Molybdenum rods are produced by a production line setup. It is desirable to check whether the process is in control, i.e. equal to 2.2 inches. Let X = length of such a rod. Assume X is approximately normally distributed where the mean and variance are unknown. Take n = 400 sample rods, with sample average length 2 inches with a standard deviation of 0.5 inches. Using α=0.05, find the rejection region and test statistic of the necessary test to be held.

Answer: Test statistic = -8 and rejected region is  (-∞,-1.966) and (1.966, ∞)Step-by-step explanation:Since we have given that X is the length of a rod.Sample Mean = 2 inchesStandard deviation = 0.5 inchesn = 400Hypothesis are [tex]H_0:\mu=2.2\\\\H_1:\mu\neq 2.2[/tex]Test statistics would be [tex]t=\dfrac{\bar{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}}\\\\t=\dfrac{2-2.2}{\dfrac{0.5}{\sqrt{400}}}\\\\t=-8[/tex]Degrees of freedom = n-1 =400-1 =39and [tex]\alpha =0.05\\\\\dfrac{\alpha }{2}=\dfrac{0.05}{2}=0.025[/tex]Using the t-distribution table, we get that critical value z = 1.966Since the two tail test will be applied to this, So, the acceptance region will be (-1.966, 1.966)Hence, the rejected region will be (-∞,-1.966) and (1.966, ∞)