Statistical Process Control - Assignment Example

?Question A SPC or Statistical Process Control which was developed by Walter A. Shewhart in the early part of the 20th century; is a set of statistical techniques that help an organization to ensure that it’s processes are operating to their maximum potential capacity. It is used to verify that the end product or service is within the requisite norms, and helps identify outlier situations and excessive wastage. It helps in identifying units that need to be rejected. The variation that is detected by the SPC can be of different types. The Common Cause Variations are those instances that are affected by naturally occurring situations and the attributes of the products or raw materials themselves. The Special Cause variations are the ones that need to be attended to, as these are variations caused by issues or problems with the manufacturing process; like the wear and tear of appliances and changes in material quality. Slight variations in product are expected due to the natural cause variations; and SPC can be used to identify when action needs be taken to rectify the processes; and also to identify when the variations are not significant enough to require attention. It is thus also a means of saving resources that may otherwise have been spent in addressing a small issue that is not cost effective. Constant mapping of product quality helps in identifying points in time that could be cause for concern by isolating times of excessive variation. When such a situation is spotted, Pareto Diagrams – amongst other techniques can help in identifying the causes of these variations; and the feasibility of responding to them. A Pareto chart is a chart that maps both the individual categories under survey as well as the cumulative total of the categories. Each category is represented as a bar; and the categories are ranked from the one causing most variation to the one causing the least. The cumulative total is represented as a line graph; and is most often curved to form a concave line. The left vertical axis represents the frequencies and the right vertical axis, the cumulative percentage. The horizontal axis represents the categories under study. The chart thus helps us identify the categories that are most in need of being addressed. It also helps us identify which categories it would not be cost effective to focus on. There are two types of Pareto charts – the Counts chart, and the Cost chart. The Counts chart helps in understanding which of the units have the most flaws; while the Cost chart helps in identifying the category that is incurring the most cost. At times, there may be different categories implicated by the two charts for a single analysis; so the organization then has to prioritze which they would like to address first. An example would be a shirts manufacturing unit; which manufactures shirts in a variety of price ranges. In a given month; the SPC finds that reject units have significantly gone up; and further analysis is conducted. The Pareto charts show that three types of shirts are most to blame for this increase in wastage – a simple cotton shirt, a linen shirt and a high end silk shirt. It is also identified that while far more units of the cotton shirt are being rejected; the costs incurred from the rejected units of the silk shirts is significantly higher. The organization can now choose which to focus on. It is decided that immediate attention should be focused on rectifying the problem with the silk shirt production; and then the quality control team should focus on the problems with the cotton shirt production. The charts also helped in identifying that the problems with the linen shirt would not be cost effective to address in the given situation; and so the company chooses not to address those problems; but monitor the production to ensure that there are no further issues. Question 1: B i. The probability of any randomly selected phone call having a response rate of less than the target rate of 19 minutes is found to be 0.9772. This means that 97.72% of the times, the ambulance responses have been in less than the target time of 19 minutes. ii. The probability of any randomly selected phone call from the sample data having a response time of more than 19 minutes is 0.0228. This means that about 2.28% of the time, the ambulance response has taken more than the targeted time of 19 minutes. iii. On the basis of the data under study, it may be gather that the ambulance response time is 19 minutes or less for more than 95% of the time. Thus, the requirements are met more than adequately. At this point it is essential to note that these analyses may be affected by sampling errors; in that if the sample is skewed towards the lower side of the data set (sample has a positive skew) it may not be fully representative of the entire data. Before taking these sample results for granted, it is important to check if the sample chosen is truly representative of the population. Question 1: C Cp or process capability is the extent to which the process meets the specifications set down. A calculation of the Cp is possible by identifying the upper limit and the lower limit of the specification, finding the difference, and then dividing this across the breadth of six sigma’s (Standard deviations). This can be represented mathematically as The Cp is attempts to verify if 99.7% of all production fits within the specifications laid down. Thus, the Cp acts as a ratio of the specifications to the actual function of the process in a given period. If the Cp is equal to 1, it means that the production matches to specifications perfectly, if Cp is more than 1, it means that the production falls within the specifications, with space left for variations; and if the