Q1. Give source wise classification of determinate error. Explain and illustrate each class?

Errors are classified into two types:

• Determinate errors (or systematic errors)

• Indeterminate errors (or random errors)

• DETERMINATE ERRORS

Those types of errors for which atleast in principle a definite cause of source can be observed are known as determinate type of error. With proper precaution, elimination of these type of error is possible. Determinate error are generally unidirectional. They can be of considerable magnitude. As a result these error can affect the measurement being made. Determinate error are reproducible and to a certain extent in some cases, predictable as well.

Types of determinate errors:-

• Instrumental errors : use of a measuring device or an instrument forms an integral part of the measurement. The instrumental error may arise due to the instrument itself or due to the effect of the environmental factors on the instrument

• Uncertainty in the last digit of a measurement: due to the least count of the instrument. Least count of an instrument is defined as the minimum value that it can measure each measuring device possesses a least count. Least count of an instrument introduces an uncertainty in the last digit of the measured value. This uncertainty is inherent and cannot be avoided.

• Poorly calibrated glassware and instruments.

• Improper response: every instrument is made to work under optimum conditions. For proper response of the instrument it is necessary. When optimum conditions are not maintained, the response may not be proper. It is assumed that under optimum conditions, the instrument behaves ideally.

With improper and non-ideal conditions, the possibility of errors creeping in cannot be ruled out.

• Methodic errors: Methodic errors are errors associated with a particular method. They are part and parcel of the method and hence cannot be eliminated but can be minimized. E.g

• Solubility of a salt: when a salt gets precipitated, the solution of the salt. Thus in gravimetric analysis when a metal ion is precipitated as a sparingly soluble salt its complete removal from the solution is not possible. This will lead to error.

• Addition of excess amount of the titrant: the volumetric analysis the end point and the equivalence point differ. The difference is known as titration error. The titration error will be the error associated with the titration method. The error will get eliminated if the end point and the equivalence point coincide. This will not be possible in each and every case. Hence attempt are made to minimise the titration error.

• Incomplete decomposition: it is an e.g. of an incomplete reaction. Many times the precipitate is obtained as one salt and weighed as some other salt. This involves conversion of one salt to another.

• Co-precipitation and post precipitation: in gravimetric analysis these are common errors. Post precipitation of one salt on the surface of the other. Whereas in co- precipitation another salt gets incorporated in the precipitate that is getting formed.

• Operational errors: Operational errors are the errors which are due to the analyst and not because of methods or procedures. The operational errors are mostly physical in nature and occurs when sound analytical technique is not followed. Hence these type of errors can be definitely eliminated. E.g.

weighing of the crucible before cooling, loss of precipitate during filtration, blowing the last drop in the nozzle of the pipette, improve recording of a measurement, under washing or over washing of precipitates.

• Personal errors: this error arises due to the physical limitations of the analyst. Colour blindness is an example of physical limitation of the analyst.

Magnitude wise classification of determinate errors:-

Magnitude wise determinate errors are classified as constant and proportionate errors

• Constant errors and additive errors

The constant errors are those in which absolute error is independent of sample size however relative error increase as the sample size is decreased.

• Proportionate errors

The proportionate errors are those in which absolute error increases in direct proportion to the sample size however relative error remain constant. Thus in proportionate error relative error remains constant irrespective of numerical values measured.

OR

Q2) explain the difference between absolute error and relative error/constant and proportionate error?

Absolute error Relative errorThe difference between the true value and the measured value with regards to the sign is the absolute error.

the absolute or mean expressed as a percentage of the true value is the relative error.

E= xi- xtWhere xi -> measured value Xt-> true value

R.E= Absolute error/ true valuei.e. R.E= xi-xt/ xt

Titration error in titrimetric analysis Presence of impurities in gravimetric analysis.

Constant error Proportionate errorThe absolute value of constant error is independent of sample size.

The absolute value of a proportionate error is dependent of sample size

Absolute error remains same but relative error increases as the sample size is decreased.

