Question

An article reported the following data on oxidation-induction time (min) for various commeroal oils 85 101 130 160 180 195 133 145 211 105 145 153 152 139 87 99 93 119 129 (a) Calculate the sample variance and standard deviation. (Round your answers to three decimal places.) s2- min2 》 the observations were r expressed in hours, what would be t e resulting va ues o the sample variance and samp e standard deviation? Answer without actually performing he reexpression Round your answer to three decimal places.) S2 hr2 hr

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Answer #1

Following table shows the calculations:

Sample size: n=19

Mean:

--22. 2561 19 i-L . 134.789 7n

Variance:

s^{2}=rac{sum left ( x-ar{x} ight )^{2}}{n-1}=1283.064 ext{ min}^{2}

Standard deviation:

Σ(z-i)2 095.1535.820mi 19 1 35.820 min

(b)

To convert min in hours we need to divide each data value by 60. So,

Variance:

-21283.064 s2 = Σ (z-i)2 ,-1283.064 0356 hours* rl 602

Standard deviation:

Σ(z-z)2ー /230951579 . 35.820 r-23095.1579 35.820 0.597 hours 19 1 60

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