Unit3
Statistical techniques1
Professor Bowley defines the average as
“Statistical constants which enable us to comprehend in a single effort the significance of the whole”
An average is a single value that is the best representative for a given data set.
Measures of central tendency show the tendency of some central values around which data tend to cluster.
The following are the various measures of central tendency
1. Arithmetic mean
2. Median
3. Mode
4. Weighted mean
5. Geometric mean
6. Harmonic mean
The arithmetic mean or mean
The arithmetic mean is a value which is the sum of all observation divided by a total number of observations of the given data set.
If there are n numbers in a dataset then the arithmetic mean will be
If the numbers along with frequencies are given then mean can be defined as
Example1: Find the mean of 26, 15, 29, 36, 35, 30, 14, 21, 25 .
Sol.
Example2: Find the mean of the following dataset.
x  20  30  40 
f  5  6  4 
Sol.
We have the following table
x  f  Fx 
20  5  100 
30  6  180 
40  7  160 
 Sum = 15  Sum = 440 
Then Mean will be
The direct method to find mean
Example: Find the arithmetic mean of the following dataset
Sol.
We have the following distribution
Class interval  Mid value (x)  Frequency (f)  Fx 
010  05  3  15 
1020  15  5  75 
2030  25  7  175 
3040  35  9  315 
4050  45  4  180 