The relative error remains constant but absolute error increases as the sample size is increased

Absolute error Relative error

The absolute error is the magnitude of the differences between exact value and approximation.

The relative error is the absolute error divided by the magnitude of exact value.

Absolute values have the same units as the quantities measured.

Relative values are ratios, and have no units. The ratios are commonly expressed as fractions

Absolute error can defined when the true value is zero.

Relative error is undefined when the true value is zero.

Absolute Accuracy Error

Example: 25.13 mL - 25.00 mL = +0.13 mL absolute error.

Relative Accuracy Error

Example: (( 25.13 mL - 25.00 mL)/25.00 mL) x 100% = 0.52% relative error.

Example: For professional gravimetric chloride results we must have less than 0.2% relative error.

Explain the meaning of significance number:

The term significant figures refers to the number of important single digits (0 through 9 inclusive) in the coefficient of an expression in scientific notation . The number of significant figures in an expression indicates the confidence or precision with which an engineer or scientist states a quantity.

Significant figures are arrived at by rounding off an expression after a calculation is executed. . In any calculation, the number of significant figures in the solution must be equal to, or less than, the number of significant figures in the least precise

expression or element.

The significant figures of a number are those digits that carry meaning contributing to its precision.

Decimal expressionScientific notation Sig. figs.

1,222,000.00 1.222 x 10 6 4

1.22200000 x 10 6 9

0.00003450000 3.45 x 10 -5 3

3.450000 x 10 -5 7

-9,876,543,210 -9.87654 x 10 9 6

-9.876543210 x 10 910

-0.0000000100 -1 x 10 -8 1

-1.00 x 10 -8 3

Example : Consider the following product:

2.56 x 10 67 x -8.33 x 10 -54To obtain the product of these two numbers, the coefficients are multiplied, and the powers of 10 are added. This produces the following result:

2.56 x (-8.33) x 10 67+(-54)

= 2.56 x (-8.33) x 1067-54 = -21.3248 x 10 13The proper form of common scientific notation requires that the absolute value of the coefficient be larger than 1 and less than 10. Thus, the coefficient in the above expression should be divided by 10 and the power of 10 increased by one, giving:

-2.13248 x 10 14 because both multiplicands in the original product are specified to only three significant figures, a scientist or engineer will round off the final expression to three significant figures as well, yielding:

-2.13 x 10 14 as the product

Identifying significant figures:

Specifically, the rules for identifying significant figures when writing or interpreting numbers are as follows:

• All non-zero digits are considered significant. For example, 91 has two significant figures (9 and 1), while 123.45 has five significant figures (1, 2, 3, 4 and 5).

• Zeros appearing anywhere between two non-zero digits are significant. Example: 101.1203 has seven significant figures: 1, 0, 1, 1, 2, 0 and 3.

• Leading zeros are not significant. For example, 0.00052 has two significant figures: 5 and 2.

• Trailing zeros in a number containing a decimal point are significant. For example, 12.2300 has six significant figures: 1, 2, 2, 3, 0 and 0. The number 0.000122300 still has only six significant figures (the zeros before the 1 are not significant). In addition, 120.00 has five significant figures since it has three trailing zeros. This convention clarifies the precision of such numbers; for example, if a measurement precise to four decimal places (0.0001) is given as 12.23 then it might be understood that only two decimal places of precision are available. Stating the result as 12.2300 makes clear that it is precise to four decimal places (in this case, six significant figures).

• The significance of trailing zeros in a number not containing a decimal point can be ambiguous. For example, it may not always be clear if a number like 1300 is precise to the nearest unit (and just happens coincidentally to be an exact multiple of a hundred) or if it is only shown to the nearest hundred due to rounding or uncertainty. Various conventions exist to address this issue:

• A bar may be placed over the last significant figure; any trailing zeros following this are insignificant. For example, 1300 has three significant figures (and hence indicates that the number is precise to the nearest ten).

• The last significant figure of a number may be underlined; for example, "2000"

has two significant figures.

• A decimal point may be placed after the number; for example "100." indicates specifically that three significant figures are meant.

