An article reported the following data on oxidation-induction time (min) for var

Question

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

(a) Calculate the sample variance and standard deviation. (Round your answers to three decimal places.)

(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.)

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Explanation / Answer

Ans:

a)Sample variance=22978.11/18=1276.561

Sample standard deviation=sqrt(1276.561)=35.729

b)Sample variance=1276.561/3600=0.355

Sample standard deviation=35.729/60=0.595

x (x-mean)^2 1 88 2179.02 2 101 1134.34 3 130 21.90 4 160 641.10 5 180 2053.90 6 195 3638.50 7 131 13.54 8 145 106.50 9 212 5978.38 10 105 880.90 11 145 106.50 12 151 266.34 13 152 299.98 14 138 11.02 15 87 2273.38 16 99 1273.06 17 92 1821.58 18 119 245.86 19 129 32.26 Total= 2559.00 22978.11 Mean= 134.6842 1276.561 Variance 35.729 Std dev

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