An article reported the following data on oxidationinduction

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.)

Solution

(a) Variance is calculated in three steps :

1. calculate the mean

2. substract the each value from the mean and square it

3. Take the mean of the squares

Mean = (86+102+130+160+180+195+133+145+213+105+145+153+152+138+87+99+94+119+129)/19 = 135

Now substracting each value from 135 and squaring it, and then taking the mean,

(49²+ 33²+5²+25²+45²+50²+2²+10²+78²+30²+10²+18²+17²+3²+48²+36²+41²+16²+6²)/19 = 22048/19 = 1160.421 minutes

Standard deviation= (variance)^1/2

                                         = (1160.42)^1/2 = 34.065 minutes

(b) variance (in hours)= 1160.421/60 = 19.430 hours

Standard deviation(in hours) = 34.065/60 = 0.567 hours