The maintenance costs of fixing all 3 HP motors is normally distributed with a s
ID: 3180348 • Letter: T
Question
The maintenance costs of fixing all 3 HP motors is normally distributed with a standard deviation of $3. A sample of 25 of these 3 HP motors were chosen from the maintenance center located in Cleveland, and their mean maintenance cost was found to be $20. Alpha = 0.05 A. State the Null and Alternative Hypothesis when we are testing the claim that the mean of all repairs of fixing all 3 HP motors is different than $21 per hour. B. If the actual mean of the population of hourly maintenance costs of fixing all 3 HP motors is equal to $ 23, calculate the Type II error probability beta for the information given in part A.Explanation / Answer
a)
µ = mean of all repairs of fixing all 3 HP motors
H0 : µ = 21 (Null Hypothesis)
H1 : µ 21 (Alternative Hypothesis)
This is a two tailed test as the claim is that the mean is not equal to 21.
b)
2 tail test with = 0.05
probability of a Type II Error: = 1 – power
n = 25
= 3
SE = /sqrt(n) = 3/sqrt(25) = 3/5 = 0.6
= 21
We next compute the lower and upper bounds of sample means for which the null hypothesis = 21 would not be rejected.
Zcrit = 1.960
Lower bound : (x – )/ SE = -1.96
(x – 21)/0.6 = -1.96
Xlower = 19.824
Upper Bound: (x – )/ SE = 1.96
(x – 21)/0.6 = 1.96
Xupper = 22.176
Therefore, so long as the sample mean is between 19.824 and 22.176 in a hypothesis test, the null hypothesis will not be rejected. Since we assume that the actual population mean is 20, we can compute the lower tail probabilities of both end points.
P(X < 19.824) = P(Z < (19.824 – 20)/0.6) = P(Z < -0.2933) = 0.384548
P(X < 22.176) = P(Z < (22.176 – 20)/0.6) = P(Z < 3.62667) = 0.999856
The difference is the probability of type 2 error = 0.999856 – 0.384548 = 0.615209 = 61.52%
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