To round to n significant figures:

If the first non-significant figure is a 5 followed by other non-zero digits, round up the last significant figure (away from zero). For example, 1.2459 as the result of a calculation or measurement that only allows for 3 significant figures should be written 1.25.

If the first non-significant figure is a 5 not followed by any other digits or followed only by zeros, rounding requires a tie-breaking rule. For example, to round 1.25 to 2 significant figures:

Round half up (also known as "5/4") rounds up to 1.3. This is the default rounding method implied in many disciplines if not specified.

Round half to even, which rounds to the nearest even number, rounds down to 1.2 in this case. The same strategy applied to 1.35 would instead round up to 1.4.

Replace non-significant figures in front of the decimal by zeros.

Relationship to accuracy and precision in measurement:

Q.3 SIGNIFICANT FIGURES:All measurements are approximations—no measuring device can give perfect measurements without experimental uncertainty. By convention, a mass measured to 13.2 g is said to have an absolute uncertainty of plus or minus 0.1 g and is said to have been measured to the nearest 0.1 g. In other words, we are somewhat uncertain about that last digit—it could be a "2"; then again, it could be a "1" or a "3". A mass of 13.20 g indicates an absolute uncertainty of plus or minus 0.01 g.

WHAT IS A "SIGNIFICANT FIGURE"?

The number of significant figures in a result is simply the number of figures that are known with some degree of reliability. The number 13.2 is said to have 3 significant figures. The number 13.20 is said to have 4 significant figures.

RULES FOR DECIDING THE NUMBER OF SIGNIFICANT FIGURES IN A MEASURED QUANTITY:

(1) All nonzero digits are significant:

1.234 g has 4 significant figures,1.2 g has 2 significant figures.

(2) Zeroes between nonzero digits are significant:

1002 kg has 4 significant figures,3.07 mL has 3 significant figures.

(3) Leading zeros to the left of the first nonzero digits are not significant; such zeroes merely indicate the position of the decimal point:

0.001 oC has only 1 significant figure,0.012 g has 2 significant figures.

(4) Trailing zeroes that are also to the right of a decimal point in a number are significant:

0.0230 mL has 3 significant figures,0.20 g has 2 significant figures.

(5) When a number ends in zeroes that are not to the right of a decimal point, the zeroes are not necessarily significant:

190 miles may be 2 or 3 significant figures,50,600 calories may be 3, 4, or 5 significant figures.

The potential ambiguity in the last rule can be avoided by the use of standard exponential, or "scientific," notation. For example, depending on whether the number of significant figures is 3, 4, or 5, we would write 50,600 calories as:

5.06 × 104 calories (3 significant figures)5.060 × 104 calories (4 significant figures), or5.0600 × 104 calories (5 significant figures).

By writing a number in scientific notation, the number of significant figures is clearly indicated by the number of numerical figures in the 'digit' term as shown by these examples. This approach is a reasonable convention to follow.

WHAT IS AN "EXACT NUMBER"?

Some numbers are exact because they are known with complete certainty.Most exact numbers are integers: exactly 12 inches are in a foot, there might be exactly 23 students in a class. Exact numbers are often found as conversion factors or as counts of objects.

Exact numbers can be considered to have an infinite number of significant figures. Thus, the number of apparent significant figures in any exact number can be ignored as a limiting factor in determining the number of significant figures in the result of a calculation.

RULES FOR MATHEMATICAL OPERATIONS:

In carrying out calculations, the general rule is that the accuracy of a calculated result is limited by the least accurate measurement involved in the calculation.

(1) In addition and subtraction, the result is rounded off to the last common digit occurring furthest to the right in all components. Another way to state this rule is as follows: in addition and subtraction, the result is rounded off so that it has the same number of digits as the measurement having the fewest decimal places (counting from left to right). For example,

100 (assume 3 significant figures) + 23.643 (5 significant figures) = 123.643,

which should be rounded to 124 (3 significant figures). Note, however, that it is possible two numbers have no common digits (significant figures in the same digit column).

