Respuesta :
Answer:
a) Upper Control Limit = 0.10
Lower Control Limit = 0.01
b) The tech assistance process is stable and in control.
Step-by-step explanation:
Given - After a number of complaints about its tech assistance, a
computer manufacturer examined samples of calls to determine
the frequency of wrong advice given to callers. Each sample
consisted of 100 calls.
SAMPLE 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Number of errors 5 3 5 7 4 6 8 4 5 9 3 4 5 6 6 7
To find - a. Determine 95 percent limits.
b. Is the tech assistance process stable (i.e., in control)
Proof -
z = 95% confidence interval
= 1.96
⇒z = 1.96
Proportion of defects,P = total defectives/total observations
= 0.0544
⇒P = 0.0544
Now,
Q = 1-P
= 0.9456
⇒Q = 0.9456
Now,
Average sample size, N = 100
Standard deviation = [tex]\sqrt{\frac{P.Q}{N} }[/tex]
= 0.0227
Now,
Upper Control Limit = P + z(Standard deviation)
= 0.0988
⇒Upper Control Limit = 0.0988 ≈ 0.10
And
Lower Control Limit = P - z(Standard deviation )
= 0.0099
⇒Lower Control Limit = 0.0099 ≈ 0.01
∴ we get
Upper Control Limit = 0.10
Lower Control Limit = 0.01
b.)
Now,
it is clear that the fraction defective values are wit in upper an lower control limits.
So, The tech assistance process is stable and in control.
