Statistics Analysis Paper - 275 Words

Statistics Analysis Paper (Statistics Project Sample) Content: 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 1SSTOTAL= (4,724.5119+4,627.9762+4,721.0206+4,788.8186) à ¢Ã¢â€š ¬ (119.05+117.7+119+119.8)2 /12=18,862.3273- 18,845.65021=16.6770917SSAMONG= (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=5.553822SSWITHIN=16.6770917-5.553822=11.1232697DF (among) =3Ms (among)=SSamong/3=5.553822/3=1.851274DF (within) =8Ms (within) =SSwithin /8=11.1232697/8=1.390408713F=ms among/ mswithin=1.851274/1.390408713=1.331460298F must be at least 1.3315 to reach alpha0.05.Therefo re there exists significant difference in the performance of the four grades of gas.QUWSTION 2F=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=6MSE= (4-1)32 +(4-1)62 /(19-2)=7.941176471F=6/7.941176471=0.755555555With 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 3ENTRANCE 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=125H0: entrance are equally utilizedH1: entrances not equally utilized