Question

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).

Answer 1

`RMS= √(1-r^2) s_y =√(1-0.2^2) 20000=19595.92`

Explanation:

Data given and previous concepts

`\bar X = 64` represent the sample mean for the height

`s_x= 2.5` represent the sample deviation for the height

`\bar y= 21000` represent the sample mean for the income

`s_y = 20000` represent the sample deviation for the income

r=0.2 represent the correlation coefficient

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:

`r=(n(\sum xy)-(\sum x)(\sum y))/(√([n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]))`

Solution to the problem

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

`y= \beta_o +\beta_1 X`

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

In order to estimate the coefficients
`\beta_0 ,\beta_1` we can use least squares estimation.

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

`\beta_1 = r (s_y)/(s_x) =0.2 (20000)/(2.5)=1600`

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

`\beta_o = \bar y - \beta_1 \bar x= 21000-(1600*64)=-81400`

And then our linear model would be given by:

`y= 1600 X- 81400`

We can estimate the RMS with the following formula:

`RMS= √(1-r^2) s_y =√(1-0.2^2) 20000=19595.92`