After a number of complaints about its directory assistance, a telephone company

Question

After a number of complaints about its directory assistance, a telephone company examined samples of calls to determine the frequency of wrong numbers given to callers. Each sample consisted of 100 calls.

a. Determine 95 percent limits for the fraction of errors chart. (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 process stable (i.e., in control)?

Yes

No

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

Explanation / Answer

p = total number of errors / total number of observations samples

= (5+3+5+7+4+6+8+4+5+9+3+4+5+6+6+7) / (16 * 100) = 87 / 1600 = 0.0544

At 95% limits, the z value is 1.96. Hence, UCL = p + z *P and LCL = p – z *P

Where P = square root of (p*(1-p)/n)

Thus, P = square root of (0.0544 * (1- 0.0544) / 100)

             = square root of ((0.0544 * 0.9456)/100) = square root of (0.0005144064) = 0.0227

Hence, UCL = p + z *P = 0.0544 + (1.96 * 0.0227) = 0.098892 or (0.0989) = 0.1

LCL = p - z *P = 0.0544 - (1.96 * 0.0227) = 0.009908 or (0.0099) = 0.01

Since all the below fractions (number of errors per 100 calls) are within the control limits by having UCL = 0.1 and LCL = 0.01, the process is stable.

Sample 1 = 5/100 = 0.05

Sample 2 = 3/ 100 = 0.03

Sample 3 = 5/100 = 0.05

Sample 4 = 7/100 = 0.07

Sample 5 = 4 /100 = 0.04

Sample 6 = 6/100 = 0.06

Sample 7 = 8/100 = 0.08

Sample 8 = 4 / 100 = 0.04

Sample 9 = 5/100 = 0.05

Sample 10 = 9/100 = 0.09

Sample 11 = 3/100 = 0.03

Sample 12 = 4/100 = 0.04

Sample 13 = 5/100 = 0.05

Sample 14 = 6 /100 = 0.06

Sample 15 = 6/ 100 =0.06

Sample 16 = 7/100 = 0.07

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