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.

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

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 = [tex]df_{treat} = 2 - 1 = 1[/tex]

There are 10 different treatment measurement = [tex]df_{tot} = 10 - 1 = 9[/tex]

[tex]df_{res} = 9 - 1 = 8[/tex]

[tex]df_{treat} + df_{res} = df_{tot}[/tex]

The estimated effects are;

[tex]\hat A_1 = 24 - 14 = 10[/tex]

[tex]\hat A_2 = 4 - 14 = -10[/tex]

[tex]SS_{treat} = 10^2 \times 5 + (-10)^2 \times 5 =1000[/tex]

[tex]\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[/tex]

[tex]\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[/tex]

[tex]S S_{res} = \sum_{i} S S_{row}_i = 778 + 18 = 796[/tex]

[tex]SS_{tot}[/tex] = (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

[tex]MS_{treat} = \dfrac{SS_{treat} }{df_{treat} } = \dfrac{1000}{1} = 1000[/tex]

[tex]MS_{res} = \dfrac{SS_{res} }{df_{res} } = \dfrac{796}{8} = 99.5[/tex]

F- value is given by the relation;

[tex]F = \dfrac{MS_{treat} }{MS_{res} } = \frac{1000}{99.5} = 10.05[/tex]

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; [tex]F = 10.05 > F_{1,8}^{Krit}(5\%) = 5.3177[/tex], we reject the null hypothesis.