Answer to Question #334559 in Statistics and Probability for salman mahad

Question

The article “A Thin-Film Oxygen Uptake Test for the Evaluation of Automotive Crankcase Lubricants” reported the following data on oxidation-induction time (min) for various commercial oils: 87 103 130 160 180 195 132 145 211 105 145 a. Calculate the sample variance and standard deviation. 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.

Answer 1

Sample mean

$$\bar{x}=\dfrac{1}{11}(87+ 103+ 130 +160+ 180$$

 

$$+195 +132+ 145+ 211+ 105+ 145)=\dfrac{1593}{11}$$

 Sample variance

$$Var(X)=s^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n-1}$$

$$=\dfrac{1}{11-1}((87-\dfrac{1593}{11})^2+(103-\dfrac{1593}{11})^2$$

$$+(130-\dfrac{1593}{11})^2+(160-\dfrac{1593}{11})^2$$

$$+(180-\dfrac{1593}{11})^2+(195-\dfrac{1593}{11})^2$$

$$+(132-\dfrac{1593}{11})^2+(145-\dfrac{1593}{11})^2$$

$$+(211-\dfrac{1593}{11})^2+(105-\dfrac{1593}{11})^2$$

$$+(145-\dfrac{1593}{11})^2)\approx542.763636\ min^2$$

$$s=\sqrt{s^2}=\sqrt{542.763636}\approx 23.297288\ min$$

b.

$$Var(cX)=s_h^2=c^2Var(X)$$

$$Var(Y)=(\dfrac{1}{60})^2Var(X)$$

$$=\dfrac{1}{3600}(542.763636)=0.1507677\ hours^2$$

$$s_h=\sqrt{s_h^2}=\dfrac{1}{60}s\approx0.388288\ hours$$