Mb0040 statistics for management spring2015_assignment- SMU_MBA-Solved-Assignment
- 1. ASSIGNMENT, Drive SPRING 2015 Program MBA, Semester 1 Subject code & Name- MB0040-Statistics for Management Book ID B1731, Credit & Marks 4 CREDIT, 60 MARKS Answerall questions. Q. 1. Distinguish between Classification and Tabulation. Explain the structure and components of a Table with an example. Q. 2. (a) Explain Arithmetic mean. (b) The mean wage is Rs. 75 per day, SD wage is Rs. 5 per day for a group of 1000 workers and the same isRs. 60 and Rs.4.5 for the othergroupof 1500 workers.Find the meanandstandarddeviation for the entire group. Q. 3. Mr. Arun and Mr. Bhandari play a game. If Mr. Arun picks up an even number from 1 to 6, Mr. Bhandari will pay him double the amount equal to picked up number. If Mr. Arun picks up an odd number then he has to pay amount equal to double the picked up number. What is Mr. Arun’s expectation? Q. 4. The probability that an employee will get an occupational disease is 20%. In a firm having five employees, what is the probability that: i) None of the employees get the disease ii) Exactly two will get the disease iii) More than four will contract the disease Q. 5. Microsoftestimatedthatoutof 10,000 potential software buyers,35% waittopurchase the new OS WindowsVista,untilanupgrade hasbeenreleased.Afteranadvertisingcampaigntoreassure the publicwasreleased,Microsoftsurveyed3000buyersandfound950 whoare still skeptical.At5% level of significance, can the company conclude that the population of skeptical people had decreased? Q. 6. Explain Chi-square test and the conditions for applying chi-square test. Q. No 1. Distinguish between Classification and Tabulation. Explain the structure and components of a Table with an example. Ans.1- Meaningof ClassificationandTabulation are asfollows:- Classification: - According to Secrist, “Classification is the process of arranging data into sequences and groups according to their common characteristics or separating them into different but related parts”.AccordingtoStocktonand Clark,“The processof groupinglarge numberof individualfactsand observations, on the basis of similarity among the items is called Classification”. Tabulation:- Tabulationfollowsclassification.Itisalogical orsystematiclistingof relateddata inrows andcolumns.The rowof atable representsthe horizontalarrangementof dataandcolumnrepresents the vertical arrangementof data.The presentationof dataintablesshouldbe simple,systematicand unambiguous.
- 2. The objectives of tabulation are to: Simplify complex data Highlight important characteristics Present data in minimum space Facilitate comparison Bring out trends and tendencies Facilitate further analysis Differences between Classification and Tabulation Table depicts the few differences between classification and tabulation. Table: Differences between Classification and Tabulation Structure and Components of a Table with an example:- Table andfigure depictthe partsof a table alongwiththe explanationof eachtab(tabs from1 to 10).
- 3. Tab 1: Table number Table numberistoidentifythe tableforreference.Whenthereare manytablesinananalysis, then table numbers are helpful in identifying the tables. Tab 2: Title Title indicatesthe scope andthe nature of contentsin a concise form.In otherwords,title of a table givesinformationaboutthe data containedinthe bodyof the table.Title shouldnot be lengthy. Tab 3 and Tab 4: Captions Captions are the headings and subheadings describing the data present in the columns. Tab 5 and Tab 6: Stubs Stubs are the headings and subheadings of rows. Tab 7: Body of the table Body of the table contains numerical information. Tab 8: Totals The sub-totals for each separate classificationand a general total for all combined classes shouldbe givenat the bottomor rightside of the figureswhose totalsare taken.Rulingand spacingseparate columnsand rows.However,totalsare separatedfrom mainbody by thick lines. Tab 9: Head note Head note is given below the title of the table to indicate the units of measurement of the data and is enclosed in brackets. Tab 10: Source note Source note indicatesthe source fromwhichdatais taken.The source note relatedtotable isplaced at the bottomonthe lefthandcorner.
