UNIT 5

Statistics

Q1) Fit a least square line for the following data. Also find the trend values and show that ∑(Y–)=0 ∑(Y–)=0.

A1):

The equation of least square line Y=a +bX 

Normal equation for ‘a’ ∑Y=na+b                    25=5a+15b —- (1)

Normal equation for ‘b’ ∑XY = a∑X+b∑X2     88=15a+55b —-(2)

Eliminate a a from equation (1) and (2), multiply equation (2) by 3 and subtract from equation (2).

Eliminate  a  from equation (1) and (2), multiply equation (2) by (3) and subtract from equation (2). Thus we get the values of  a and b 

Here    a=1.1  and b=1.3 , the equation of least square line becomes

 Y=1.1+1.3X

Q2) Using least square method to fit a straight line of the following data

Solution:

First we calculate for the given data

Now we calculate

Calculate the slope

  m = = -131/118.4

calculate the y-intercept

use the formula to calculate the y-intercept

b =

   = 7-(-1.1*6.4)

The required line equation is

Y= -1.1x+14.0

Q3)

Solution:

∑Y i =Na+b∑X i +c∑

∑X i Y i =a∑X i +b∑+c∑

 ∑ Y i =a∑ +b∑+c∑

The required parabola is of the form y= ax2+bx+c

∴74=9a+b(0)+60c∴9a+60c=74…(i)

51=a(0)+60b+0c  ∴60b=51   ∴b=5160 =0.85411=60a+0b+708 c∴60a+708c=411…(ii) 

Solving (i) and (ii) simultaneously, we get
a=10.004 , c=-0.267

The Equation of parabola is therefore,

y=10.004+0.85X−0.267X 2

 =10.004+0.85(x−5)−0.267(x−5) 2 

=10.004+0.85x−4.25−0.267(x 2 −10x+25)

=10.004+0.85x−4.25−0.267x 2 +2.67x−6.675

∴  y = −0.921+3.52x−0.267x 2 

Q4) Find the least square approximation of  degree two to the data

Solution:

the normal equations are:

Here,

n = 5,

by substituting all the above values in normal equations we get,

30 = 5a+10b+30c

120=10a+30b+100c

434=30a+100b+354c

By solving the above equations we get

 a = -4, b=2,c=1.

Therefore the required polynomial is

Y= -4x+2x+x2 and errors =0

Q5) Determine the constants a and b by the method of least square such that

Solution:

The given relation is

Taking logarithms on both sides we get,

log y = log a+ bx…..(1)

let,

log y = Y

x = X

log a = A

b = B

now we have,

….(2)

….(3)

Now we need to find

The normal equations to fit the straight line is

Y = logey

Y= ln(y)

17.025 = 5A +30B…..(4)

122.150 = 30A+220B….(5)

By solving 4 and 5 we get

30A +180B = 102.15…(4)

30A+220B = 122.150…(5)

We get a =   0.405   ,b = 0.5

A =log a

 a = 1.499

since we have  X=x and Y=y

log y=Y,

And we  know y= aebx

Y = (1.499)e0.5x is the required exponential curve.

Q6) Fit the curve of the form y= aebx for the following data

Solution:

The given relation is

Taking logarithms on both sides we get,

log y = log a+ bx logee…..(1)

the required normal equations are,

….(2)

….(3)

We have n=3

The normal equations become

3A +6b = 7.8750

6A + 20 b = 18.4456

By solving the above two equations we get

A = 1.361  and b = 0.3415

Since A =logea a = e1.361 = 6.9317

The curve of the fit is

Thus,the required  equation is,

Q 7)  Ten students got the following percentage of marks in Economics and Statistics

Calculate the of correlation.

Solution:. Let the marks of two subjects be denoted by and respectively.

Then the mean for marks and the mean ofy marks

and are deviations ofx’s and ’s from their respective means, then the data may be arranged in the following form:

Q8: Find the correlation betweenx and , when the lines ofregression are: and

Solution. Let the line of regression ofx on be

Then, the line ofregressionofy on is

and

which is not possible. So our choice of regression line is incorrect.

The regression line ofx on is

And, the regression line ofy on is

And

Hence the correlation coefficient between and is

Q9) The following regression equations were obtainedfrom a correlation table:

Find the value of the correlation coefficient,

(b) the mean and

(c) the mean of

Solution.

(a)  From (1),

(b) From (2),

From (3) and (4)

Coefficient of correlation

(b) (1) and (2) pass through the point .

   (5)

   (6)

On solving (5) and (6), we get

Q10) Compute Spearman’s rank correlation coefficient r for the following data:

Solution:

Q11) If X and Y are uncorrelated random variables, the of correlation between and

Solution.

Let and

Then

Now

Similarly

Now

Also

(As and are not correlated, we have )

Similarly