English
Tamil Nadu Board of Secondary EducationHSC Commerce Class 11

Obtain the two regression lines from the following data N = 20, ∑X = 80, ∑Y = 40, ∑X2 = 1680, ∑Y2 = 320 and ∑XY = 480.

Advertisements
Advertisements

Question

Obtain the two regression lines from the following data N = 20, ∑X = 80, ∑Y = 40, ∑X2 = 1680, ∑Y2 = 320 and ∑XY = 480.

Sum
Advertisements

Solution

N = 20, ∑X = 80, ∑Y = 40, ∑X2 = 1680, ∑Y2 = 320 and ∑XY = 480

`bar"X" = (sum"X")/"N" = 80/20` = 4

`bar"Y" - (sum"Y")/"N" = 40/20` = 2

byx = `("N"sum"XY" - (sum"X")(sum"Y"))/("N"sum"X"^2 - (sum"X")^2)`

= `(20(480) - (80)(40))/(20(1680) - (80)^2)`

= `(9600 - 3200)/(33600 - 6400)`

= `6400/27200`

= 0.235

= 0.24

Regression line of Y on X

`"Y" - bar"Y" = "b"_"yx"("X" - bar"X")`

Y − 2 = 0.24 (X − 4)

Y = 0.24X − 0.96 + 2

Y = 0.24X + 1.04

bxy = `("N"sum"XY" - (sum"X")(sum"Y"))/("N"sum"Y"^2 - (sum"Y")^2)`

= `(20(480) - (80)(40))/(20(320) - (40)^2)`

= `(9600 - 3200)/(6400 - 1600)`

= `6400/4800`

= 1.33

Regression line of X on Y

`"X" - bar"X" = "b"_"xy"("Y" - bar"Y")`

X – 4 = 1.33 (Y – 2)

X = 1.33Y – 2.66 + 4

X = 1.33Y + 1.34

shaalaa.com
Regression Analysis
  Is there an error in this question or solution?
Chapter 9: Correlation and Regression Analysis - Exercise 9.2 [Page 226]

APPEARS IN

Samacheer Kalvi Business Mathematics and Statistics [English] Class 11 TN Board
Chapter 9 Correlation and Regression Analysis
Exercise 9.2 | Q 4 | Page 226

RELATED QUESTIONS

From the data given below:

Marks in Economics: 25 28 35 32 31 36 29 38 34 32
Marks in Statistics: 43 46 49 41 36 32 31 30 33 39

Find

  1. The two regression equations,
  2. The coefficient of correlation between marks in Economics and Statistics,
  3. The mostly likely marks in Statistics when the marks in Economics is 30.

Given the following data, what will be the possible yield when the rainfall is 29.

Details Rainfall Production
Mean 25`` 40 units per acre
Standard Deviation 3`` 6 units per acre

Coefficient of correlation between rainfall and production is 0.8.


The following data relate to advertisement expenditure (in lakh of rupees) and their corresponding sales (in crores of rupees)

Advertisement expenditure 40 50 38 60 65 50 35
Sales 38 60 55 70 60 48 30

Estimate the sales corresponding to advertising expenditure of ₹ 30 lakh.


You are given the following data:

Details X Y
Arithmetic Mean 36 85
Standard Deviation 11 8

If the Correlation coefficient between X and Y is 0.66, then find

  1. the two regression coefficients,
  2. the most likely value of Y when X = 10.

A survey was conducted to study the relationship between expenditure on accommodation (X) and expenditure on Food and Entertainment (Y) and the following results were obtained:

Details Mean SD
Expenditure on Accommodation (₹) 178 63.15
Expenditure on Food and Entertainment (₹) 47.8 22.98
Coefficient of Correlation 0.43

Write down the regression equation and estimate the expenditure on Food and Entertainment, if the expenditure on accommodation is ₹ 200.


The equations of two lines of regression obtained in a correlation analysis are the following 2X = 8 – 3Y and 2Y = 5 – X. Obtain the value of the regression coefficients and correlation coefficients.


When one regression coefficient is negative, the other would be


The lines of regression of X on Y estimates


X and Y are a pair of correlated variables. Ten observations of their values (X, Y) have the following results. ∑X = 55, ∑XY = 350, ∑X2 = 385, ∑Y = 55, Predict the value of y when the value of X is 6.


The following information is given.

Details X (in ₹) Y (in ₹)
Arithmetic Mean 6 8
Standard Deviation 5 `40/3`

Coefficient of correlation between X and Y is `8/15`. Find

  1. The regression Coefficient of Y on X
  2. The most likely value of Y when X = ₹ 100.

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×