Question

A researcher collected data on the hours of TV watched per day from a sample of five people of different ages. Here are the results:i Age TV Hrs 1 43 1 2 30 6 3 22 4 4 20 3 5 5 6 1. Calculate the least squares estimated regression equation using simple linear regression. 2. What is the independent variable in this study?a) {y}b) agec) tv hoursd) Ie) 53) Create an ANOVA table. Using α=.05.

Answer 1

1. The least squares regression is y = -0.1015·x + 6.51

2. The independent variable is b) age

Please see attached table

Explanation:

The least squares regression formula is given as follows;

`\frac{\sum_(i = 1)^(n)(x_(i) - \bar{x})* \left (y_(i) - \bar{y} \right ) }{\sum_(i = 1)^(n)(x_(i) - \bar{x})^(2)}`

We have;

`\bar x` = 24

`\bar y` = 4

`\Sigma (x_i - \bar x) (y_i - \bar y)` = -79

`\Sigma (x_i - \bar x)^2` = 778

`\therefore \hat \beta =\frac{\sum_(i = 1)^(n)(x_(i) - \bar{x})* \left (y_(i) - \bar{y} \right ) }{\sum_(i = 1)^(n)(x_(i) - \bar{x})^(2)} = (-79)/(778) = -0.1015`

The least squares regression is y = -0.1015·x + α

∴ α = y -0.1015·x = 6 - (-0.1015 × 5) = 6.51

The least squares regression is thus;

y = -0.1015·x + 6.51

2. The independent variable is the age b)

3. Steps to create an ANOVA table with α = 0.05

The overall mean = (43 + 30 + 22 + 20 + 5 + 1 + 6 + 4 + 3 + 6 )/10 = 14

There are 2 different treatment =
`df_(treat) = 2 - 1 = 1`

There are 10 different treatment measurement =
`df_(tot) = 10 - 1 = 9`

`df_(res) = 9 - 1 = 8`

`df_(treat) + df_(res) = df_(tot)`

The estimated effects are;

`\hat A_1 = 24 - 14 = 10`

`\hat A_2 = 4 - 14 = -10`

`SS_(treat) = 10^2 * 5 + (-10)^2 * 5 =1000`

`\sum_(i)\SS_(row)_i = \sum_(i)\sum_(j) (y_(ij) - \bar y)= [(1 - 4)^2 + (6 - 4)^2 + (4 - 4)^2 + (3 - 4)^2 + (6 - 4)^2] = 18`

`\sum_(i) S S_(row)_i = \sum_(i)\sum_(j) (y_(ij) - \bar y) ^2= [(43 - 24)^2 + (30 - 24)^2 + (22 - 24)^2 + (20 - 24)^2 + (5 - 24)^2] = 778`

`S S_(res) = \sum_(i) S S_(row)_i = 778 + 18 = 796`

`SS_(tot)` = (43 - 14)² + (30 - 14)² + (22 - 14)² + (20 - 14)² + (5 - 14)² + (1 - 14)1² + (6 - 4 )² + (3 - 14)² + (6 - 14)² = 1796

`MS_(treat) = (SS_(treat) )/(df_(treat) ) = (1000)/(1) = 1000`

`MS_(res) = (SS_(res) )/(df_(res) ) = (796)/(8) = 99.5`

F- value is given by the relation;

`F = (MS_(treat) )/(MS_(res) ) = (1000)/(99.5) = 10.05`

We then look for the critical values at degrees of freedom 1 and 8 at α = 0.05 on the F-distribution tables 5.3177

Hence;
`F = 10.05 > F_(1,8)^(Krit)(5\%) = 5.3177`, we reject the null hypothesis.