Question

An article reported the following data on oxidation-induction time (min) for various commercial oils: 86 101 130 160 180 195 133 145 213 105 145 151 154 135 8799 95 119 129 (a) Calculate the sample variance and standard deviation. (Round your answers to three decimal places.) 2 min 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.) hr2 hr

Answer 1

a) Using Excel we get,

sample variance = s² = 1284.981 min²

sample standard deviation = s = 35.846 min

b) The observations were reexpressed in hours.

We know, 1 hour = 60 min

=> 1 min = 1/60 hour

Therefore, the new standard deviation would be 1/60 of the original and the variance would be 1/3600 of the original.

sample variance=1284.981/(60)² =1284.981/3600=0.357

=> sample variance = s² = 0.357 hr²

^{sample standard deviation = 35.846/60 = 0.597}

^{=> sample standard deviation = s = 0.597 hr}

^{86 101 130 160 180 195 133 145 213 105 145 151 154 135 Excel Commands: 2 VAR(A1 A19) Sample Variance: Sample standard deviation: STDEV(A1:A19) 10 12 13 14 15 16 17 18 19 20 21 22 sample variance 23 24 sample standard deviation 87 95 119 129 1284.918129 35.84575468}