Statistics 1

Business Mathematics and Statistic (523)

First Assignment

1

Q.1 (a)

The following table presents the number of new business  opened   in   Pakistan   during   the   third   quarter   of   1990.  Construct a pie chart to represent these data.

Ans.

To construct the pie chart, we will have to calculate the  percentage for each type of new business opened to the  total   number   of   businesses   opened   during   the   third  quarter of 1990 and then angle of each representation will  determine in the following manner: The total number of new businesses opened = 31728

Nature of  Business Retailing Services Construction Wholesaling Manufacturing Other Total:

No. of  Business 10,724 4,886 4,315 3,776 2,760 5,267 31,728

Angle

33.8/100 x 360 =  15.4/100 x 360 = 

13.6/100 x 360 = 11.9/100 x 360 = 8.7/100 x 360 = 16.6/100 x 360 =

122° 55° 49° 43° 31° 60° 360°

Now   the   pie   chart   will   be   constructed   by   dividing   a  circular   pie   into   six   parts   of   proportional   sizes   as  indicated by the percentages representing them. The pie  chart would appear as:

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16.6%

First Assignment

Retailing 33.8%

8.7%

S ervices Construction Wholesaling

11.9%

Manufacturing 13.6%

15.4%

Other

(b)

Why   does   business   statistics   usually   involve   the  use   of  sample   data   instead   of   the   entire   population   when  making a business decision.

Ans.

Business   statistics   usually   involves   the   use   of   sample  data instead of the entire population, due to the following  reasons. •

In most cases, it is impossible to obtain the data for  the entire population. For example, if a TV network  wants   to   obtain   the   views   of   its   viewers   about   a  certain talk show, it would be practically impossible  for the network to reach all of its viewers.

•

Sampling   saves   time,   cost,   and   effort.   To   gather  data   about   every   member   of   the   population   is  usually very time consuming and costly. Usually, it  is   so   costly   that   the   benefits   expected   to   be  achieved from analyzing the data are not worth the  cost involved. Therefore, a random sample, which is  representative of the population, is chosen.

•

In   some   cases,   inspection   or   testing   of   a   good   or  product   destroys   its   usefulness.   For   example,   if   a  firm manufactures bullets, it will not be feasible to 

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Business Mathematics and Statistic (523)

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First Assignment

test each and every bullet by firing it because then  they   would   become   useless.   Therefore,   in   such  cases, sampling is unavoidable.

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Q.2

A particular industrial product is shipped in lots of 20.  Testing to determine whether an item is defective is costly  hence the manufacturer samples production rather than  using   a   100%   inspection   plan.     A   sampling   plan  constructed   to   minimize   the   number   of   defective   units  shipped   to   customers   calls   for   sampling   5   items   from  each lot and rejecting the lot if more than one defective  units is observed.  (If rejected, each item in the lot is then  tested).   If a lot contains 4 defectives units, what is the  probability that it will be accepted.

Ans. n=5 Probability of acceptance ­ p =1/5 Probability of rejection ­ q = P(r ≤ 1) =  P(0)

4/5

P(0)+P(1)

= C n pr q n-r

r = C 50 (.2)0 (.8)5

= 0.3277 P(1)

= C n pr q n-r

r = C15 (.2)1 (.8) 4

= 0. 4096 = 0.3277 + 4096  = 0.7373

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Q.3(a)

First Assignment

A decline in the demand for a specific brand of a product  may   result   from   either   a   decline   in   demand   for   the  product, in general, or a reduction in the percentage of  the   purchasers   selecting   this   specific   brand   over   its  competitors.   Suppose that to remain competitive in the  marketplace,   a   company   producing   brand   “A”   must  maintain at least an equal share of the market with its  three major competitors.  IF 100 consumers are randomly  selected   and   interviewed   regarding   their   brand  preferences, what is the probability of observing sample  proportion preferring brand “A” as small as 0.15 or less if  in fact one­fourth or 25 per cent of the consumers prefer  brand “A”.

