Definition and Methods of Quantitative Investigation - Case Study Example

QUANTITATIVE RESEARCH METHODS Our advice: choose a more quantitative method when most of the following conditions apply: The research is confirmatory rather than exploratory i.e. this is a frequently researched topic, and (numerical) data from earlier research is available.. You are trying to measure a trend (almost impossible with qualitative research). There is no ambiguity about the concepts being measured, and only one way to measure each concept. The concept is being measured on a ratio or ordinal scale. Quantitative Research "There's no such thing as qualitative data. Everything is either 1 or 0" - Fred Kerlinger The aim of qualitative analysis is a complete, detailed description. In quantitative research we classify features, count them, and construct statistical models in an attempt to explain what is observed. Recommended during latter phases of research projects. Researcher knows clearly in advance what he/she is looking for. All aspects of the study are carefully designed before data is collected. Researcher uses tools, such as questionnaires or equipment to collect numerical data. Data is in the form of words, pictures or objects. Data is in the form of numbers and statistics. Quantitative data is more efficient, able to test hypotheses, but may miss contextual detail. Researcher tends to remain objectively separated from the subject matter. Fast growing economy, volatile market conditions and the stiff open market competition have rendered research and development mandatory. Organizations, in order to retain their market share, need to keep themselves abreast about, not only the, product and services provided by them but also by their competitors. They also need to keep a close watch on the scientific and technological advances being made and how these can be incorporated within their structures. In order to accomplish the abovementioned tasks, a series of detailed investigations have to be carried out by these organizations. These investigations are scientific in nature and study the various quantitative variables available. Hence, establishing relationships among these variables based on their properties. This scientific investigation and analysis of the quantitative data is known as the quantitative research. Through quantitative research, numerical data is collected periodically, classified into statistical models, interpreted and analyzed. This research is a continuous process and it assists the organizations to solve the real world problems. Quantitative research using statistical methods typically begins with the collection of data based on a theory or hypothesis, followed by the application of descriptive or inferential statistical methods. Causal relationships are studied by manipulating factors thought to influence the phenomena of interest while controlling other variables relevant to the experimental outcomes. (1) The branch of statistics forms the backbone of this research technique. Statistical analysis helps decode complex problems and aids in decision-making. The measures of central tendency, the measures of variability, the coefficient of correlation, the measures of dispersion etc are the statistical tools, which aid the quantitative research. To further explain and exhibit the role of statistics in the real world problem solving, we take the help of the following case studies. CASE I: Andrews Bacon & Curtis ltd (ABC ltd) Andrews Bacon & Curtis ltd is company based in Oostende, Belgium, which deals in perishable food items. They supply these goods to their customers through air cargo from the Oostende airport and also through refrigerated transport, to the plants located within 100 kms area. XYZ Supermarkets located at 1650 kms from their nearest plant, are their main clients and purchase weekly in lots of 250 units each. Their orders range from 1000 units to 2000 units per week and observing the relationship between the two, the trend is expected to continue. An airfreight agent has approached ABC ltd with an offer to serve them with reduced airfares, if the company promises a minimum guaranteed stock, which would be transported through the air cargo on a fixed flight, every week. For the contract period, ABC will have to pay the freight on the guaranteed stock, whether or not it is shipped. Studying the pattern of dealings with XYZ supermarkets, ABC ltd deduce that a weekly shipment of 1000 units can certainly be