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- Linear Regression and Correlation Kamal Hamid 15 You run a regression analysis on a bivariate set of data (n 73). You obtain the regression equation = 1.5422+-1.366 with a correlation coefficient of r = 0.45 (which is signifi average) for the explanatory variable will give you a value of 80 on the res cant at α = 0.01). You want to predict what value (on What is the predicted explanatory value?
Run a regression analysis on the following bivariate set of data with y as the response variable. 48.9 59.8 3.2 32.9 19.2 39.3 29.6 475 44.3 53 24.3 37.3 47.2 37.6 46.9 17 37.7 33.6 50.2 27.3 40.1 36.5 48.7 43.9 51.2 Verify that the correlation is significant at an a - (on average) for the explanatory variable will give you a value of 24.2 on the response variable. What is the predicted explanatory value? (Report answer accurate to one decimal place.) ints possible: 1
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Answer #1

#1 )  

We are given regression equation :

y = 1.542x - 1.366

Therefore , x =  y+1.366 1.542

We are given value of response variable y = 80 and asked to find x

x = 80 1.366 1.542

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

1 48.9 2 3.2 3 19.2 4 29.6 5 44.3 7 37.3 8 37.6 9 17 10 33.6 11 27.3 12 36.5 13 43.9 14 59.8 32.9 39.3 47.5 53 24.3 47.2 46.9 37.7 50.2 40.1 48.7 51.2 -CORREL(A1:A13,B1:B13) 10.9688

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  alpha = 0.05, and we have n = number of pairs of x and y = 13

2-tailed testing 0.878 0.729 0.810.917 0.669 0.754 0.875 0.6210.707 0.834 0.5820.666 0.798 0.805 0.959 6 9 10 549 0.6320.705 0.5210.602 0.735 0.497 0.576 0.708 12 13 0.4760.553 0.684 14 15 0.458 0.532 0.661 0.441 0.514 0.641

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.

1 48.9 2 3.2 3 19.2 4 29.6 5 44.3 7 37.3 8 37.6 9 17 10 33.6 11 27.3 12 36.5 13 43.9 59.8 32.9 39.3 47.5 53 24.3 47.2 46.9 37.7 50.2 40.1 48.7 51.2 -SLOPE(B1:B13,A1:A13) 0.5272 INTERCEPT(B1:B13,A1:A13) 29.6225

Therefore slope a = 0.5272 and intercept b = 29.6225

The regression equation is Y = 0.5272X + 29.6225

X = y - 29.6225 0,5272

We are given Y = 24.2 and asked to find X

X = 24.2 - 29.6225 0,5272

X = -10.3

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