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 Click here for the Excel Data File a. Determine 95 percent limits. (Do not round your intermediate calculations. Round your final answers to 2 decimal places. Round up any negative control limit value to zero.) b. Is the tech assistance process stable (i. e., in control)

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.

the perimeter is increased by a factor of 2

it costs $20 to interview each person for a survey. find the cost to obtain a +- 3% margin of error.

step-by-step explanation:

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