(2) In multiplication and division, the result should be rounded off so as to have the same number of significant figures as in the component with the least number of significant figures. For example,

3.0 (2 significant figures ) × 12.60 (4 significant figures) = 37.8000

which should be rounded to 38 (2 significant figures).

RULES FOR ROUNDING OFF NUMBERS:

(1) If the digit to be dropped is greater than 5, the last retained digit is increased by one. For example,12.6 is rounded to 13.

(2) If the digit to be dropped is less than 5, the last remaining digit is left as it is. For example,12.4 is rounded to 12.

(3) If the digit to be dropped is 5, and if any digit following it is not zero, the last remaining digit is increased by one. For example,12.51 is rounded to 13.

(4) If the digit to be dropped is 5 and is followed only by zeroes, the last remaining digit is increased by one if it is odd, but left as it is if even. For example,11.5 is rounded to 12, 12.5 is rounded to 12.This rule means that if the digit to be dropped is 5 followed only by zeroes, the result is always rounded to the even digit. The rationale for this rule is to avoid bias in rounding: half of the time we round up, half the time we round down.

GENERAL GUIDELINES FOR USING CALCULATORS:

When using a calculator, if you work the entirety of a long calculation without writing down any intermediate results, you may not be able to tell if an error is made. Further, even if you realize that one has occurred, you may not be able to tell where the error is.

In a long calculation involving mixed operations, carry as many digits as possible through the entire set of calculations and then round the final result appropriately. For example,

(5.00 / 1.235) + 3.000 + (6.35 / 4.0)=4.04858... + 3.000 + 1.5875=8.630829...

The first division should result in 3 significant figures. The last division should result in 2 significant figures. The three numbers added together should result in a number that is rounded off to the last common significant digit occurring furthest to the right; in this case, the final result should be rounded with 1 digit after the decimal. Thus, the correct rounded final result should be 8.6. This final result has been limited by the accuracy in the last division.

Most modern calcualtors allow you to carry all the results of intermediate calcuations in the display when performing a complex series of calcuations. By doing this, you can retain the results of each individual calculation step, and avoid having to re-enter intermediate results (a practice that may encourage rounding too soon). In this manner, you can completely avoid truncation errors introduced by rounding intermediate results.

Warning: carrying all digits through to the final result before rounding is critical for many mathematical operations in statistics. Rounding intermediate results when calculating sums of squares can seriously compromise the accuracy of the result.

SAMPLE PROBLEMS ON SIGNIFICANT FIGURES:

Instructions: print a copy of this page and work the problems. When you are ready to check your answers, go to the next page.

1.    37.76 + 3.907 + 226.4 = ?

2.    319.15 - 32.614 = ?

3.    104.630 + 27.08362 + 0.61 = ?

ANSWER KEY TO SAMPLE PROBLEMS ON SIGNIFICANT FIGURES:

 1.    37.76 + 3.907 + 226.4 = 268.1

 2.    319.15 - 32.614 = 286.54

 3.    104.630 + 27.08362 + 0.61 = 132.32

Q.4 STANDARD DEVIATION:Definition:It is defined as the square root of the means of square of individual deviation.It is denoted by ‘S’.

Where ‘S’ is the standard deviation for observation less than 20.When number of observations are more than 20, the term σ (sigma) is used and it is defined as

OR The Standard Deviation is a measure of how spread out numbers are.

STANDARD DEVIATION:

1-The Standard Deviation is a better way to measure variation. First, we look at

the difference between each data value and the mean:. Then, to make sure all the distances are positive, we square that difference:. Next, we add up these differences for all of the observations: . Finally, we divide by (n –1) and take the square root to in some way undo the squaring from before. This gives the formula: . Standard Deviation is denoted s, and it is in the same units as the observations in the data set.

2-The Variance is the square of the standard deviation, s2. Since its units are in squared units of the original observations, its value is harder to interpret than standard deviation. As a result, it is not used as much.

