical evidence at 1% level to accept that    the mean net contents exceeds 12 oz.Step-by-step explanation:Given that a random sample of ten containers is selected, and the net contents (oz) are as follows: 12.03, 12.01, 12.04, 12.02, 12.05, 11.98, 11.96, 12.02, 12.05, and 11.99.We find mean = 11.015Sample std deviation = 3.157a) [tex]H_0: \bar x= 12 oz\\H_a: \bar x >12[/tex](Right tailed test)Mean difference /std error = test statistic[tex]\frac{11.015-12}{\frac{3.157}{\sqrt{10} } } \\=-0.99[/tex]p value =0.174Since p >0.01, our alpha, fail to reject H0Conclusion:There is no statistical evidence at 1% level to accept that    the mean net contents exceeds 12 oz.