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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.

Respuesta :

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

[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 =39

and [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 = 1.966

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, ∞)

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