 Sum = 28  Sum = 760 
Short cut method to find mean
Suppose ‘a’ is assumed mean, and ‘d’ is the deviation of the variate x form a, then
Example: Find the arithmetic mean of the following dataset.
Class  010  1020  2030  3040  4050 
Frequency  7  8  20  10  5 
Sol.
Let the assumed mean (a) = 25,
Class  Midvalue  Frequency  x – 25 = d  fd 
010  5  7  20  140 
1020  15  8  10  80 
2030  25  20  0  0 
3040  35  10  10  100 
4050  45  5  20  100 
Total 
 50 
 20 
Step deviation method for mean
Where
Median
Median is the midvalue of the given data when it is arranged in ascending or descending order.
1. If the total number of values in the data set is odd then the median is the value of item.
NoteThe data should be arranged in ascending r descending order
2. If the total number of values in the data set is even then the median is the mean of the item.
Example: Find the median of the data given below
7, 8, 9, 3, 4, 10
Sol.
Arrange the data in ascending order
3, 4, 7, 8, 9, 10
So there total 6 (even) observations, then
Median for grouped data
Here,
Example: Find the median of the following dataset
Sol.
Class interval  Frequency  Cumulative frequency 
0  10  3  3 
10 – 20  5  8 
20 – 30  7  15 
30 – 40  9  24 
40 – 50  4  28 
So that median class is 2030.
Now putting the values in the formula
So that the median is 28.57
Mode
A value in the data which is most frequent is known as a mode.
Example: Find the mode of the following data points
Sol. Here 6 has the highest frequency so that the mode is 6.
Mode for grouped data
Here,
Example: Find the mode of the following dataset
Sol.
Class interval  Frequency 
0  10  3 
10 – 20  5 
20 – 30  7 
30 – 40  9 
40 – 50  4 
Here the highest frequency is 9. So that the modal class is 4050,
Put the values in the given data
Hence the mode is 42.86
Note
Mean – Mode = [Mean  Median]
Geometric Mean
If are the values of the data, then the geometric mean
Harmonic mean
The harmonic mean is the reciprocal of the arithmetic mean
It can be defined as
Note
1.
2.
The rth moment of a variable x about the mean x is usually denoted by is given by
The rth moment of a variable x about any point a is defined by
The relation between moments about mean and moment about any point:
where and
In particular
Note. 1. The sum of the coefficients of the various terms on the righthand side is zero.
2. The dimension of each term on the right‐hand side is the same as that of terms on the left.
MOMENT GENERATING FUNCTION
The moment generating function of the variate about is defined as the expected value of and is denoted .
where , ‘ is the moment of order about
Hence coefficient of or
again )
Thus the moment generating function about the point moment generating function about the origin.
Skewness
The word skewness means lack of symmetry
The examples of the symmetric curve, positively skewed, and negatively skewed curves are given as follows
1. Symmetric curve
2. Positively skewed
3. Negatively skewed
To measure the skewness we use Karl Pearson’s coefficient of skewness.
Then the formula is as follows
Note the value of Karl Pearson’s coefficient of skewness lies between 1 to +1.
Kurtosis
It is the measurement of the degree of peakedness of a distribution
Kurtosis is measured as
Calculation of kurtosis
The second and fourth central moments are used to measure kurtosis.
We use Karl Pearson’s formula to calculate kurtosis
Now, three conditions arise
1. If , then the curve is mesokurtic.
2. If , then the curve is platykurtic
3. if , then the curve is said to be leptokurtic.
Example: If the coefficient of skewness is 0.64. The standard deviation is 13 and mean is 59.2, then find the mode and median.
Sol.
We know that
So that
And we also know that
Example: Calculate Karl Pearson’s coefficient of skewness of marks obtained by 150 students.
Sol. The mode is not well defined so that first we calculate mean and median
Class  f  x  CF  fd  
010  10  5  10  3  30  90 
1020  40  15  50  2  80  160 
2030  20  25  70  1  20  20 
3040  0  35  70  0  0  0 
4050  10  45  80  1  10  10 
5060  40  55  120  2  80  160 
6070  16  65  136  3  48  144 
7080  14  75  150  4  56  244 
Now,
And
Standard deviation
Then
Example. The first four moments about the working mean 28.5 of distribution are 0.2 94, 7.1 44, 42.409, and 454.98. Calculate the moments about the mean. Also, evaluate and comment upon the skewness and kurtosis of the distribution.
Solution. The first four moments about the arbitrary origin 28.5 are
, which indicates considerable skewness of the distribution.
, which shows that the distribution is leptokurtic.
Example. Calculate the median, quartiles, and the quartile coefficient of skewness from the following data:
Weight (lbs)  7080  8090  90100  100110  110120  120130  130140  140=150 
No. of persons  12  18  35  42  50  45  20  8 
Solution. Here total frequency
The cumulative frequency table is
Weight (lbs)  7080  8090  90100  100110  110120  120130  130140  140=150 
Frequency  12  18  35  42  50  45  20  8 
Cumulative Frequency  12  30  65  107  157  202  222  230 
Now, N/2 =230/2= 115th item which lies in the 110 – 120 group.
Median or
Also, is 57.5th or 58th item which lies in the 90100 group.
Similarly 3N/4 = 172.5 i.e. is 173rd item which lies in the 120130 group.
Hence quartile coefficient of skewness =
Method of Least Squares
Let (1)
Be the straight line to be fitted to the given data points
Let be the theoretical value for
Then,
For S to be minimum
On simplification equation (2) and (3) becomes
The equation (3) and (4) are known as Normal equations.
On solving ( 3) and (4) we get the values of a and b
(b)To fit the parabola
The normal equations are
On solving three normal equations we get the values of a,b and c.
Example. Find the best values of a and b so that y = a + bx fits the data given in the table
x  0  1  2  3  4 
y  1.0  2.9  4.8  6.7  8.6 
Solution.
y = a + bx
x  y  xy  
0  1.0  0  0 
1  2.9  2.0  1 
2  4.8  9.6  4 
3  6.7  20.1  9 
4  8.6  13.4  16 
x = 10  y ,= 24.0  xy = 67.0 
Normal equations, y= na+ bx (2)
On putting the values of
On solving (4) and (5) we get,
On substituting the values of a and b in (1) we get
Example. Find the leastsquares approximation of second degree for the discrete data
X  2  1  0  1  2 
Y  15  1  1  3  19 
Solution. Let the equation of seconddegree polynomial be
X  y  xy  
2  15  30  4  60  8  16 
1  1  1  1  1  1  1 
0  1  0  0  0  0  0 
1  3  3  1  3  1  1 
2  19  38  4  76  8  16 
x=0  y=39  xy=10 
Normal equations are
On putting the values of x, y, xy, have
On solving (5),(6),(7), we get,
The required polynomial of the second degree is
Example: Find the straight line that best fits the following data by using the method of least square.
X  1  2  3  4  5 
y  14  27  40  55  68 
Sol.
Suppose the straight line
y = a + bx…….. (1)
Fits the best
Then
x  y  xy  
1  14  14  1 
2  27  54  4 
3  40  120  9 
4  55  220  16 
5  68  340  25 
Sum = 15  204  748  55 
Normal equations are
Put the values from the table, we get two normal equations
On solving the above equations, we get
So that the best fit line will be (on putting the values of a and b in equation (1))
Seconddegree parabolas and more general curves
To fit the seconddegree parabola
The normal equations will be
Note Change of scale
We change the scale if the data is large and given at equal intervals.
As
Example: Fit the seconddegree parabola of the following data by using the method of least squares.
X  1929  1930  1931  1932  1933  1934  1935  1936  1937 
Y  352  356  357  358  360  361  361  360  359 
Sol.
By taking u = x – 1933 and v = y – 357
Then equation becomes
Putting the values from the table in normal equations
We get
11 = 3A + 0B + 60C or 11 = 9A + 60C
51 = 0A + 60B + 0C or B = 17 / 20
9 = 60A + 0B + 708C or 9 = 60A + 708C
On solving, we get
On solving the above equation, we get
Example: Fit the curve by using the method of least square.
X  1  2  3  4  5  6 
Y  7.209  5.265  3.846  2.809  2.052  1.499 
Sol.
Here
Now put
Then we get
x  Y  xY  
1  7.209  1.97533  1.97533  1 
2  5.265  1.66108  3.32216  4 
3  3.846  1.34703  4.04109  9 
4  2.809  1.03283  4.13132  16 
5  2.052  0.71881  3.59405  25 
6  1.499  0.40480  2.4288  36 
Sum = 21 
 7.13988  19.49275  91 
Normal equations are
Putting the values form the table, we get
7.13988 = 6c + 21b
19.49275 = 21c + 91b
On solving, we get
b = 0.3141 and c = 2.28933
c =
Now put these values in equations (1), we get
When two variables are related in such a way that a change in the value of one variable affects the value of the other variable, then these two variables are said to be correlated and there is a correlation between two variables.
Example Height and weight of the persons of a group.
The correlation is said to be a perfect correlation if two variables vary in such a way that their ratio is constant always.
Scatter diagram
Karl Pearson’s coefficient of correlation
Here and
Note
1. Correlation coefficient always lies between 1 and +1.
2. Correlation coefficient is independent of the change of origin and scale.
3. If the two variables are independent then the correlation coefficient between them is zero.
Correlation coefficient  Type of correlation 
+1  Perfect positive correlation 
1  Perfect negative correlation 
0.25  Weak positive correlation 
0.75  Strong positive correlation 
0.25  Weak negative correlation 
0.75  Strong negative correlation 
0  No correlation 
Example: Find the correlation coefficient between age and weight of the following data
Age  30  44  45  43  34  44 
Weight  56  55  60  64  62  63 
Sol.
X  y  ())  
30  56  10  100  4  16  40 
44  55  4  16  5  25  20 
45  60  5  25  0  0  0 
43  64  3  9  4  16  12 
34  62  6  36  2  4  12 
44  63  4  16  3  9  12 
Sum= 240 
360 
0 
202 
0 
70

