Respuesta :
- (A) If the acceptance region is defined as 98.5 ≤ x¯ ≤ 101.5, find the type I error probability α is 0.09296.
- (B) For the case in which the true mean heat evolved is 103, β is 0.04648.
What is heat?
- The transfer of kinetic energy from one medium or object to another, or from an energy source to a medium or object, is referred to as heat.
- Energy can be transferred in three ways: radiation, conduction, and convection.
- When kings weren't hiding in oak trees or being beheaded in the 1600s, they enjoyed a game of horse racing.
- And when they took their horse for a gallop to prepare it for a race, they called it heat because they were obviously warming the horse up.
To find the answers to the given situation:
Use n = 5, everything else held constant:
(A) To find the type I error probability α:
- P( X ≤ 98.5) + P( X >101.5)
- P(X-100/(2/√5) ≤ 98.5-100/(2/√5)) + P(X-100/(2/√5)>101.5-100/(2/√5))
- P(Z ≤ −1.68) + P(Z > 1.68)
- P(Z ≤ −1.68) + (1 − P(Z ≤ 1.68))
- = 0.04648 + (1 − 0.95352) = 0.09296
So, if the acceptance region is defined as 98.5 ≤ x¯ ≤ 101.5, find the type I error probability α has is 0.09296.
(B) To find β:
- β = P(98.5 ≤ X ≤ 101.5 when µ = 103)
- P(98.5-103/(2/√5) ≤ X-103/(2/√5) ≤ 101.5-103/(2/√5))
- P(−5.03 ≤ Z ≤ −1.68) = P(Z ≤ −1.68) − P(Z ≤ −5.03)
- = 0.04648 − 0 = 0.04648
So, for the case in which the true mean heat evolved is 103, β is 0.04648.
Therefore, answers to both the questions are shown.
Know more about the heat here:
https://brainly.com/question/19666326
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The correct question is given below:
The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 100, and the standard deviation is 2. You wish to test H0 : µ = 100 versus H1 : µ 6= 100 with a sample size of n = 5 specimens.
(a) If the acceptance region is defined as 98.5 ≤ x¯ ≤ 101.5, find the type I error probability α.
(b) Find β for the case in which the true mean heat evolved is 103.
