Business Math Homework for session 1 Exam. questions Question 1 1. Quota sampling. A technique which is popular in market research. Such a method uses a team of interviewers, each with an appropriate number of subjects to interview.
It is quite wide-spread method, especially among social marketing. For example, some clothes producers make such surveys using street interviewers to find out preferences of their customers. 2. Cluster sampling. A non-random method of sampling, used in cases when there is no sampling frame in evidence. Main idea is to select one or several areas, where you will inquire all necessary information.For example, such a practice is used by companies which are producing different kinds of goods for different areas.
3. Stratification. A method which splits a group of people into different parts of groups, depending on which kind of investigation do you have. It is a basis of quota sampling and widely spread nearly at all surveys. Researchers of stratify interviewed groups by gender, age, nationality and race. 4. Simple random sampling.
A sampling method where nearly every person has the same chance to be chosen for the investigation. Used in cases when it is necessary to find out basic attitude to smth among huge group of people.For example, “Do you like Coca-Cola? ” to check the loyalty to the brand. Question 2 a) A postal questionnaire is far more cheaper method of collecting data than the personal interview. This method could collect much more samples for a short time.
At the same time it has low response rate, but even if you would get answers from just 10% of people, it would cost you less time and money than with personal interview method. The personal interview method is good when researcher has some complicated questions to ask and it is possible to get complex answer only after several connected questions during the interview. ) Simple random sampling is quite good because it is simple. Every person has same chances to be chosen for the focus group. However it cant guarantee that chosen group would represent all necessary public straits.
At the same time, quota sampling cant be randomized and it is possible that final data would be less representative. However, if the data was classified and gathered well, the information could have very high quality. For example, inquiry questioning of film-goers after the new production. Question 3 a) (i) (ii) Other goods: cars, yachts, tv setsOther services: health services, tourism, intertainment b) Question 4 (a) i. Absolute error – difference between an estimated value and its genuine value. Only a maximum error could be calculated.
For example: 21. 571 is the True value 20. 000 is the Recorded Value. Thus: (True value) – (Recorded Value) = Absolute error (21.
571) – (20. 000) = 1. 571 ii. Relative error – absolute error, expressed as a % of the given estimated value. For example, 1,571/21,571 x 100%= 7,28% iii. Compensating error – happens when “fair” rounding has been accomplished.When numbers, subject to compensating errors, are added, the total relative error should be approximately 0. iv.
Biased errors – happens when rounding is carried out in one direction. For examples, when we are rounding down some data to the closest high/low number. | Minimum| Estimate| Maximum| Time/Wage rate| 145hours$4/hr| 150 hours$4/ hr| 155 hours$4. 4/hr| Labour costMaterial cost| $580$2550| $600$2600| $682$2650| Total cost| $3130| $3200| $3332| Quote| $4000| $4000| $4000| Profit| $668| $800| $870| Question 5 a) Largest value – 469, smallest – 347, therefore, range – 122 5 classes – 122/5=24. 345-369 – number of weeks 16 370-394 – number of weeks 8 395-419 – 4 weeks 420-444 – 1 week 445-469 – 11 weeks Total – 40 b) Cumulative frequency needs to be plotted against class upper bounds Question 6 a) A component time is best represented by a line diagram. b) Component bar charts enable comparisons between components across the years to be made easily, showing also yearly totals. The main disadvantages is the fact that actual values cannot be determined c) Overall, there has been a steady increase in the number of policies issued each year.Household policies have shown a steady increase over the five year period at the expence of Motor, which have steadily decreased.
Life has shown a very small increase over the period except for a small dip in 1981. Other policies have remained fairly steady, fluctuating only slightly around 6 000. Question 7 Question 8 (i) (ii) Overall, the total value of the given assets has increased steadily from just under 1m in 1978 to 1,3m in 1982.
The most significant increase had debtors component, which has risen up the stock and work in progress component.The property component shows small increases, while plant and machinery shows small increases in the first four years and a decrease in the fifth year. The cash component has increased for five years period and now comparable with property. Question 9 a) Pictogram. A representation that is easy to understand for a non-sophisticated audience. However, it cannot represent data accurately or be used for any further statistical work. Simple bar chart.
The most common form of representing data which can be used for time series or qualitative frequency distributions.It is easy to understand and can represent data accurately. Pie chart. Chart, which can have a lot of impact. Used in cases when the classes need to be compared in relative terms.
Involve technical calculations. Simple line diagram. The simplest and most popular form of representing time series. They are easy to understand and represent data accurately. b) A pie chart is one of the charts that could be drawn for the given data and is shown below: Question 10 The company can make and sell 10 000 +/- 2000 units in the year The selling price will lie in the range 50 +/- 5 per unitThus the maximum revenue is 12000 x 55 = 660 000 The minimum revenue is 8 000 x 45 = 360 000 Estimated revenue – 10 000 x 50 = 500 000 The maximum error from the estimated revenue is 660 000 – 500 000 = 160 000 And relative error = 150 000 / 500 000 x 100% = 32% The ranges of costs are: Costs| Min| Max| materials| 147 000| 153 000| wages| 95 000| 105 000| marketing| 45 000| 55 000| miscellaneous| 45 000| 55 000| Total| 332 000| 368 000| Maximum error from the estimated costs – 18 000 And relative error = 80 000 / 350 000 x 100% = 5,1% Max contribution = 660 000 – 332 000 = 328 000Minimum contribution = 360 000 – 368 000 = -8000 The estimated contribution = 150 000 The maximum error from the estimated contribution is 328 000 – 150 000 = 178 000, relative error 178 000 / 150 000 x 100% = 118,7% Max contribution/ unit of 328 000 arises when 12 000 units are made and sold Contribution/Unit = 328 000 / 12 000 = 27.
33/unit Min contribution of – 8000 arises when 8 000 units are made and sold, therefore contr. /unit is -1 The estimated contribution/unit = 15, max error is 15-(-1)= 16; relative error = relative error as 16/15×100%=106,7%