32 
Karl Pearson’s coefficient of correlation
Here the correlation coefficient is 0.27.which is the positive correlation (weak positive correlation), this indicates that as age increases, the weight also increases.
Shortcut method to calculate correlation coefficient
Here,
Example: Find the correlation coefficient between the values X and Y of the dataset given below by using the shortcut method
X  10  20  30  40  50 
Y  90  85  80  60  45 
Sol.
X  Y  
10  90  20  400  20  400  400 
20  85  10  100  15  225  150 
30  80  0  0  10  100  0 
40  60  10  100  10  100  100 
50  45  20  400  25  625  500 
Sum = 150 
360 
0 
1000 
10 
1450 
1150 
Shortcut method to calculate correlation coefficient
Example. Psychological tests of intelligence and Engineering ability were applied to 10 students. Here is a record of ungrouped data showing intelligence ratio (I.R) and Engineering ratio (E.R). Calculate the coefficient of correlation.
Student  A  B  C  D  E  F  G  H  I  J 
I.R.  105  104  102  101  100  99  98  96  93  92 
E.R.  101  103  100  98  95  96  104  92  97  94 
Solution. We construct the following table
Student  Intelligence ratio x  Engineering ratio y y  XY  
A  105 6  101 3  36  9  18 
B  104 5  103 5  25  25  25 
C  102 3  100 2  9  4  6 
D  101 2  98 0  4  0  0 
E  100 1  95 3  1  9  3 
F  99 0  96  2  0  4  0 
G  98 1  104 6  1  36  6 
H  96 3  92 6  9  36  18 
I  93 6  97 1  36  1  6 
J  92 7  94 4  49  16  28 
Total  990 0  980 0  170  140  92 
From this table, the mean of x, i.e. and mean of y, i.e.
Substituting these value in the formula (1)p.744 we have
Example. The correlation table given below shows that the ages of husband and wife of 53 married couples living together on the census night of 1991. Calculate the coefficient of correlation between the age of the husband and that of the wife.
Age of husband  Age of wife  Total  
1525  2535  3545  4555  5565  6575  
1525  1  1          2  
2535  2  12  1        15  
3545    4  10  1      15  
4555      3  6  1    10  
5565        2  4  2  8  
6575          1  2  3  
Total  3  17  14  9  6  4  53  
Solution.
Age of husband  Age of wife x series  Suppose  
1525  2535  3545  4555  5565  6575 
Total f  
Years  Midpoint x  20  30  40  50  60  70  
Age group  Midpoint y 

