An article reported the following data on oxidation-induction time (min) for various commercial oils: 88 104...

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An article reported the following data on oxidation-inductiontime (min) for various commercial oils: 88 104 130 160 180 195 132145 213 105 145 151 153 136 87 99 95 119 129 (a) Calculate thesample variance and standard deviation. (Round your answers tothree decimal places.) s2 = min2 s = min (b) If the observationswere reexpressed in hours, what would be the resulting values ofthe sample variance and sample standard deviation? Answer withoutactually performing the reexpression. (Round your answer to threedecimal places.) s2 = hr2 s = hr

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4.2 Ratings (655 Votes)

Sol:

X X - mean (X-mean)2 X2
88 -47.0526 2213.95014 7744
104 -31.0526 964.265928 10816
130 -5.05263 25.5290859 16900
160 24.94737 622.371191 25600
180 44.94737 2020.26593 32400
195 59.94737 3593.68698 38025
132 -3.05263 9.31855956 17424
145 9.947368 98.9501385 21025
213 77.94737 6075.79224 45369
105 -30.0526 903.160665 11025
145 9.947368 98.9501385 21025
151 15.94737 254.31856 22801
153 17.94737 322.108033 23409
136 0.947368 0.89750693 18496
87 -48.0526 2309.0554 7569
99 -36.0526 1299.79224 9801
95 -40.0526 1604.2133 9025
119 -16.0526 257.686981 14161
129 -6.05263 36.634349 16641
2566 -2.8E-14 22710.9474 369256

s^2=sample variance=22710.9474/19-1=1261.719 min^2

s=sample standard deviation=sqrt(1261.719)=35.521 min

Solutionb:

1 hour=60 min

1min=1/60 hr

s^2=sample variance=22710.9474/19-1=1261.719 min^2=1261.719 (1/60 )^2hr^2=0.3504775=0.350 hour^2

s=sample standard deviation=sqrt(1261.719)=35.521 min=35.521/60=0.5920167 hour=0.592 hour


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