Question

The composition of a flake of paint found on the clothes of a hit-and-run victim was compared with at of paint from the car suspected of causing the accident. Both samples were measured by inductively coupled plasma - mass spectrometry (ICP-MS) paint from clothes:%Ti=4.0, 4.2 ,4.4, 4.6 Paint from car:%T=4.5,5.3,5.5,5.0,4.9 a. Report the mean value, standard deviation and relative standard deviation in % Ti of paint from each source to an appropriate number of significant Calculated the pooled standard deviation of the ICP-MS method and state the number of degree of freedom in the pooled precision. (3 pts) figures Using an appropriate statistical approach, determine at the 95% confidence level whether the suspected car could have been involved in the accident. State what test or approach you are using. show all work, including all calculated and all critical values clearly state your conclusion The same spectroscopic method, applied to titanium-free paint, yielded the follow mg: % Ti = 0.2, -0.5,-0.2,0.6,0.4,-0.3

Answer 1

(a). Paint from clothes -

	x	x - $\bar{x}$	(x - $\bar{x}$)2
	4.0	- 0.3	0.09
	4.2	- 0.1	0.01
	4.4	0.1	0.01
	4.6	0.3	0.09
$\Sigma$	17.2		0.2

Total number of data points, N = 4

Mean value of %T of paint from clothes:

$\bar{x}$ = $\Sigma$ x / N

= (4.0 + 4.2 + 4.4 + 4.6) / 4

= 4.3

Standard deviation, S.D = $\sqrt{\Sigma (x - \bar{x})^2) / N}$

= $\sqrt{0.2 / 4}$

= $\sqrt{0.05}$

= 0.22

Relative standard deviation = S.D / $\bar{x}$ * 100

= 0.22 / 4.3 * 100

= 5.1 %

Paint from Car -

	x	x - $\bar{x}$	(x - $\bar{x}$)2
	4.5	- 0.54	0.2916
	5.3	0.26	0.0676
	5.5	0.46	0.2116
	5.0	- 0.04	0.0016
	4.9	- 0.14	0.0196
$\Sigma$	25.2		0.592

Mean value of %T of paint from car:

$\bar{x}$ = $\Sigma$ x / N

= 25.2 / 5

= 5.0

Standard deviation, S.D = $\sqrt{\Sigma (x - \bar{x})^2) / N}$

= $\sqrt{0.592 / 5}$

= 0.34

Relative standard deviation = S.D / $\bar{x}$ * 100

= 0.34 / 5.0 * 100

= 6.8%

(b). S1 = 0.22 , N1 = 4

S2 = 0.34,   N2 = 5

Pooled standard deviation,

= $\sqrt{}$[(4-1)*0.22² + (5-1)*0.34²] / (4+5-2)

= $\sqrt{}$[( 3*0.0484 + 4*0.1156 ) / 7]

= $\sqrt{}$(0.1452 + 0.4624) / 7

= $\sqrt{}$0.0868

= 0.29