 20  10  0  10  20  30  
2  1  0  1  2  3  
1525  20  20  2  4 1  2 1 



 2  4  8  6 
2535  30  10  1  4 2  12 12  0 1 


 15  15  15  16 
3545  40  0  0 
 0 4  0 10  0 1 

 15  0  0  0 
4555  50 



 0 3  6 6  2 1 
 10  10  10  8 
5565  60 




 4 2  16 4  12 2  8  16  32  32 
6575  70 





 6 1  18 2  3  9  27  24 
Total f  3  17  14  9  6  4  53 = n  16  92  86  
6  17  0  9  12  12  10  Thick figures in small sqs. for Check: From both sides  
12  17  0  9  24  36  98  
8  14  0  10  24  30  86 
With the help of the above correlation table, we have
Spearman’s rank correlation
When the ranks are given instead of the scores, then we use Spearman’s rank correlation to find out the correlation between the variables.
Spearman’s rank correlation coefficient can be defined as
Example: Compute the Spearman’s rank correlation coefficient of the dataset given below
Person  A  B  C  D  E  F  G  H  I  J 
Rank in test1  9  10  6  5  7  2  4  8  1  3 
Rank in test2  1  2  3  4  5  6  7  8  9  10 
Sol.
Person  Rank in test1  Rank in test2  d =  
A  9  1  8  64 
B  10  2  8  64 
C  6  3  3  9 
D  5  4  1  1 
E  7  5  2  4 
F  2  6  4  16 
G  4  7  3  9 
H  8  8  0  0 
I  1  9  8  64 
J  3  10  7  49 
Sum 


 280 
Example. Ten participants in a contest are ranked by two judges as follows:
x  1  6  5  10  3  2  4  9  7  8 
y  6  4  9  8  1  2  3  10  5  7 
Calculate the rank correlation coefficient
Solution. If
Hence,
Example. Three judges A, B, C give the following ranks. Find which pair of judges has a common approach
A  1  6  5  10  3  2  4  9  7  8 
B  3  5  8  4  7  10  2  1  6  9 
C  6  4  9  8  1  2  3  10  5  7 
Solution. Here n = 10
A (=x)  Ranks by B(=y)  C (=z)  xy  y  z  zx 
 
1  3  6  2  3  5  4  9  25 
6  5  4  1  1  2  1  1  4 
5  8  9  3  1  4  9  1  16 
10  4  8  6  4  2  36  16  4 
3  7  1  4  6  2  16  36  4 
2  10  2  8  8  0  64  64  0 
4  2  3  2  1  1  4  1  1 
9  1  10  8  9  1  64  81  1 
7  6  5  1  1  2  1  1  4 
8  9  7  1  2  1  1  4  1 
Total 