ExampleConsider the data set 1 2 3 4 5. We see that . Let’s find the range, standard deviation, and variance.Range = Maximum – Minimum = 5 – 1 = 4.To find standard deviation, we can use a table to calculate each part separately.

xi We see that , so:1 1 – 3 = -2 42 2 – 3 = -1 13 3 – 3 = 0 04 4 – 3 = 1 15 5 – 3 = 2 4

Sum --- --- 10

Finally, the variance is .Now, consider the data set 1 1 3 5 5. We see that again. Let’s find the range, standard deviation, and variance.Range = Maximum – Minimum = 5 – 1 = 4.To find standard deviation, we can use a table to calculate each part separately.

xi We see that , so:1 1 – 3 = -2 42 1 – 3 = -2 43 3 – 3 = 0 0

4 5 – 3 = 2 45 5 – 3 = 2 4

Sum --- --- 16

In this data set, the variance is . So we see that the second data set has more variation than the first one, which makes sense. The observations are spread further away from the mean than in the first set of data.

Some NotationIn a sample, the standard deviation is denoted s and the variance s2.In a population, we call the standard deviation σ (sigma) and the variance σ2 (sigma squared).Formulas

Here are the two formulas, explained at Standard Deviation Formulas  if you want to know more:

The "Population Standard Deviation":  

The "Sample Standard Deviation":

 

Looks complicated, but the important change is to divide by N-1 (instead of N) when calculating a Sample Variance.

A quick note about the standard deviation formula: Often, people wonder why we divide by n – 1 in the formula instead of n, which is the number of observations. The reason is because of something called degrees of freedom. We already need to know the meanfor the standard deviation formula. Therefore, if we know only n – 1 of the observations, we could figure out the last one, since the mean tells you the sum of the observations. As a result, it turns out that dividing by n – 1 will make the standard deviation an unbiased estimator, meaning that as the sample size increases, s will not consistently overestimate or underestimate the true population

standard deviation.

The number of significant figures roughly corresponds to precision, not accuracy

SEEMA TAMBE

1.EXPLAIN HOW STANDARD DEVIATION (S) IS CALCULATED?

Ans: The standard deviation (S) is defined as the sum of squares of deviations of individual measurement from the mean of the set is divided by the number of the degree of freedom and then the square root of this term is obtained

Standard deviation (S) is expressed as

Standard deviation (S)can be estimated from spread or range (w)of the set by using equation

S=w/d

Where d is statistical factor

S=Standard deviation can be estimated by using equation

S=W/√N

Where N is Total number of the measurement

2.Q-Test

Ans:The Q-test is one of the most statistically correct criteria used for rejecton of doubtfull data

Scope:It doesnot require large no of measurement in the set

Applications:First arrange all the measured value of the set in their increasing order

Find the spread or range of set thus arranged

Find difference between doubtfull measurement and its neighbouring measurement

sDivide these difference by spread of the set

This gives Q’s calculation:

T test

A statistical examination of two population means. A two-sample t-test examines whether two samples are different and is commonly used when the variances of two normal distributions are unknown and when an experiment uses a small sample size. For example, a t-test could be used to compare the average floor routine score of the U.S. women's Olympic gymnastic team to the average floor routine score of China's women's team.

Chi 2 test

Pearson's chi-squared test is used to assess two types of comparison: tests of goodness of fit and tests of independence.

• A test of goodness of fit establishes whether or not an observed frequency distribution differs from a theoretical distribution.

• A test of independence assesses whether paired observations on two variables, expressed in a contingency table, are independent of each other (e.g. polling responses from people of different nationalities to see if one's nationality is related to the response).

The procedure of the test includes the following steps:

• Calculate the chi-squared test statistic, , which resembles a normalized sum of squared deviations between observed and theoretical frequencies (see below).

• Determine the degrees of freedom, df, of that statistic, which is essentially the number of frequencies reduced by the number of parameters of the fitted distribution.

• Compare to the critical value from the chi-squared distribution with df degrees of freedom, which in many cases gives a good approximation of the distribution of .

