Question

An article reported the following data on oxidation-induction time (min) for various commercial oils:87 105 130 160 180 195 135 145 213 105 145151 152 136 87 99 92 119 129(a) Calculate the sample variance and standard deviation. (Round your answers to three decimal places.)s^2 = ________. min^2s = ________. 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.)s^2 =______ hr^2s = ______hr

Answer 1

Explanation:

Mean = (87 + 105 + 130 + 160 + 180 + 195 + 135 + 145 + 213 + 105 + 145 + 151 152 + 136 + 87 + 99 + 92 + 119 + 129)/19 = 129

Variance = (summation(x - mean)²/n

Standard deviation = √(summation(x - mean)²/n

n = 19

Variance = [(87 - 129)^2 + (105 - 129)^2 + (130 - 129)^2+ (160 - 129)^2 + (180 - 129)^2 + (195 - 129)^2 + (135 - 129)^2 + (145 - 129)^2 + (213 - 129)^2 + (105 - 129)^2 + (145 - 129)^2 + (151 - 129)^2 + (152 - 129)^2 + (136 - 129)^2 + (87 - 129)^2 + (99 - 129)^2 + (92 - 129)^2 + (119 - 129)^2 + (129 - 129)^2]/19 = 23634/19 1243.895 min

Standard deviation = √1243.895 = 35.269 min

60 minutes = 1 hour

Converting the variance to hours,

Each division would have been divided by 60². 60² can be factorized out

Variance = 23634/60² = 6.565 hours

Converting the standard deviation to hours, it becomes

√6.565 = 2.562 hours