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Statistics د:محمد عبد الجليل Redox theory أ.د.حريه Redox application أ.د.حسن عزقل Complexometry أ.د.سميحه. STATISTICS. Error : deviation from the absolute value .

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Statisticsد:محمد عبد الجليل

  • Redox theoryأ.د.حريه
  • Redox applicationأ.د.حسن عزقل
  • Complexometry أ.د.سميحه
slide4

Error: deviation from the absolute value .

  • Absolute error (E) : the difference between an observed or measured value (O) and the true value (T) , with regard to the sign and it is reported in the same units as the measurement.
  • E = O – T
  • Mean error: the difference between the average of several measurements and the true value (T).
  • Relative Error: absolute or mean error (E) expressed as a percentage (%) of the true value (T)
slide5

Example,1:

  • If a 2.62 g sample of material analyzed to be 2.52 g.
  • The absolute error (E) = 2.52 – 2.62 = -0.10 g.
  • Example,2:
  • In the titration of 10 ml of 0.1 N NaOH with laboratory prepared 0.1 N HCl, the true value is 9.9 ml, and we have:
  • 10.1, 9.9, 9.8, 10.2, 10.1 observed values,
  • So, Mean = summation of observed values / their number

= (10.1 + 9.9 + 9.8 + 10.2 + 10.1) / 5 = 10.02 ml.

  • And mean error = 10.02 – 9.9 = 0.12 ml.
slide6

In example,1:

Relative Error = (-0.10/2.62) x 100% = - 3.8 %

  • In example,2:

Relative Mean Error= ( 0.12/9.9) x 100% = 1.21%

slide7

Types of Errors:

  • (A) Determinate or systemic (constant) errors:
  • can be determined, (can be avoided)
  • (B) Indeterminate (random, accidental or chance) errors:

cannot be determined or corrected.

slide8

Accuracy:agreement of a measurement with the true value.

  • Determination of accuracy:

Absolute method

  • Accuracy is determined from the relative error;
  • In example,1: Relative Error = (-0.10/2.62) x 100% = -3.8 %
  • And accuracy = 100.0 – 3.8 = 96.2 %.
  • In example,2: Relative Mean Error = ( 0.12/9.9) x 100% = 1.21%
  • And accuracy = 100.00 – 1.21 = 98.79 %.
slide9

Example 3:

  • In practical exam of volumetric analysis, three students get the following results:
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Precision :The agreement between several measurements of the same substance.

  • Mean (X):It is the arithmetic average of all measured values.
  • The range (w): the"spread":

It is the difference between the highest measurement and the lowest one.

  • The median: It is the measurement in the middle of the arranged measurements where the numbers of higher and lower measurements are equal.
slide11

standard deviation (s):

  • Variance =The square of the standard deviation = S2
  • Relative standard deviation (RSD)
  • Coefficient of variation (C.V.) :
slide12

Example:

  • Analysis of a sample of iron ore gave the following % values:

7.08, 7.21, 7.12, 7.09, 7.16, 7.14, 7.07, 7.14, 7.18, 7.11.

  • Calculate the mean, standard deviation, the variance and coefficient of variation;
  • Find also the median and the range for these data.
slide14

Variance (S2) = 0.002

  • (b) The arranged data are:
  • 7.07, 7.08, 7.09, 7.11, 7.12, 7,14, 7,14, 7.16, 7.18, 7.21
  • The median is : (7.12 + 7.14) / 2 = 7.13
  • The range is : 7.21 – 7.07 = 0.14
rejection of a result the q test
Rejection of a result (The Q test):
  • The Q test is used to determine if an “outlier” is due to a determinate error or due to indeterminate error.
  • If it is due to a determinate error, it should be rejected.
  • If it is not due to a determinate error, then it falls within the expected random error and should be retained.
  • The ratio Q is calculated by arranging the data in decreasing order of numbers.
slide16

The difference (a) between the suspect number (the outlier) and its nearest neighbour number is divided by:

  • the range (w), which is the difference between the highest number and the lowest number,
slide17

The ratio is compared with the tabulated values of Q (see the Table).

  • If Q measured is equal or greater than the tabulated value, the suspected observation can be rejected.
  • If it is smaller than the tabulated value, the suspected observation is retained
slide18

Example:

  • The following set of chloride analysis on separate aliquots of serum were reported; 103, 106, 107 and 114 meq/L. one value appears suspect.
  • Determine if it may be rejected or not.
slide19

Answer:

  • The suspected result is 114 meq/L.
  • It differs from the nearest neighbor by (a) : 114 – 107 = 7 meq/L.
  • The range (w) is : 114 – 103 = 11 meq/L.
  • Therefore, Q = a/w= 7/11 = 0.64
  • The tabulated Q value (4 observations, 95% confidence level) is: 0.829
  • Since the calculated Q value is less than the tabulated Q value,
  • the suspected no. (114 meq/L) retained.
slide20

Significant figures

  • ‘digit’ = 0, 1, 2, ………..8,9
  • A significant figure = is a digit which denotes the amount of quantity in the place in which it stands.
  • The digit 0 is a significant figure except when it is the first figure in a number.
  • In 1.2680 g and 1.0062 g 5
  • the zero is significant,
  • but in the quantity 0.0025 kg 2
  • the zero is not significant, because 0.0025 kg = 2.5 g.
slide21

1 g means that it is between 0.9 and 1.1 g

  • 1.0 g means that it is between 0.99 and 1.01 g
  • 1.00 g means that it is between 0.999 and 1.001g
  • Take 10.0 ml of Zn2+ sample, add 10 ml of NH3-buffer
  • Weigh 1.000 g of powdered drug sample, add 2 g of hexamine reagent …..
  • 1 kg of tomato xxxxxxx 1.000 kg of gold !!!
  • volume which is known to be between 20.5 ml and 20.7 ml should be written as 20.6 ml; but not as 20.60.
  • 20.60 ml indicates that the value lies between 20.59 ml and 20.61 ml.
slide22

Rejection quotient, Q,

at different confidence levels