guaranteed, but they would like to consider 1250 units per week as their target. It has been observed that on an average ABC deliver 6 lots per week, with 4 lots as the minimum and 8 lots as the maximum. Going through projected figures, we ascertain that, if a deal be made with the airfreight company, the normal freight of 1200 euros/ lot would be reduced down to 900 euros/ lot for the guaranteed lot. Hence there will be a saving of 300 euros /lot. But if the number of lots to be transported exceeds the minimum guaranteed lot, then same rate i.e. 1200 euros/ lot would apply to each exceeding lot. ABC earns a gross profit of 4500 euros/ lot, which means that the net profit on guaranteed lots will be: 4500 – 900 = 3600 euros. Also, profit on lots send, exceeding the minimum guarantee limit, would be: 4500 – 1200 = 3300 euros. On the other hand, loss suffered, on the lots, which were guaranteed but not send, would be 900 euros. Here presented is the 180 week record of the transactions between the two companies: Weekly Delivery (250 lot)(x) Occurrences F Percentage of Total 4 14 7.78 5 36 20.00 6 80 44.44 7 34 18.89 8 16 8.89 30 180 100 Average lots sent per week = A.M = fx/f =180/30 = 6 As an average the company can send six lots a week. There is a chance of 9 times to send the lots less than six. The probable loss in that period if the company guarantees 6 lots per week. 2x900x14 = 25,200 euros 1x900x36 = 32,400 euros. The total probable loss = 25,200 + 32,400 = 57600 The probable profit by giving guarantee of a minimum of 6 lots 6x300x80 = 144000 6x300x34 =61200 6x300x16 =28800 Total probable profit by giving minimum shipping guarantee is 234000 euros. Probable loss from the shipments more than 6 a week. 1x300x34 = 10200 2x300x16 = 9600 Net probable profit 214200 euros. The minimum probable profit by giving minimum guarantee of 6 lots a week = Total probable profit – total probable loss = 214200 – 57600 =156600 Euros. So the company can give a guaranteed shipment of 6 lots a week for the freight carrier in order to increase net profit by 156600 euros. CASE II: Harry Loft house Company (HLC) Harry Loft house, the owner with two operators, works in the small engineering shop comprising of a number of lathes, metal working machines and small presses. For the next 6 months, HLC has contracted to do a series of work for a local manufacturer and will be requiring two of their machines, the drill press and the reamer. The drilling operations are to be carried on first, followed by the tasks on the reamer. The two operators are paid hourly, for the work done by them and for the waiting time as well. The tasks to be done are numbered by Harry and then carried out in the same order. The operations on the drill machine first and then the reamer. Due to this arrangement, the operator on the drill machine always has work to do, while the other has to wait till the first job is done and passed on to him. To reduce the waiting time, and hence the time lost, we study the operation time for each job including 1 minute which is used for setting up each operation. Job 1 2 3 4 5 6 7 8 9 10 Drill Press Time 50 45 60 15 65 45 10 60 50 50 Reamer Time 55 55 30 15 35 45 20 45 75 80 To elaborate, the schedule of jobs is presented, showing actual working schedules: Drill Press Reamer Job Time Start End Time Start End Lost Time 1 .50 1 50 55 51 105 50 2 .45 51 95 55 106 160 0 3 .60 96 155 30 161 190 0 4 .15 156 170 15 191 205 0 5 .65 171 235 35 236 270 30 6 .45 236 280 45 281 325 10 7 .10 281 290 20 326 345 0 8 .60 291 350 45 351 395 5 9 .50 351 400 75 401 475 5 10 50 401 450 80 476 555 0 450 455 As seen above nearly 2 hours have been lost Drill Press Reamer Job Time Start End Time Start End Lost time 1 10 1 10 20 21 40 10 2 15 11 25 15 41 55 0 3 45 26 70 55 71 125 15 4 45 71 115 45 126 170 0 5 50 116 165 55 171 225 0 6 50 166 215 75 226 300 10 7 65 216 280 35 301 335 20 8 60 281 340 45 341 385 6 9 60 341 400 30 401 430 15 10 50 401 450 80 456 535 26 Instead of 120 minutes the loss was decreased to 96 minutes saving 24 minutes. We study the schedule of the previous day, to compare both setup: Drill Press Reamer Job Time Start End Time Start End Time Lost 1 40 1 40 60 41 100 40 2 20 41 60 10 101 110 0 3 80 61 140 60 141 200 30 4 35 141 175 40 201 240 0 5 60 176 235 50 241 290 0 6 65 236 300 70 301 370 10 7 30 301 330 40 371 410 0 8 15 331 345 25 411 435 0 9 60 346 405 50 436 485 0 10 55 406 460 60 486 545 0 460 465 80 The schedule of the day can be