Ans.

n = 100 p 0 = 0. 25 q 0 = 0. 75

and

q0 = 1 - p 0

P(P ≤ 0.15) = ? So Z =

Z =

Z =

) P - p0 p 0q 0 n 0.15 - 0.25 0.25 × 0.75 100 0.15 - 0.25 0.0433

P(P ≤ 0.15) = P(Z≤ ­2.31) = 0.5­P(­2.31≤ Z≤ 0) = 0.5­0.4896 So

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P(P ≤ 0.15) = 0.0104

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7

(b)

A   sample   of   n=100   employees   from   a   company   was  selected   and   the   annual   salary   for   each   was   recorded.  The mean and standard deviation of their salaries were  found to be  X = Rs . 7,750 and s = Rs . 900 Construct   the   95   per   cent   confidence   interval   for   the  population average salary µ.

Ans.

X - Z

∝

σ 2

n

7750 - Z.025 7750 - Z.025

<µ < X+Z 900 100 900

∝

σ 2

n

< µ < 7750 + Z .025

900 100

100

7750 - (1.96)

900 100

7750 - 176. 4 = 7573.6 and 7750 + Z .025

900

100 = 7750 + 176.4 = 7926.4 7573.6 < µ

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< 7926.4

# 7

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First Assignment

8

Q.4 (a)

The   following   data   have   been   obtained   for   a   random  sample of ten electronics firms, where X represents age of  the company in years and  Y  represents annual sales in  millions   of   dollars.     Determine   the   correlation   between  age of the firm and sales. X

Y

X2

Y2

XY

3 10 5 6 12 15 9 2 9 7 78

2.5 6 2.5 3.5 6 6.5 6 1.5 5.5 5 45

9 100 25 36 144 225 81 4 81 49 754

6.25 36.00 6.25 12.25 36.00 42.25 36.00 2.25 30.25 25.00 232.50

7.5 60.0 12.5 21.0 72.0 97.5 54.0 3.0 49.5 35.0 412.0

The coefficient of correlation: r=

∑ (X − X)(Y − Y) [∑ ( X − X ) (Y − Y ) 2

2

]

( ∑ X )(∑ Y ) ( ) n (∑ X ) −

∑ (X − X)(Y − Y) = ∑ XY − 2

∑(X − X)

2

∑(Y − Y)

2

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= ∑ X2

= ∑ Y2 −

n

( ) ∑Y

2

n

# 8

Business Mathematics and Statistic (523)

First Assignment

9

∑ X ) (∑ Y ) ( = 412 ∑ XY − n

∑X

2

( ) − ∑X

2

n

= 754 −

782

-

78 × 45 10

= 61

= 145.6

10

∑ Y) ( −

2

∑( Y − Y) = ∑ Y 2

r =

r =

2

n

= 232.50 −

452 10

= 30

61 145.6 × 30 61 66. 09

r = 0.923

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10

(b)

Describe   a   situation   in   your   area   of   interest   in   which  simple correlation analysis would be useful.   Use real or  realistic data and do a complete correlation analysis.

Ans.

The   data   given   below   corresponds   to   the   months­wise  sales   and  marketing  expenses  of   the  ABM   Data   System  (Pvt)   Limited   which   is   renowned   dealer   in   Computer  business. In   the   following,   the   correlation   coefficient   analysis   is  made between the total sales (X) and marketing expenses  (Y):

Month of  Year 1995

Total sales (X)

Mktg.  Exp. (Y)

X2

Y2

XY

January

820000

77900

672400000000

6068410000

63878000000

February

830000

82000

688900000000

6724000000

68060000000

March

720000

72000

518400000000

5184000000

51840000000

April

705000

69500

497025000000

4830250000

48997500000

May

980000

99275

960400000000

9855525625

97289500000

June

875000

86000

765625000000

7396000000

75250000000

July

1100000

98000

1210000000000

9604000000

107800000000

August

920000

99750

846400000000

9950062500

91770000000

September

700000

89000

490000000000

7921000000

62300000000

October

620000

70000

384400000000

4900000000

43400000000

November

860000

80000

739600000000

6400000000

68800000000

December

870000

76575

756900000000

5863730625

66620250000

10000000

1000000

8530050000000

84696978750

846005250000

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Business Mathematics and Statistic (523)

First Assignment

11

r =

[N ∑ X

r =

N ∑ XY − ( ∑ x )( ∑ y ) 2

− ( ∑ X )2

][N ∑ Y

2

(

)

− ( ∑ Y )2

]

