Question

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 1

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 inches

Standard deviation = 0.5 inches

n = 400

Hypothesis are

$H_0:\mu=2.2\\\\H_1:\mu\neq 2.2$

Test statistics would be

$t=\dfrac{\bar{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}}\\\\t=\dfrac{2-2.2}{\dfrac{0.5}{\sqrt{400}}}\\\\t=-8$

Degrees of freedom = n-1 =400-1 =39

and $\alpha =0.05\\\\\dfrac{\alpha }{2}=\dfrac{0.05}{2}=0.025$

Using the t-distribution table, we get that critical value z = 1.966

Since 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, ∞)