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 =
= 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.
To solve this problem you must apply the proccedure shown below:
1- You must apply the following formula:
Where is the residual,
is the observed value
and
is the predicted value
.
2- You only need to substitute the into the equation
and then, you must apply the formula for calculate the residual:
The answer is: The residual is
Formula of interest =PRT divided by 100
Answer:
They are parallel because their slopes are equal
Step-by-step explanation:
