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 87 103 130 160 180 195 131 145 215 105 145 152 154 137 87 99 92 119 129 (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.) hr

Explanation / Answer

a) Sample Variance will be = 1306.333

b) Sample Standard deviation will be = 36.143

b) Multiply Sample Variance by 60*60, we get 4702800

Multiply Sample Standard deviation by 60 , we get= 36.143*60 = 2168.594

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