1. For women age 25–34 in the HANES5 sample, the relationship between height and income can be summarized as follows: average height ≈ 64 inches, SD ≈ 2.5 inches average income ≈ $21,000, SD ≈ $20,000, r ≈ 0.2 The slope of the regression equation is . The y-intercept is The predicted income for a woman who is 67 inches tall is . The predicted income for a woman who is 60 inches tall is . The RMS error for these predictions is (round your answer to 2 digits).

The correlation coefficient is a "statistical measure that calculates the strength of the relationship between the relative movements of two variables". It's denoted by r and its always between -1 and 1.

And in order to calculate the correlation coefficient we can use this formula:

[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]

Solution to the problem

Let's suppose that we have the following linear model:

[tex]y= \beta_o +\beta_1 X[/tex]

Where Y is the dependent variable (income) and X the independent variable (height). [tex]\beta_0[/tex] represent the intercept and [tex]\beta_1[/tex] the slope.

In order to estimate the coefficients [tex]\beta_0 ,\beta_1[/tex] we can use least squares estimation.

We have an useful formula in order to estimate the slope for the linear model given by:

[tex]\beta_1 = r \frac{s_y}{s_x} =0.2 \frac{20000}{2.5}=1600[/tex]

Now we can find the intercept for the linear model with the following formula:

[tex]\beta_o = \bar y - \beta_1 \bar x= 21000-(1600*64)=-81400[/tex]

And then our linear model would be given by:

[tex]y= 1600 X- 81400[/tex]

We can estimate the RMS with the following formula:

[tex]RMS= \sqrt{1-r^2} s_y =\sqrt{1-0.2^2} 20000=19595.92[/tex]