Cp is less than 1, it means that the production quality exceeds the limits set, and defects spill out on either side of the set specifications. Thus, higher the Cp, more standardized is the product quality and more is there potential for variation within the limits. This can be represented as a diagram thus: Cpk is the index used when the mean of the studied production does not coincide with the mean value between the upper and lower specifications. A problem that can arise with this type of a situation is that there will be far more defects concentrated at one end of the continuum as compared to the other; and this can cause wastage. The Cpk corrects for the lack of centering, and adjusts the understanding of the production specifications by identifying the end at which there is more wastage and adjusting from the other end so that the calculation is now centered. Cpk can be calculated using the formula Process variation is the inherent variation that exists in the product due to situational factors, quality of raw materials, and other such unforeseeable conditions. The Cp is an attempt to measure this variation and ensure that it does not spill beyond the set specifications leading to wastage. Given the example of an organization that manufactures shirts; there are certain minimum and maximum quality standards that are set in order to ensure that the product manufactured is of good quality and yet cost effective in its production. The quality of raw materials and the factor of human error does create a certain amount of variation within the units of the final product. The organization sets down a minimum and maximum quality limit so that there is a check on the quality of the final products. An analysis shows that this distance between theses limits is mathematically defined into 20 units. The Cp and the Cpk calculated are both 2; meaning that the Cp is centered between the specifications, and the quality of the end product is well within the limits set down. Thus, it was established that 99.7% of all the product manufactured fit the specified requirements; and wastage was at a minimum. Question 2: A i. The Control Chart drawn on the basis of the Mean number of weeks taken per examined case. ii. The Control Chart drawn on the basis of the maximum number of weeks taken per day’s sample. Question 2: B It was decided that charts for the mean and the maximum scores should be drawn; as the mean number of week would help ascertain the general experience of the patients per day’s sample; and the maximum number of weeks chart would help identify the number of days when at least one instance required more time that the stipulated 18 week period. The control chart drawn for the mean number of weeks it has taken each patient in the sample to complete the process shows that week 9 is a definite outlier; but other points also fall beyond the 18 week target. There is a constant fluctuation in the data gathered; and 30% of the means fall beyond the stipulated 18 weeks target. The control chart for the maximum number of week taken shows a similar erratic movement; and only in one week does the maximum score still meet the 18 week criteria. In 9 out of ten cases, there is at least one patient who has been in the process for more that 18 weeks. Given that this is a health industry; delays can prove detrimental to the wellbeing of the patients. Analysis of the charts shows that day four had the lowest number of weeks patients spent in the process of getting treatment; while day 9 recorded the maximum time taken. While all other days’ data fell into the expected range; most of the data did not correspond to the requirements set down by the NHS constitution. On all but one days, there was at least one person who had required more than 18 weeks to receive treatment; and for three different days, the mean number of weeks required were more 18. Since the trends for both charts were similar; we can conclude that the means are representative of the day’s data; and specific cases have not skewed the data. It is necessary to repeat the analysis with more data to ascertain whether the problem is truly as big as it seems. A 10 day period if data collection is recommended; with data on 15 -20 cases per day to be included at the least. This will help ensure that the results of this analysis are not exaggerated by faulty sampling. It is also important to identify the factors that are causing the delays identified. Overcrowding; lack of equipment, and lack of space may be causes; and they need to be examined and addressed. Hiring staff or machinery; devising alternative means of assessment and administering treatment need to be evaluated and set into practice. Once this is done, data needs to be collected for another 10 day period; preferably for more than 10 – 15 cases per day; so that the data is more representative and helps identify if the anomalies that still exist. Question 3: A Given the data at hand, it is possible to tabulate the data thus: Postage Charged (1) Not charged (0) Pictures Displayed (1) N = 50 Mean = 3.4 S.D. = 4.9 N = 50 Mean = 4.1 S.D. = 3.1 Not displayed (0) N = 50 Mean = 2.1 S.D. = 0.7 N = 50 Mean = 3.3 S.D. = 0.9 The main effect calculated for the variable of postage was found to be 5.17 which was significant beyond the 0.03 level of significance [ F (1,196) = 5.17; p < 0.03, two tailed]. This shows that there was a significant difference in amounts for which books were bought between the conditions where people were expected to pay postage and the one where people were not. The means show us that more expenditure was made when people were not expected to pay for postage. The main effect for pictures was found to be 6.31, and was significant beyond the 0.02 level of significance [F (1,196.0) = 6.31; p Read More