# find the correlation coeff, the equation for the regression line and predict the final grade for a student whose absence score is 3.5 chart # of absence / ...

find the correlation coeff, the equation for the regression line and predict the final grade for a student whose absence score is 3.5

chart

# of absence / 0, 1, 2, 3, 4, 5

final grade / 100 92 83 78 66 51

please show work

### 1 Answer | Add Yours

The correlation coefficient `rho = (E[(X-mu_x)(Y-mu_y)])/(sigma_xsigma_y)`

where X is the absence measure and Y is the final grade, `mu` is the population mean and `sigma` is the population standard deviation.

We estimate `rho` with `hat rho = (Sigma (x-bar x)(y-bary))/((n-1)hat sigma_x hat sigma_y)`

where `bar x`, `bar y` are the sample means and `hat sigma_x, hat sigma_y` are the sample standard deviations

`bar x = Sigma (x / n) = 15/6 = 2.5` and `bar y = Sigma(y/n) = 470/6 = 78.3`

`Sigma((x-bar x)(y- bar y)) = Sigma(x y) - n bar x bar y = `

`Sigma (0*100 + 1*92 + 2*83 + 3*78 + 4*66 + 5*51) - 6(2.5)(78.3) = `

`1011 - 1175 = -164`

`hat(sigma_x)^2 = Sigma((x-bar x)^2/(n-1)) = 17.5/5 = 3.5` `hat (sigma_y)^2 = Sigma((y-bar y)^2/(n-1)) = 1577/5 = 315.5`

So correlation coefficient `hat rho = -164/ ((n-1)sqrt((3.5)(315.5))) = -164/(5sqrt(1104)) = -0.987` **answer**

The regression line is `y = alpha + beta (x-bar x)` for standardised `x`

We estimate `alpha` by `hat alpha = bar y = 78.3`

and `beta` by `hat beta = (Sigma(x-bar x)(y-bar y))/(Sigma(x-bar x)^2) = -164/17.5 = -9.37`

So we estimate the regression line `y = 78.3 -9.37(x-2.5)` **answer**

When absence x = 3.5, we predict the grade y to be 78.3 -9.37(3.5-2.5) = 78.3 - 9.37(1) = 69.0 **answer**