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

An article reported the following data on oxidation-induction time (min) for various commercial oils: 88 104 130 160 180 195 132 145 213 105 145 151 153 136 87 99 95 119 129 (a) Calculate the sample variance and standard deviation. (Round your answers to three decimal places.) s2 = min2 s = min (b) If the observations were reexpressed in hours, what would be the resulting values of the sample variance and sample standard deviation? Answer without actually performing the reexpression. (Round your answer to three decimal places.) s2 = hr2 s = hr

Solutions

Expert Solution

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

orchestra answered 9 months ago

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