A process is using the following approach to monitor its performance (assume the
ID: 3175402 • Letter: A
Question
A process is using the following approach to monitor its performance (assume the process follows normal distribution with sigma equals 2.5).
The inspector take a sample (with 5 observations) every 20 minutes.
The average of these 5 observations will be calculated and z test will be performed to determine if the process is centered at 96 psi (the desired target of the process).
If the sample average is greater than 100 psi, the process will be stopped for adjustment.
Question 1 - On a given day, the inspector collected the following five samples: 102, 103.6, 100, 96 and 95. Do you think the process is still centered at 96 psi? Why?
Question 2 - If the process center is shifted to 98 psi, what is the probability that the shift will be detected by the next two samples (i.e. the process will be stopped for adjustment)? For Question 2, assume the process follows normal distribution.
I NEED HELP WITH QUESTION TWO OF THIS PROBELM
use excel to show your work and aplha 0.05 for any hypothesis test
Explanation / Answer
Solution As specified in the question, Q2 is solved below.
Back-up Theory
If a random variable X ~ N(µ, 2), i.e., X has Normal Distribution with mean µ and variance 2, then
Z = (X - µ)/ ~ N(0, 1), i.e., Standard Normal Distribution ………………………..(1)
P(X or t) = P[{(X - µ)/ } or {(t - µ)/ }] = P[Z or {(t - µ)/ }] .………(2)
X bar ~ N(µ, 2/n),…………………………………………………………….…….(3),
where X bar is average of a sample of size n from population of X.
So, P(X bar or t) = P[Z or {(n)(t - µ)/ }] …………………………………(4)
Probability values for the Standard Normal Variable, Z, can be directly read off from
Standard Normal Tables or found using Excel Function………………………..(5)
The process will be stopped for adjustment in the next two samples of 5 each if either shows an average greater than 100. The process center is shifted to 98 psi => µ = 98. We are already given that sigma equals 2.5.
Let X = psi. Then, X ~ N(98, 2.5). So, P(Xbar > 100) = P{Z > (5)(100 - 98)/2.5}
= P(Z > 1.78888) = 1 - P(Z 1.78888) = 1 – 0.9632 = 0.0368.
Now, process will be stopped for adjustment in the next two samples, if first sample stops or first does not and second stops.
So, required probability = P(first sample stops) + P(first does not and second stops)
= 0.0368 + (1 – 0.0368)(0.0368) = 0.0722 ANSWER
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