Unit - 4

Statistical techniques

Q1) Define positive correlation.

A1)

Correlation between two variables is said to be positive if the values of the variables deviate in the same direction i.e. if the values of one variable increase (or decrease) then the values of other variable also increase (or decrease). For example:

1. Heights and weights of group of persons;

2. House hold income and expenditure;

3. Amount of rainfall and yield of crops

Q2) What is Karl Pearson’s coefficient of correlation?

A2)

Coefficient of correlation measures the intensity or degree of linear relationship between two variables. It was given by British Biometrician Karl Pearson (1867-1936).

Karl Pearson’s Coefficient of Correlation is widely used mathematical method is used to calculate the degree and direction of the relationship between linear related variables. The coefficient of correlation is denoted by “r”.

If X and Y are two random variables then correlation coefficient between X and Y is denoted by r and defined as-

Karl Pearson’s coefficient of correlation-

Here- and

Q3) Find the correlation coefficient between Age and weight of the following data-

A3)

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 the as age increases, the weight also increase.

Q4) Psychological tests of intelligence and of 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 co-efficient of correlation.

A4)

We construct the following table:

Student	Intelligence ratio x	Engineering ratio y	X2	Y2	XY
A B C D E F G H I J	100                      6     104                     5     102                     3     101                     2     100                     1       99                     0       98                    -1       96                    -3       93                    -6       92                    -7	101                 3   103                 5   100                 2    98                  0    95                 -3    96                 -2   104                 6    92                 -6    97                 -1    94                 -4	36 25 9 4 1 0 1 9 36 49	9 25 4 0 9 4 36 36 1 16	18 25 6 0 -3 0 -6 18 6 28
Total	990                    0	980                 0	170	140	92

From this table, mean of i.e., and mean of , i.e.,

Substituting these values in the formula (1), we have

Q5) Three judges A, B, C, give the following ranks. Find which pair of judges has common approach

A5)

Here

	Ranks by
1 6 5 10 3 2 4 9 7 8	3 5 8 4 7 10 2 1 6 9	6 4 9 8 1 2 3 10 5 7	-2 1 -3 6 -4 -8 2 8 1 -1	-3 1 -1 -4 6 8 -1 -9 1 2	5 -2 4 -2 -2 0 -1 1 -2 -1	4 1 9 36 16 64 4 64 1 1	9 1 1 16 36 64 1 81 1 4	25 4 16 4 4 0 1 1 4 1
Total			0	0	0	200	214	60

Since is maximum, the pair of judges A and C have the nearest common approach.

Q6) Compute the Spearman’s rank correlation coefficient of the dataset given below-

A6)

Person	Rank in test-1	Rank in test-2	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

Q7) Calculate Spearman rank-order correlation

A7)

Rank by taking the highest value or the lowest value as 1.

Here, highest value is taken as 1

R = 0.67

Therefore, this indicates a strong positive relationship between the ranks individuals obtained in the math and English exam.

Q8) Explain regression.

A8)

If the scatter diagram indicates some relationship between two variables and , then the dots of the scatter diagram will be concentrated round a curve. This curve is called the curve of regression. Regression analysis is the method used for estimating the unknown values of one variable corresponding to the known value of another variable.

Regression is the measure of average relationship between 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 independent variable.

It is known as predictor or regressor.

The variable whose value is predicted by independent variable is called dependent variable or regressed or explained variable.

The scatter diagram shows relationship between 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 line of regression and the regression is called linear regression.

Q9) Two variables X and Y are given in the dataset below, find the two lines of regression.

A9)

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

Standard deviation of x-

Similarly-

Correlation coefficient-

Put these values in regression line equation, we get

Regression line y on x-

Regression line x on y-

Q10) The following data give the experience of machine operators and their performance rating given by the number of good parts turned out per 100 pieces.

Obtain the two regression equations and estimate the performance rating of an operator who has put 15 years in service.

A10)

We define the variables,

X: Experience		Y: Performance rating Table of calculations:
X	Y	Xy	X2	Y2
16 12 18 4 3 10 5 12	87 88 89 68 78 80 75 83	1392 1056 1602 272 232 800 375 996	256 144 324 16 9 100 25 144	7569 7744 7921 4624 6084 6400 5625 6889

Now the two regression equations are,

(

Where,

      and  

Also,

Co(x,y)

Co(x,y)=30.75

Now we find,

Regression co-efficient od X on Y

  and

Regression xo-efficient of X on Y

Now substituting the values of x , y , bxy and byx in the regression

Equations we get,

(x-10) = 0.67(y-81) -------x on y (i)

(y-81) =1.13(x-10) ------- y on x (ii)

As the two regression equations.

Now to estimate Performance rating (y) when Experience (x) = 15, we use the regression equation of y on x

 (y-81) =1.13(15-10)

 y = 81+ 5.65 = 86.65

Hence the estimated performance rating for the operator with 15 years of

Experience is approximately 86.65 i.e. approximately 87

Q11) Find the marks of a student in the Subject of Mathematics who have scored 65 marks in Accountancy Given,

Coefficient of correlation between the marks of Mathematics and marks of Accountancy is 0.64.

A11)

We define the variables,

X: Marks in Mathematics

Y: Marks in Accountancy

Therefore, we have,

Now we want to approximate the marks in Mathematics (x), we obtain the

Regression equation of x on y, which is given by

Substituting the values, we get,

i.e.                                   

Therefore, when marks in Accountancy (Y) = 65

x- 70 = 0.57(65-80)

   x = 70-2.85 = 67.15                   i.e. 67 approx.

