Use the data on the nonconformances for a period of 10 days to complete (a) through (c).Day Sample Size Nonconformances Day Sample Size Nonconformances Full data set1623451501501501501508465214LCL =152098910150150150150150The proportion of nonconformances is the smallest on dayb. What are the LCL and UCL?-0(Round to five decimal places as needed.)15i W N 8 G1812a. On what day is the proportion of nonconformances largest? smallest?The proportion of nonconformances is the largest on day1315 Use the data on the nonconformances for a period of 10 days to complete (a) through (c).Day Sample Size Nonconformances Day Sample Size Nonconformances Full data set123451501501501501508465201415678910UCL =(Round to five decimal places as needed.)c. Are there any special causes of variation?15015015015015015 18 12 13 15O A. The nonconformance varies from day to day. Therefore, there does not appear to be any special causes of variation.B. The nonconformance varies from day to day. Therefore, there does appear to be at least one special cause of variation.OC. The proportions are not within control limits. Therefore, there does appear to be at least one special cause of variation.D. The proportions are within control limits. Therefore, there does not appear to be any special causes of variation.

Day Sample Size Non-conformances 1 150 8 2 150 14 3 150 6 4 150 15 5 150 20 6 150 15 7 150 18 8 150 12 9 150 13 10 150 15a. The proportion of Non-conformances is obtained below: p1=Number of Non-conformancesSample Size=8150≈0.05333 The table below shows the all proportions: Day Sample Size Non-conformances Proportion 1 150 8 0.05333 2 150 14 0.09333 3 150 6 0.04000 4 150 15 0.10000 5 150 20 0.13333 6 150 15 0.10000 7 150 18 0.12000 8 150 12 0.08000 9 150 13 0.08667 10 150 15 0.10000 The proportion 0.13333 is the largest proportion and that corresponds to the day 5. The proportion 0.04000 is the smallest proportion and that corresponds to the day 3. The proportion of non-conformances is the largest on day 5. The proportion of non-conformances is the smallest on day 3.b. A p-chart is used to obtain the control limits for the proportion of non-conformances. The table below shows the necessary calculations: Day Sample Size Non-conformances Proportion 1 150 8 0.05333 2 150 14 0.09333 3 150 6 0.04000 4 150 15 0.10000 5 150 20 0.13333 6 150 15 0.10000 7 150 18 0.12000 8 150 12 0.08000 9 150 13 0.08667 10 150 15 0.10000 SUM 136 0.90667 Let, pi denotes the proportion of non-conformances in the ith sub-group (Day). The pooled fraction of non-conformances is obtained below: p¯=∑piNumber of sub-groups=0.9066710=0.090667 Therefore, CL=p¯=0.090667. The values of LCL and UCL are obtained below: LCL=p¯-3p¯×1-p¯n=0.090667-3×0.090667×1-0.090667150≈0.09067-0.07033≈0.02033 UCL=p¯+3p¯×1-p¯n=0.090667+3×0.090667×1-0.090667150≈0.090667+0.07033≈0.16100c. The control limits are used to check whether the process is under control or not. If the the observations are within the control limits, the the process is in under control and there will be no special causes of variation. The LCL and UCL are 0.02033 and 0.16100 respectively. Since all the proportion of non-conformances i.e. 0.05333, 0.09333, 0.04000, 0.10000, 0.13333, 0.10000, 0.12000, 0.08000, 0.08667, 0.10000 are lies between the LCL and UCL, there does not appear to be any special causes of variation. From the above explanation it can be said that option D is the correct choice and the other options are incorrect.