Statistics Analysis Paper (Statistics Project Sample)

X1

X2

X2

X2

X3

X23

X4

X24


39.31

1545.2761

36.69

1346.1561

38.99

1520.2201

40.04

1603.2016


39.87

1589.6169

40

1600.0000

40.02

1601.6004

39.89

1591.2121


39.87

1589.6189

41.01

1681.8201

39.99

1599.2001

39.93

1594.4049










Sum

SX1=119.05

SX21=4,724.5119

SX2=117.7

SX22=4,627.9762

SX3=119

SX23=4,721.0206

SX4=119.86

SX24=4,788.8186

(SX)2

14,172.9025


13,853.29


14,161


14,366.4196


QUESTION 1
SSTOTAL= (4,724.5119+4,627.9762+4,721.0206+4,788.8186) – (119.05+117.7+119+119.8)2 /12
=18,862.3273- 18,845.65021=16.6770917
SSAMONG= (14,172.9025/3 +13,853.29/3+14161/3+14,366.4196/3) - (119.05+117.7+119+119.8)2 /12
=4,724.300833+4,617.763333+4,720.333333+4,788.806533- 18,845.65021
=18,851.20403-18,845.65021=5.553822
SSWITHIN=16.6770917-5.553822=11.1232697
DF (among) =3
Ms (among)=SSamong/3=5.553822/3=1.851274
DF (within) =8
Ms (within) =SSwithin /8=11.1232697/8=1.390408713
F=ms among/ mswithin=1.851274/1.390408713=1.331460298
F must be at least 1.3315 to reach alpha<0.05.
Therefore there exists significant difference in the performance of the four grades of gas.
QUWSTION 2
F=Variation among sample means/variation within individual observation in the same sample





Dispenser

mean

Standard deviation

n


A

203

3

10


B

205

6

9







MSG=3(203-204)2 +3(205-204)2 /1=6
MSE= (4-1)32 +(4-1)62 /(19-2)=7.941176471
F=6/7.941176471=0.755555555
With the level of significance at 0.05, F is greater than the level of significance. Therefore the variability of dispenser B exceeds that of dispenser A.
QUEASTION 3
ENTRANCE

FREQUENCY

(x-mean)

(x-mean)2


NORTH

135

10

100


EAST

145

20

400


SOUTH

115

-10

100


WEST

105

-20

400


TOTAL

500


1000


Mean=500/4=125
H0: entrance are equally utilized
H1: entrances not equally utilized
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