MBA in Management


 Statistics for Managers


Final Exam Paper


1.                   Samples of light bulbs were obtained from two suppliers and tested ‘to destruction’ in the laboratory. The following results were obtained for the length of life:


 


Length of Life (hours)


 


700-899


900-1099


1100-1299


1300-1499


Total


Supplier A


14


16


26


12


68


Supplier B


5


36


21


5


67


 


(a)   Which supplier’s bulbs have greater average length of life?


For easier computation, the class intervals were assigned numbers. Interval 700-899 is 1; 900-1099 is 2; 1100-1299 is 3; and 1300-1499 is 4.


SUPPLIER A      = 14 (1) + 16 (2) + 26 (3) + 12 (4) / 68


                                          = 2.53 or 3


l 3 or class interval 1100-1299 is the average length of life for Supplier A.


SUPPLIER B      = 5 (1) + 36 (2) + 21 (3) + 5 (4) / 67


                                          = 2.39 or 2


l 2 or class interval 900-1099 is the average length of life for Supplier B.


ll Supplier A has bulbs that have a greater average length of life, from 1100 to 1299 hours, compared to Supplier B, whose bulbs have an average length of life is from 900-1099 hours only.


(b)   Which supplier’s bulbs are more uniform in quality?


Supplier A mean = 17                                             Supplier A standard deviation = 5.38


Supplier B mean = 16.75                           Supplier B Standard deviation = 12.89


                                               



 


5.38


 


                                                 17


                                    =          .3165 or 31.7%


 



 


                                                12.89


 


                                                16.75


                                    =          .77 or 77%


 


l The bulbs of Supplier A are more uniform in quality, as a lower coefficient of variation denotes that the bulbs are more uniform in quality.


(c)   Which supplier would you prefer to use?


I would prefer to use Supplier A based on the above computations, because their light bulbs proved to be more cost-efficient, as the bulbs have a greater average length of life and are more uniform in quality than those of Supplier B.


2.                   At the end of an MBA course in business statistics, the final examination grades have a mean of 70.2 and a standard deviation of 12.3. There were 210 students on the course. Assuming that the distribution of these grades (all whole numbers) in normal, find:


(a)   the percentage of grades that should exceed 90



Z    = 90 – 70.2


                              12.3


                  = 1.61


P (to the right of Z = 1.61)


      = .5 – .4463


      = .054 or 5.4%


l There is a 5.4% percentage of grades that should exceed 90.


(b)   the percentage of grades less than 35



Z    = 35 – 70.2


                              12.3


                  = -2.86


P (to the left of Z = -2.86)


      = .5 – .4979


      = .0021 or .21%


l There is a .21% percentage of grades less than 35.


(c)   the number of failures (pass = 70 per cent)



 



 


69.9 – 70.2   


 


   12.3


              Z        = .024


        


      P (to the right of Z = .024)


  P           = .5 – .0080


    = .492 or 49.2%


    49.2 % x 210 = 103.32 or 104


l There are 104 students out of 210 students who failed the final examination.


(d)               the lowest distinction mark if at most the highest 5 per cent of grades are to be awarded distinctions.



Z = .45 = 1.645



 


x – 70.2


 


 12.3


 


x = (12.3) (1.645) + 70.2


 = 90.43


 


l The lowest distinction mark in the highest 5 percent of grades is 90.43.


3.                  , researcher for the Colombian Coffee Corporation is interested in determining the rate of coffee usage per household in the United Sates. She believes that yearly consumption per household is normally distributed with an unknown mean and a standard deviation of about 2 pounds.


a)                 If  takes a sample of 16 households and records their consumption of coffee for one year, what is the probability that the sample mean is within one pound of the population mean?



Z       = .1915 x 2


                     = .383 or 38.3%


l The probability that the sample mean is within one pound of the population mean is 38.3%.


b)                  How large a sample must she take in order to be 98 percent certain that the sample mean is within one pound of the population mean?