Substitute the value of a and b in the equation. Regression line of X on Y is      

Q12) Find the means values of x,y and r from the two regression equations. 3x+2y-26=0 and sx+y-31=0. Also find sx when sy = 3.

A12)

The two regression equations are,

3x+2y-26=0 -------- (i)

6x+y-31=0 ----------(ii)

Now for x and y we solve the two equations as the simultaneous equations.

On solving (1) and (2), we get-

x = 4 and y = 7.

Now to find ‘r’ we express the equations in the form y=a+bx

So, from eqns (i) and (ii)

        and         

Since, b1 < b2 (i.e., b1 is smaller in number irrespective of sign + or -)

... Equation (i) is regression of on and

Hence, eqn (ii) is regression of on and - 1/6 = - 0.16

Now we find,

Note: The sign of ‘r’ is same as the sign of regression coefficients

Now to find 6x when 6y = 3, we use the formula,

Hence means

Q13) Calculate the regressive coefficients from the data given below:

	Series	Series
Average	25	22
Standard deviation	4	2	r=0.8

A13)

The coefficient of regression of y on x is

The coefficient of regression of y on x is

Q14) From the following regression equation, find means x , y ,and d ‘r’ 3x-2y-10 = 0, 24x-25y+145 = 0

A14)

The two regression equations are,

3x-2y-10=0(i)

24x-25y+145=0(ii)

Now for x and y we solve the two equations as the simultaneous equations.

Therefore, by (i) x 8 and (ii) x1, we get

24x-16y-80 = 0

24x-16y-80 = 0

24x-25y+145 = 0

9y-225=0y = 25

Putting y = 25 in eqn (i), we get

3x-2(25)-10 = 0

3x – 60=0

X = 20

Hence x= 20 and y= 25.

Now to find ‘r’ we express the equations in the form y=a+bx So, from eqns (i) and (ii)

B1=1.5   b2=0.96

Equation (ii) is regression of y on x and byx = 0.9

Hence eqn (i) is regression of x on y and  bxy = 1/1.5 =0.67

Now we find, r =  √ bxyXbyxi.e. r=√0.67x0.96= + 0.84

Q15) Fit the straight-line curve to the following data.

A15)

First drawing the table,

X	Y		XY
75	82	5625	1
80	78	6400	4
93	86	8349	9
65	72	4225	16
87	91	7569	25
71	80	5041
98	95	9605
68	72	4624
84	89	7056
77	74	5929

The normal equation are:

Σy = aΣx + nb

And

Σxy = aΣx2 + bΣx.

Substituting the values, we get,

819 = 798a + 10b

66045 = 64422a + 798b

Solving, we get

a = 0.9288 and b = 7.78155

Therefore, the straight line equation is:

y = 0.9288x + 7.78155.

Q16) Find the arithmetic mean of the following dataset.

A16)

Let the assumed mean (a) = 25,

Q17) Calculate the first and third quartile of the following data-

A17)

Here N/4 = 28/4 = 7

The 7th observation falls in the class 10-20. So, this is the first quartile class. 3N/4 = 21th observation falls in class 30-40, so it is the third quartile class.

For first quartile l = 10, f = 5, C = 3, N = 28

We know that-

For third quartile l = 30, f = 9, C = 15

Q18) Find the quartile deviation of the following data.

A18)

We have N/4 = 28/4 = 7 and 7th observation falls in the class 10-20.

This is the first quartile class. Similarly, 3N/4 = 21 and 21st observation falls in the interval 30-40. This is the third quartile class.

By using the formula of quartile deviation, we will find -

Therefore-

Q = ½ × (Q3 – Q1) = (36.67 – 18) / 2 = 9.335

Q19) The students of statistics got the marks as below- 16, 24, 13, 18, 15, 10, 23

Find the mean deviation from mean.

A19)

X	x-17	|x- |
16	-1	1
24	7	7
13	-4	4
18	1	1
15	-2	2
10	-7	7
23	6	6
Sum = 119		28

Then

Hence

Q20) Suppose batsman A has mean 50 with SD 10. Batsman B has mean 30 with SD 3. What do you infer about their performance?

A20)

A has higher mean than B. This means A is a better run maker.

However, B has lower CV (3/30 = 0.1) than A (10/50 = 0.2) and is consequently more consistent.

Q21) Suppose a series of 100 data points has mean 50 and variance 20. Another series of 200 data points has mean 80 and variance 40. What is the combined variance of the given series?

A21)

The mean of the combined series-

Therefore, d1= 50 – 70 = –20 and d2 = 80 – 70 =10

Variance of the combined series

Q22) Calculate the standard deviation of the following frequency distribution-

A22)

Weight	Item (f)	X	d = x – 67	f.d
60 – 62	5	61	-6	-30	180
63 – 65	18	64	-3	-54	162
66 – 68	42	67	0	0	0
69 – 71	27	70	3	81	243
72 – 74	8	73	6	48	288
Total	100			45	873

Q23) Calculate the coefficient of skewness from the following data:

A23)

Here total frequency

The cumulative frequency table is

Now, N/2 =230/2= 115th item which lies in 110 – 120 group.

Median or

Also, is 57.5th or 58th item which lies in 90-100 group.

Similarly 3N/4 = 172.5 i.e. is 173rd item which lies in 120-130 group.

Hence quartile coefficient of skewness =

Q24) For a distribution Karl Pearson’s coefficient of skewness is 0.64, standard deviation is 13 and mean is 59.2 Find mode and median.

A24)

It is given that:

Coeff. Of skewness = 0.64, σ = 13 and Mean = 59.2

Therefore by using formula

Mode = 59.20 – 8.32 = 50.88

Mode = 3 Median – 2 Mean

50.88 = 3 Median - 2 (59.2)