 0  0  0  200  214  60 
Since is maximum, the pair of judge A and C have the nearest common approach.
Regression
Regression is the measure of the average relationship between the independent and dependent variable
Regression can be used for two or more than two variables.
There are two types of variables in regression analysis.
1. Independent variable
2. Dependent variable
The variable which is used for prediction is called the independent variable.
It is known as a predictor or regressor.
The variable whose value is predicted by an independent variable is called the dependent variable or regressed or explained variable.
The scatter diagram shows the relationship between the independent and dependent variable, then the scatter diagram will be more or less concentrated round a curve, which is called the curve of regression.
When we find the curve as a straight line then it is known as the line of regression and the regression is called linear regression.
Note regression line is the best fit line that expresses the average relation between variables.
Equation of the line of regression
Let
y = a + bx ………….. (1)
is the equation of the line of y on x.
Let be the estimated value of for the given value of .
So that, According to the principle of least squares, we have the determined ‘a’ and ‘b’ so that the sum of squares of deviations of observed values of y from expected values of y,
That means
Or
…….. (2)
Is the minimum.
Form the concept of maxima and minima, we partially differentiate U with respect to ‘a’ and ‘b’ and equate to zero.
Which means
And
These equations (3) and (4) are known as the normal equation for a straight line.
Now divide equation (3) by n, we get
This indicates that the regression line of y on x passes through the point
.
We know that
The variance of variable x can be expressed as
Dividing equation (4) by n, we get
From equation (6), (7), and (8)
Multiply (5) by, we get
Subtracting equation (10) from equation (9), we get
Since ‘b’ is the slope of the line of regression y on x and the line of regression passes through the point (), so that the equation of the line of regression of y on x is
This is known as the regression line of y on x.
Note
are the coefficients of regression.
2.
Example: Two variables X and Y are given in the dataset below, find the two lines of regression.
x  65  66  67  67  68  69  70  71 
y  66  68  65  69  74  73  72  70 
Sol.
The two lines of regression can be expressed as
And
x  y  xy  
65  66  4225  4356  4290 
66  68  4356  4624  4488 
67  65  4489  4225  4355 
67  69  4489  4761  4623 
68  74  4624  5476  5032 
69  73  4761  5329  5037 
70  72  4900  5184  5040 
71  70  5041  4900  4970 
Sum = 543  557  36885  38855  37835 
Now
And
The standard deviation of x
Similarly
Correlation coefficient
Put these values in the regression line equation, we get
Regression line y on x
Regression line x on y
A regression line can also be found by the following method
Example: Find the regression line of y on x for the given dataset.
X  4.3  4.5  5.9  5.6  6.1  5.2  3.8  2.1 
Y  12.6  12.1  11.6  11.8  11.4  11.8  13.2  14.1 
Sol.
Let y = a + bx is the line of regression of y on x, where ‘a’ and ‘b’ are given as
We will make the following table
x  Y  Xy  
4.3  12.6  54.18  18.49 
4.5  12.1  54.45  20.25 
5.9  11.6  68.44  34.81 
5.6  11.8  66.08  31.36 
6.1  11.4  69.54  37.21 
5.2  11.8  61.36  27.04 
3.8  13.2  50.16  14.44 
2.1  14.1  29.61  4.41 
Sum = 37.5  98.6  453.82  188.01 
Using the above equations we get
On solving these both equations, we get
a = 15.49 and b = 0.675
So that the regression line is –
y = 15.49 – 0.675x
Note – Standard error of predictions can be found by the formula given below
Difference between regression and correlation
1. Correlation is the linear relationship between two variables while regression is the average relationship between two or more variables.
2. There are only limited applications of correlation as it gives the strength of the linear relationship while the regression is to predict the value of the dependent variable for the given values of independent variables.
3. Correlation does not consider dependent and independent variables while regression considers one dependent variable and other independent variables.
References
1. Erwin Kreyszig, Advanced Engineering Mathematics, 9thEdition, John Wiley & Sons, 2006.
2. N.P. Bali and Manish Goyal, A textbook of Engineering Mathematics, Laxmi Publications.
3. P. G. Hoel, S. C. Port, and C. J. Stone, Introduction to Probability Theory, Universal Book Stall.
4. S. Ross, A First Course in Probability, 6th Ed., Pearson Education India,2002.