OtherPapers.com - Other Term Papers and Free Essays
Search

Quantitative Methods

Essay by   •  September 5, 2012  •  Essay  •  1,034 Words (5 Pages)  •  1,576 Views

Essay Preview: Quantitative Methods

Report this essay
Page 1 of 5

CONTENT

Question 1 3

Question 2 8

Question 3 9

Question 4 10

Question 5 13

BIBLIOGRAPHY 15

QUESTION 1

Class Interval of iPods Freq f Class Midpt, x fx fx2

20-50 6 35 210 44100

50-80 12 65 780 608400

80-110 14 95 1330 17728900

110-140 4 125 500 250000

140-170 4 155 620 384400

=40 =3440 =19015800

1.1

1.2.1

1.2.2

1.2.3.

1.3.1 P = (k/100)(n)

= (65/100)(40)

= 26th position

65th percentile = L+[(P-cf)/f (U-L)]

= 80.5+[(26-18)/14 (110.5-80.5)]

= 80.5 + 17.14

= 97.14 or 97

1.3.2 Q1 = 1(N)th value

= (1(40))/4 th value

= 10th value

10th value is in the interval 50.5 - 80.5

Therefore Q1 is (50.5 - 80.5)

Q1 = L+ h/f(iN/4-c)

= 50.5+ 30/12(1(40)/4-6)

= 50.5 + 10

= 60.5 or 61

Q3 = 3(N)th value

= (3(40))/4 th value

= 30th value

30th value is in the interval 80.5 - 110.5

Therefore Q3 is (80.5 - 110.5)

Q3 = L+ h/f(iN/4-c)

= 80.5+ 30/14(3(40)/4-18)

= 80.5 + 25.71

= 106.21

The Quartile Deviation = (Q3 - Q1)/2

= (106.2-60.5)/2

= 22.85 or 23

QUESTION 2

2.1 Mean (μ) = X/N

= 10833/12

= 902.75

Standard Deviation (σ) = √(〖(X-M)〗^2/(N-1))

=√(143420/(12-1))

= √13038.21

= 114.185

2.2 = P(X < 800)

= P(X -)/ < (800 - 902.75)/114.185

= P (Z< -0.899855)

= .1841

2.3 1 billion = 1000 million

μ = 902.75

σ = 114.185

standardise x to z = (x - μ) / σ

P(x > 1000) = P (z > (1000-902.75) / 114.185)

= P (z > 0.8517) = .1977

2.4 We know from the standard normal that

P (Z < z) = 0.95 = P(Z > z) = 0.05

when z = 1.644854

x = μ + z * σ

x = 902.75 + 1.644854 * 114.185

x = 1090.568 or 1091

QUESTION 3

H0: There is no evidence of a change in the market shares of car rental firms

H1: There is evidence of a change in the market shares of car rental firms

 = 0.05

Reject H0 if the calculated X2 > 15.507

df = (k-1)(r-1) = (5-1)(3-1) = 8

CONTINGENCY TABLE:

CONTINGENCY TABLE

f0 fe x^2=  ((f0 - fe)2)/fe

30 28.91 0.041

25 24.78 0.002

20 16.84 0.593

14 17.47 0.691

6 6.99 0.140

29 30.13 0.042

23 25.83 0.309

18 17.55 0.012

20 18.21 0.176

9 7.28 0.404

32 31.96 0.000

30 27.39 0.248

15 18.61 0.701

21 19.31 0.147

7 7.73 0.068

299 299 3.575

The

...

...

Download as:   txt (5.8 Kb)   pdf (192.1 Kb)   docx (13.1 Kb)  
Continue for 4 more pages »
Only available on OtherPapers.com