Que> In replicate analysis of a metal ore the following results are obtained

sample 1 2 3 4 5 6% of metals

40.13 40.11 40.14 40.10 40.18 40.16

Calculate mean, average deviation ,and standard deviation

Ans>

Mean=Σxi/n

=40.13+40.11+40.14+40.10+40.18+40.16/6

=40.13

Obr. X XT (X-XT ) (X-XT )²

1 40.10 -O.03 0.0009 2 40.11 -O.O2 0.0004 3 40.13 40.13 0 0 4 40.14 O.O1 0.0001 5 40.16 O.O3 0.0009 6 40.18 0.05 0.0025

O.14 0.0048

Average deviation =Σdi/n =0.14/6=0.0233

Std deviation=√Σ(x-xT )²/n-1

=√0.0048/6-1=0.03098

Que> in this six determination of ion from samples, each of which contain 320 miligram of ion. The following results were obtained.

Sample no

1 2 3 4 5 6

Fe(found in mg)

120 138 158 168 162 178

Mean=Σxi/n

=120+138+158+168+162+178/6

=154

Median= Arrange in ascending order

120,138,158,162,168,178

=158+162/2 =160

Obr no x xֿ (x-xֿ) (x-xֿ)²

1 120 -34 11562 138 -16 2563 158 154 4 164 162 8 645 168 14 1966 178 24 576

100 2264

Average deviation=Σdi/n

100/6=16.66

Standard deviation=√Σ(x-x ֿ)²/n-1

=√2264/√6-1

=47.581/2.236

=21.279

Range= Xmax-Xmin

=178-120

=58

The normality of solution is determined by four separate titration .The result being 0.1541, 0.1549, 0.1539, 0.1543 . Calculate mean,median ,range,average deviation,relative average deviation from mean,standard deviation,variance,coefficient of variation. CALCULATION: 1. MEAN(x¯) =(0.1541+0.1549+0.1539+0.1543 )/4= 0.6172/4=0.1543 2. By arranging the observed value in ascending order we get, 0.1539, 0.1541, 0.1543, 0.1549 Since there are even number of values therefore ,the meadian is the mean of the two middle values MEADIAN(m¯) = (0.1541+0.1543)/2= 0.1542

3.RANGE = 0.1549 - 0.1539 = 0.0010

4.DEVIATION FROM MEAN = (Observed value -mean) d1 = |0.1541 - 0.1543| = 0.0002 d2 = |0.1549 -0.1543| = 0.0006 d3 = |0.1539 -0.1543| = 0.0004 d4 =|0.1543 - 0.1543| = 0.0000

5.AVERAGE DEVIATION FROM MEAN =(0.0002 + 0.0006 + 0.0004 + 0.0000) /4 = 0.0012 /4 = 0.0003

4.RELATIVE AVERAGE DEVIATION FROM MEAN = ( Average deviaton from mean)/Mean = 0.0003/ 0.1543 = 0.00194

5.DEVIATION FROM MEADIAN = (Observed value -meadian ) d1 =| 0.1541 - 0.1542| = 0.0001 d2 =|0.1549 - 0.1542|= 0.0007 d3 =|0.1539 - 0.1542|= 0.0003 d4 =|0.1543 - 0.1542|= 0.0001

6.AVERAGE DEVIATION FROM MEADIAN = (0.0001+0.0007+0.0003+0.0001) /4 = 0.0012 / 4 = 0.0003

7.RELATIVE AVERAGE DEVIATION FROM MEADIAN = (Average deviation from meadian) /Meadian = 0.0003 / 0.1542 = 0.00194

8.STANDARD DEVIATION i=4 Ʃ=(x - x¯)2 =(0.0012)2 = 0.00000144

i=1 =√0.00000144 /(N -1)

=√0.00000144/(4-1) =√0.00000144/3

=√0.00000048

= ±6.9282*10-4

9. VARIANCE = ( S)2

= (6.9282*10-4)2

= 47.9999*10-8

10.COEFFICIENT OF VARIATION = (S / X¯) *100 = (6.9282*10-4 / 0.1543)*100 = (44.9008*10-4)*100 =0.4490 %