- 4. Q. No 2. (a) Explain Arithmetic mean. (b) The mean wage is Rs. 75 per day, SD wage is Rs. 5 per day for a group of 1000 workers and the same isRs. 60 and Rs.4.5 forthe othergroupof1500 workers.Findthe meanandstandard deviation for the entire group. Ans.2- (a) Arithmeticmean: Arithmeticmeanisdefinedasthe sumof all valuesdividedbynumberof values andis representedby X. Arithmeticmeanisalsocalled‘average’.Itismostcommonlyusedmeasures of central tendency.ArithmeticMeanof a seriesisthe value obtainedbyaddingall the observations of a series and dividing this total by the number of observations. There are two types of Arithmetic Mean: 1) Simple arithmetic Mean 2) Weighted arithmetic Mean (b) Solution: We have by data, Combined Mean: Combined Standard deviation:
- 5. Q. No 3. Mr. Arun and Mr. Bhandari play a game. If Mr. Arun picks up an evennumber from 1 to 6, Mr. Bhandari will pay him double the amount equal to pickedup number.If Mr. Arun picks up an odd numberthen he has to pay amount equal to double the pickedup number. Whatis Mr. Arun’s expectation? Ans.3- Let Xi be the random variable and P(Xi) be its probability. The probabilities are indicated in table. Required Values for Calculating Meanand Variance for the Data NO. (Xi) P(Xi) Xi P(Xi) 1 -2 1/6 - 2/6 2 4 1/6 4/6 3 -6 1/6 -1 4 8 1/6 8/6 5 -10 1/6 -10/6 6 12 1/6 12/6 Total 1 1 Expectation of Mr. Arun is E(Χ) = Χ Ρ(Χ ) = 1.
- 6. Q. 4. The probability that an employee will get an occupational disease is 20%. In a firm having five employees, what is the probability that: i) None of the employees get the disease ii) Exactly two will get the disease iii) More than four will contract the disease Ans.4- Solution: - Given that: p 20/1000.2 q 10.2 0.8 i) The probability that none of the employees get the disease is given by: Therefore, the probability that none of the employees get the disease is 0.3277. ii) The probability that exactly two employees will get the disease is given by: Therefore, the probability that exactly two employees will get the disease is 0.2048. iii) The probability that more than four employees will get the disease is given by: Therefore, the probability that more than four employees will get the disease is 0.00032.
- 7. Q. 5. Microsoftestimatedthat out of 10,000 potential software buyers,35% wait to purchase the new OS WindowsVista,until an upgrade has beenreleased.Afteran advertisingcampaign to reassure the publicwas released,Microsoftsurveyed3000 buyers and found950 who are still skeptical.At 5% level ofsignificance,can the company conclude that the populationof skeptical people had decreased? Ans.5:- Solution: The procedure is explained in the following steps: 1. Null hypothesis Ho: p = .35 Alternate hypothesis H1: p < 0.35 2. Level of significance = 0.05 Ztab = - 1.645 and R: z < -1.645 3. Teststatistics 4. Given p = 950/3000 = 19/60 = 0.317, p = 0.35, q = 1-p = 1- 0.35 = 0.65, N=10,000, n = 3000 5. Conclusion:Since Zcal (-4.52) < Ztab (-1.645) and isin the rejectionregion,Hoisrejected.At 5% level of significance,we conclude that the proportionof skeptical people hassignificantly decreased.
- 8. Q. 6. ExplainChi-square test and the conditionsfor applyingchi-square test. Ans.6- Meaningof Chi-Square test:- The Chi-square testisone of the most commonlyused non-parametrictests instatistical work.The GreekLetter 2 is used to denote this test. 2 describe the magnitude of discrepancybetween the observedandthe expectedfrequencies.The value of 2 is calculatedas: Where, O1, O2, O3….On are the observed frequencies and E1, E2, E3…En are the corresponding expected or theoretical frequencies. Conditions for applying the Chi-Square test: - The following are the conditions for using the Chi-Square test: 1. The frequencies used in Chi-Square test must be absolute and not in relative terms. 2. The total number of observations collected for this test must be large. 3. Each of the observationswhichmake upthe sample of thistestmustbe independentof eachother. 4. As 2 test is based wholly on sample data, no assumption is made concerning the population distribution. In other words, it is a non-parametric-test. 5. 2 testis whollydependentondegreesof freedom.Asthe degreesof freedomincrease,the Chi- Square distribution curve becomes symmetrical. 6. The expected frequency of any item or cell must not be less than 5, the frequencies of adjacent items or cells should be polled together in order to make it more than 5. 7. The data should be expressed in original units for convenience of comparison and the given distribution should not be replaced by relative frequencies or proportions. 8. Thistestisusedonlyfordrawinginferencesthroughtestof the hypothesis,soitcannotbe usedfor estimation of parameter value.