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Statistics / Probability and Statistics for Engineers and the Scientists 9 / Chapter 1 / Problem 51E

Probability and Statistics for Engineers and the Scientists | 9th Edition | ISBN: 9780321629111 | Authors: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye

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The article “A Thin-Film Oxygen Uptake Test for the

Chapter 2, Problem 51E

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QUESTION:

The article “A Thin-Film Oxygen Uptake Test for the Evaluation of Automotive Crankcase Lubricants” (Lubric. Engr., 1984: 75–83) reported the following data on oxidation-induction time (min) for various commercial oils:

87 103 130 160 180 195 132 145 211 105 145

153 152 138 87 99 93 119 129

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.

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QUESTION:

The article “A Thin-Film Oxygen Uptake Test for the Evaluation of Automotive Crankcase Lubricants” (Lubric. Engr., 1984: 75–83) reported the following data on oxidation-induction time (min) for various commercial oils:

87 103 130 160 180 195 132 145 211 105 145

153 152 138 87 99 93 119 129

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:

Step 1 of 2

Given The article “A Thin-Film Oxygen Uptake Test for the Evaluation of Automotive Crankcase Lubricants”.

Consider the data

87	145
103	153
130	152
160	138
180	87
195	99
132	93
145	119
211	129
105

Now we have to calculate the sample variance and standard deviation.

$x$	${x}_{i}^{2}$
87	7569
103	10609
130	16900
160	25600
180	32400
195	38025
132	17424
145	21025
211	44521
105	11025
145	21025
153	23409
152	23104
138	19044
87	7569
99	9801
93	8649
119	14161
129	16641
$\Sigma{}{x}_{i}$ = 2563	$\Sigma{}{x}_{i}^{2}$ = 368501

So we have to find mean.

The formula of the mean is

$\overline{x}=\frac{\Sigma{}{x}_{i}}{n}$

Substitute the value ${\Sigma{}x}_{i}$ = 20,179 and n = 27.

$\overline{x}=\frac{2,563}{19}$

$\overline{x}=134.8947$

The formula of the sample variance is.

${s}^{2}$ = $\frac{(\Sigma{}{x}_{i}{)}^{2}\ -\frac{(\ \Sigma{}{x}_{i}\ {)}^{2}}{n}\ }{n-1}$

Substitute the value ${\Sigma{}x}_{i}$ , n and $\overline{x}$ .

${s}^{2}$ = $\frac{(368501{)}^{\ }\ -\ \frac{(\ 2563\ {)}^{2}}{19}\ }{19-1}$

${s}^{2}$ = $\frac{(368501{)}^{\ }\ -\ \frac{(\ 658969{)}^{\ }}{19}\ }{18}$

${s}^{2}$ = $\frac{(368501{)}^{\ }\ -\ 345735.2105\ }{18}$

${s}^{2}$ = $\frac{(22765.78947\ }{18}$

${s}^{2}$ = 1,264.7660

Therefore sample variance is 1,264.7660

Then we have to find standard deviation.

Standard deviation is square root of the sample variance.

We know that sample variance is 1,264.7660.

s = $\sqrt{1,264.7660\ \ \ }$

s = 35.5635

Therefore the standard deviation is 35.5635.

________________________

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