#1 )
We are given regression equation :
y = 1.542x - 1.366
Therefore , x =
We are given value of response variable y = 80 and asked to find x
x =
x = 52.7665
( Please round the value of x to given specific decimal places )
#2 ) First we need to find correlation coefficient r for the given data set , we can use excel function =CORREL (x,y) to find the value of r
Therefore correlation coefficient r = 0.9688
To check the significance of correlation we have to perform correlation test .
H0 : The significant correlation exists between X and Y
Ha : The significant correlation does not exists between X and Y
Test statistic : r = 0.9688
Critical value : We are given = 0.05, and we have n = number of pairs of x and y = 13
Critical value = 0.553
Decision rule :
If correlation coefficient r is greater than critical value , there is significant correlation exists.
If correlation coefficient r is less than critical value , there is no significant correlation exists.
Here r = 0.9688 and critical value = 0.553
As r is greater than critical value , there is significant correlation exists between the X and Y
Now we have to find the regression equation Y = ax + b
a is slope and b is intercept of the equation.
We can use excel function =SLOPE(Y,X) and =INTERCEPT(Y,X) to find slope a and intercept b respectively.
Therefore slope a = 0.5272 and intercept b = 29.6225
The regression equation is Y = 0.5272X + 29.6225
X =
We are given Y = 24.2 and asked to find X
X =
X = -10.3
- Linear Regression and Correlation Kamal Hamid 15 You run a regression analysis on a bivariate...
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