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.

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

Test statistics would be

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

and

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, ∞)

rounding to the nearest thousandth would be 452,000

use this formula:

p(a\b) = p(a and b)/p(b).

take it from here.