modified as follows by altering the order of the operations. The time was being wasted when the time taken by next drill press operation to the reamer operation is more than the time taken by reamer. For example see the time lost between 4 and 5 operations in the first table. Drill Press Reamer Job Time Start End Time Start End Lost Time 1 50 1 50 55 51 105 50 2 45 51 95 55 106 160 0 3 60 96 155 30 161 190 0 4 15 156 170 15 191 205 0 5 10 171 180 35 181 215 0 6 45 181 225 45 226 270 10 7 50 226 275 75 276 350 5 8 60 276 335 80 351 395 0 9 50 336 385 20 396 415 0 10 65 386 450 45 416 460 0 450 455 Here the time loss is reduced to 65 minutes. Care should be taken that the time taken by reaming operation should be more than the next drilling operation the loss of working hours by making the reamer to wait can be reduced. If that is not possible the difference in the time period can be minimized as much as possible. In the I table the loss was decreased by 24 minutes and in the second table the loss was decreased by 15 minutes. Total 40 minutes loss was recovered by adjusting the order of the operations thus decreasing the waiting time of the reamer. CASE III: Mycock, Halls& Paul ltd (MPH ltd Mycock, Halls & Paul ltd is an international accounting and auditing organization. They audit the accounts of various public and private sector companies. They have a widespread clientele. For this purpose, MHP have set up a ‘flying squad’, a group of their employees, who visit these clients regularly. The visits are random, in the sense, that no employee has been assigned any particular area. MHP have five main clients, namely, A, B, C, D and E. Here we compare the distance in km, between clients and the employees, in order to calculate the minimum distance between them, so that an arrangement can be allocated: BC1 BC2 BC3 BC4 BC5 A 560 280 120 280 1080 B 800 280 480 240 1200 C 400 120 160 200 840 D 320 480 280 480 840 E 520 1000 960 1040 320 BC1 is nearer to D when compared to other cities BC2 is nearer to C when compared to other cities BC3 is nearer to A when compared to other cities BC4 is nearer to C when compared to other cities BC5 is nearer to E when compared to other cities The system of travel allocation for 5 bean counters must be in such a manner to reduce the travel time. It is not possible to allocate nearest client to each bean counter as each bean counter have to visit all the five clients. In this case each bean counter can be allocated the nearest client more number of times and the remaining clients in the order of the distance. So in the case of BC1 D can be allocated 6 times a year and the remaining clients in the order of C E A B C E. this makes BC1 to travel a minimum distance. This results in 320x6 =1920 km in 6 months and in the remaining months the distance will be 400+520+560+800+400+520=3180 Total distance that need to be traveled = 5100 Km. This distance will be minimum in the case of BC1 as 6 months of the year were allotted the least distance destination. Similarly in case of BC2, C can be allocated 6 times an year and remaining clients in the order of A B D E A B. The distance to be traveled by BC2 in 6 months = 720 Km The distance to be traveled by BC2 in next 6 months = 2600 Km BC3 can be allocated A 6 times and others in the order of C D B E C D. The distance to be traveled by BC3 in 6 months 720 Km The distance to be traveled by BC3 in another 6 months in the year = 2320Km BC4 can be allocated C 6 times in year and other in the order of B A D E B A. Distance to be traveled by BC4 in 6 months of the year = 1200 Km Distance to be traveled by BC4 in another 6 months of the year =2560Km BC5 can be allocated E six times in year and remaining in the order of D C A B D C. Distance to be traveled by BC5 in 6 times a year = 1920 Km Distance to be traveled by BC5 in another 6 times a year = 2760 + 2880 = 5640 Km The distances can still be minimized by allotting certain bean counters to their nearest destination, but that is not the policy of the organization. REFERENCES The references were given in the following format. Author, year, title, publisher, edition, type of media, date retrieved, website address 1. Media wiki, 2006, Quantitative research, Wikipedia, , electronic, 23-12-06, http://en.wikipedia.org/wiki/Quantitative_research, 2. Carter Mac Namara, 2006, Basic Advice about Planning Your Research, Free management library, , electronic, 23-12-06, http://www.managementhelp.org/research/planning.htm Read More