Question

0 -14 points 1.E.051. An article reported the following data on oxidation-induction time (min) for various commercial oils: 87 101 130 160 180 195 131 145 215 105 145 151 154 138 87 99 95 119 129 (a) Calculate the sample variance and standard deviation. (Round your answers to three decimal places) min2 S- min (b) If the observations were reexpressed in hours, what would be the resulting values of the sample variance your answer to three decimal places.) and sample standard deviation? hr2 SE hr

Answer 1

a) For the given sample in minutes

87
101
130
160
180
195
131
145
215
105
145
151
154
138
87
99
95
119
129

Sample standard deviation is calculated s

ー1

where $\bar{x}=\frac{\sum x_{i}}{N}$

$\bar{x}=\frac{\sum x_{i}}{N}=135.0526$

S= 36.032

hence S²= Sample variance =1298.275

b) if Sample changed into Hrs by dividing each sample with 60 then the sample becomes

1.4500
1.6833
2.1667
2.6667
3.0000
3.2500
2.1833
2.4167
3.5833
1.7500
2.4167
2.5167
2.5667
2.3000
1.4500
1.6500
1.5833
1.9833
2.1500

Similarly using Sample standard deviation formula

ー1

and

$\bar{x}=\frac{\sum x_{i}}{N}$

Mean =2.68Hrs

and Standard deviation, s=0.601

and S²=0.361