12 × 846005250000 - 10000000 × 1000000

12 × 8530050000000 − 10000000

2

× 12 × 846005250000

- (1000000 )2

r = 0.77

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Business Mathematics and Statistic (523)

First Assignment

12

Q. 5 (a)

A penal of potential cereal product consumers was asked  to rate one of four potential new products.  Each member  of the panel rated only one of the products on a 100 point  scale,  as  compared  to  a   standard,  existing  cereal.     The  data is given below: I II III IV

16, 31, 57, 62, 67, 71, 73, 75 30, 35, 52, 60, 64, 65, 65, 67, 70, 71, 75, 82 43, 51, 53, 54, 56, 58, 61, 64, 64, 67, 68, 70, 71, 75, 79 29, 39, 46, 50, 59, 61, 62

Test the hypothesis of equal mean product scores by an  analysis of variance.  First use ∝ = 0.05, then find a ρ - value. Ans. H 0 :µ1 = µ 2 = µ 3 = µ 4 H1 : All me ans are not e qual

∝ =

0.05

I

II

III

IV

16 31 57 62 67 71 73 75

30 35 52 60 64 65 65 67 70 71 75 82

43 51 53 54 56 58 61 64 64 67 68 70 71 75

29 39 46 50 59 61 62

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79 C1   =  452 C2   =  736 C3   =  934 C4   =  346

452 8 736 X2 = 12 934 X3 = 15 346 X4 = 7 X1 =

= 56.5 = 61.33 = 62.27 = 49.43 2

Sum of square = SS total =  ∑ (X ) -

∑ (X

2

(∑ X )2 n

) = (16)2 + (31)2 + (57)2 + (62)2 ................(50)2 + (59)2 + (61)2 + (62)2 = 154080

∑X

= 452 + 736 + 934 + 346 = 2468

∑ (X )2

= (2468 )2 = 6091024

SS(T)

=

2

∑ (X ) -

= 154080−

(∑ X )2 n

(2468)2 42

= 154080 ­ 145024.381 = 9055.62 SS (Factors) + SS (Error) = SS (Total) SS (Products) = 

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 C2 C2 C2 C2   1 + 2 + 3 + 4−  K1 K2 K 3 K4   2 452 = +  8 

( )

(∑ X )

2

n 2

(736) + (934) + (346)  − 145024.381 2

12

2

15

7

 

=  145938.69 ­ 145024.381 = 914.305

SS (Error ) =

∑

X2

 2 452 = 154080 −   8 

( )

 C2 C2 C2 C2  -  1 + 2 + 3 + 4  K 1 K 2 K 3 K 4  2

(736) + (934) + (346)  2

2

12

15

7

 

= 154080 - 145938 .69 = 8141.31 SS (Factor) + SS (Error) = SS (Total) 914.305 + 8141.31 = 9055.62

ANOVA TABLE SOURCE

S.S.

d.f.

M.S.

Product

914.305

C­1=3

914.305 3 = 304.768

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Error

8141.31

n­c=38

Total

9055.62

n­1 = 41

F =

304.768 214.245

8141.31 38 = 214.245

∝

0.05

F = 1.422 F (3,38, 0.05 = 2.85) F ∝ = 0.05 is 2.85 Result The calculated value F = 1422 is less than F 0.05 = 2.85,  so we do not reject H0 at 0.05 level of significance

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(b)

First Assignment

A   production   superintendent   claims   that   there   is   no  difference in the employee accident rates for the day and  evening   shifts   in   a   large   manufacturing   plant.     The  number of accidents per day is recorded for both the day  and the evening shifts for n=100 days.   It is found that  the number of accidents  yE  per day for the evening shift  exceeded the corresponding number of accidents   yD   on  the  day   shift  on  63  of  the  100 days.   Do  these results  provide   sufficient   evidence   to   indicate   that   more  accidents tend to occur on one shift than on the other, or  1 P(yE > y D) = 2? equivalently, that 

Ans. H0 : p ( Y E > Y D ) =

1 2

H0 : p ( Y E < Y D ) ≠

1 2

n = 100 X = 63 so that ) P = 63 100 also p 0 = 0.50 q 0 = 0.50 Z =

Z =

) P - p0 p 0q 0 n 0.63 - 0.50 (0.50)(0.50 ) 100

Ζ = 2.6

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