 



 


38.3                                98


 


16                                    n


                n = 40.94 or 41


l Ms.  must take 41 samples in order to be 98 percent certain that the sample mean is within one pound of the population mean.


4.                  A chain of department stores is moving into a phase of expansion and opening several new stores. As part of the expansion planning process a project team is carrying out an investigation to find out how the sales levels at new stores might be predicted. One approach has been to use regression analysis. The average level of sales per week (y) for each existing store in the chain (fourteen department stores) together with a measure of disposable income per family in each store’s catchment area are given in Table 11.1


Table 11.1


Store number


Average sales per week (₤000s)


Y


Average disposable family income (coded)


x


1


90


301


2


97


267


3


86


397


4


84


227


5


82


273


6


80


253


7


78


203


8


75


263


9


70


190


10


68


212


11


64


157


12


61


141


13


58


119


14


52


133


Mean


74


217


The computer gives the following results after regressing sales against family income:


Variable                     Coefficient


Income                       0.17846


Constant        35.228


Correlations Coefficient = 0.92


Standard error = 4.72


 


(a)               According to the above table, what is the predictable average sales per week with average family disposable income at 221?


y = a + bx


    = 35.228 + .17846 (221)


y  = 74.67


l The predictable average sale per week assuming disposable income is at 221 is 74.67.


(b)        How good is the application of regression analysis in the above data?


The correlations coefficient and the standard error results are suffice evidence that the application of regression analysis in the above data is reliable. The data from the average sales per week is the endogenous and dependent variable in the regression analysis, while the average disposable family income is the exogenous or the independent variable, changes on it that the former reacts to. Correlation coefficient values ranges from negative one to positive one, and a value closer to one (either negative or positive) indicates higher relationship between the two variables. Two closely related variables used in a regression analysis produce more accurate results in showing how changes in the independent variable affects the dependent variable, which, in this case, is the sales per week.


(c)        What are the limitations of this method to predict the average sales per week?


For one, there are other factors affecting the computation of sales other than the disposable family income of the consumers. These other factors may come in the form of inflation and interest rates, equity markets, amount of advertising expenditures and number of competitors, which the use of a multiple linear regression would sufficiently accommodate. The other limitation is that the values that can be accurately predicted are restricted to a certain range. Thus, computations for values outside the range will be doubtful because the investigation provided no evidence as to the nature of the statistical relationship outside the existing range.


5.                  Factory A conducted a sample check of total number of apples per crate and get the following results. The Quality Manager would like to see if the data are out of control.


Sample 1


Sample 2


Sample 3


Sample 4


Sample 5


110


93


99


98


109


103


95


109


95


98


97


110


90


97


100


96


102


105


90


96


105


110


109


93


98


110


91


104


91


101


100


96


104


93


96


93


90


110


109


105


90


105


109


90


108


103


93


93


99


96


97


97


104


103


92


103


100


91


103


105


90


101


96


104


108


97


106


97


105


96


99


94


96


98


90


106


93


104


93


99


90


95


98


109


110


96


96


108


97


103


109


96


91


98


109


90


95


94


107


99


91


101


96


96


109


108


97


101


103


94


96


97


106


96


98


101


107


104


109


104


96


91


96


91


105


(a)       Calculate LCL & UCL


UCLx   = x + A2 R                                                                  UCLR  = D4 R


= 99.54 + (.58) 19.8                                                                      = 2.11 (19.8)


UCLx   = 111.02                                                                     UCLR = 41.78


 


LCLx   = x – A2 R                                                                   LCLR   = D3 R


                        = 99.54 – (.58) 19.8                                                               = 0 (19.8)


LCLx   = 88.06                                                                                   LCLR   = 0


 


(b)       Calculate Average of Mean ( x )


Sample1x= 99       Sample2x= 98           Sample3x= 101


Sample4x= 98.7            Sample5x= 101


            x          = 99 + 98 + 101 + 98.7 + 101


            x          = 99.54


(c)        Draw the X – R chart




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