Question

For Icpms: 2 7.01 +8.3-0997323 4799 as shown in following table? RMS FAAS / ppm ICPMS/ppm P 7.22 7.01 8.87 8.23 1.22 0.99 = 3,0 3.44 3.23 т 5.20 4.99 N=5. Se=EN 6. Below are two sets of analyses of Fe on the same sample by two 4 methods. Are their variances different significantly at a = 0.05? AW30 Method Concentration of Fe found / ppm 127, 125, 123, 130, 131, 126, 129 Y 130, 128, 131, 129, 127, 125 х An analysis of iron ore gives the following results: 33.78%, 33.84%, 33.60%, 33.15% 23.15% 23.6% Perform a Q-test to assess the possible outliers that show significance deviation at a = 5%.

6

Answer 1

Method X

Mean of values

= (127 + 125 + 123 + 130 + 131 + 126 + 129)/7

= 127.3

Variance :

Value	mean - value	(mean - value)2
127	127 - 127.3 = -0.3	(-0.3)2 = 0.09
125	125 - 127.3 = -2.3	(-2.3)2 = 5.29
123	123 - 127.3 = -4.3	(-4.3)2 = 18.49
130	130 - 127.3 = 2.7	(2.7)2 = 7.29
131	131 - 127.3 = 3.7	(3.7)2 = 13.69
126	126 - 127.3 = -1.3	(-1.3)2 = 1.69
129	129 - 127.3 = 1.7	(1.7)2 = 2.89
Variance	$\frac{1}{7}\sum (mean - x)^{2}$	7.06

Method Y

Mean of values

= (130 + 128 + 131 + 129 + 127 + 125)/6

= 128.33

Variance = 3.88

F value =Variance of X : Variance of Y = 7.06/3.88 = 1.82

Degree of freedom = (7-1) and (6-1) i.e 6 and 5

At 95% level of confidence, table value is 4.950

As F value is less than table value, so variances are not significantly different.

Q test

Arrange data in increasing order

33.15%, 33.60%, 33.78% , 33.84%

Considering 33.15% as outlier,

Q value = gap/range = (33.60 - 33.15)/(33.84 - 33.15)

= 0.652

Table value = 0.970

As Q value > table value, so, 33.15% is not outlier.

Considering 33.84% as outlier,

Q value = gap/range = (33.84 - 33.78)/(33.84 - 33.15)

= 0.0952

Table value = 0.970

As Q value > table value, so, 33.84% is not an outlier.

Hence, there are no outliers