Answer:
1. The least squares regression is y = -0.1015·x + 6.51
2. The independent variable is b) age
Please see attached table
Step-by-step explanation:
The least squares regression formula is given as follows;
![\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}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csum_%7Bi%20%3D%201%7D%5E%7Bn%7D%28x_%7Bi%7D%20-%20%5Cbar%7Bx%7D%29%5Ctimes%20%5Cleft%20%28y_%7Bi%7D%20-%20%5Cbar%7By%7D%20%20%5Cright%20%29%20%7D%7B%5Csum_%7Bi%20%3D%201%7D%5E%7Bn%7D%28x_%7Bi%7D%20-%20%5Cbar%7Bx%7D%29%5E%7B2%7D%7D)
We have;
= 24
= 4
= -79
= 778
![\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](https://tex.z-dn.net/?f=%5Ctherefore%20%5Chat%20%5Cbeta%20%3D%5Cdfrac%7B%5Csum_%7Bi%20%3D%201%7D%5E%7Bn%7D%28x_%7Bi%7D%20-%20%5Cbar%7Bx%7D%29%5Ctimes%20%5Cleft%20%28y_%7Bi%7D%20-%20%5Cbar%7By%7D%20%20%5Cright%20%29%20%7D%7B%5Csum_%7Bi%20%3D%201%7D%5E%7Bn%7D%28x_%7Bi%7D%20-%20%5Cbar%7Bx%7D%29%5E%7B2%7D%7D%20%3D%20%5Cfrac%7B-79%7D%7B778%7D%20%3D%20-0.1015)
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](https://tex.z-dn.net/?f=df_%7Btreat%7D%20%3D%202%20-%201%20%3D%201)
There are 10 different treatment measurement = ![df_{tot} = 10 - 1 = 9](https://tex.z-dn.net/?f=df_%7Btot%7D%20%3D%2010%20-%201%20%3D%209)
![df_{res} = 9 - 1 = 8](https://tex.z-dn.net/?f=df_%7Bres%7D%20%3D%209%20-%201%20%3D%208)
The estimated effects are;
![\hat A_1 = 24 - 14 = 10](https://tex.z-dn.net/?f=%5Chat%20A_1%20%3D%2024%20-%2014%20%3D%2010)
![\hat A_2 = 4 - 14 = -10](https://tex.z-dn.net/?f=%5Chat%20A_2%20%3D%204%20-%2014%20%3D%20-10)
![SS_{treat} = 10^2 \times 5 + (-10)^2 \times 5 =1000](https://tex.z-dn.net/?f=SS_%7Btreat%7D%20%3D%2010%5E2%20%5Ctimes%205%20%2B%20%28-10%29%5E2%20%5Ctimes%205%20%3D1000)
![\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](https://tex.z-dn.net/?f=%5Csum_%7Bi%7D%5CSS_%7Brow%7D_i%20%3D%20%5Csum_%7Bi%7D%5Csum_%7Bj%7D%20%28y_%7Bij%7D%20-%20%5Cbar%20y%29%3D%20%5B%281%20-%204%29%5E2%20%2B%20%286%20-%204%29%5E2%20%2B%20%284%20-%204%29%5E2%20%2B%20%283%20-%204%29%5E2%20%2B%20%286%20-%204%29%5E2%5D%20%3D%2018)
![\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](https://tex.z-dn.net/?f=%5Csum_%7Bi%7D%20S%20S_%7Brow%7D_i%20%3D%20%5Csum_%7Bi%7D%5Csum_%7Bj%7D%20%28y_%7Bij%7D%20-%20%5Cbar%20y%29%20%5E2%3D%20%5B%2843%20-%2024%29%5E2%20%2B%20%2830%20-%2024%29%5E2%20%2B%20%2822%20-%2024%29%5E2%20%2B%20%2820%20-%2024%29%5E2%20%2B%20%285%20-%2024%29%5E2%5D%20%3D%20778)
![S S_{res} = \sum_{i} S S_{row}_i = 778 + 18 = 796](https://tex.z-dn.net/?f=S%20S_%7Bres%7D%20%3D%20%5Csum_%7Bi%7D%20S%20S_%7Brow%7D_i%20%3D%20778%20%2B%2018%20%3D%20796)
= (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} = \dfrac{SS_{treat} }{df_{treat} } = \dfrac{1000}{1} = 1000](https://tex.z-dn.net/?f=MS_%7Btreat%7D%20%3D%20%5Cdfrac%7BSS_%7Btreat%7D%20%7D%7Bdf_%7Btreat%7D%20%7D%20%3D%20%5Cdfrac%7B1000%7D%7B1%7D%20%3D%201000)
![MS_{res} = \dfrac{SS_{res} }{df_{res} } = \dfrac{796}{8} = 99.5](https://tex.z-dn.net/?f=MS_%7Bres%7D%20%3D%20%5Cdfrac%7BSS_%7Bres%7D%20%7D%7Bdf_%7Bres%7D%20%7D%20%3D%20%5Cdfrac%7B796%7D%7B8%7D%20%3D%2099.5)
F- value is given by the relation;
![F = \dfrac{MS_{treat} }{MS_{res} } = \frac{1000}{99.5} = 10.05](https://tex.z-dn.net/?f=F%20%3D%20%5Cdfrac%7BMS_%7Btreat%7D%20%7D%7BMS_%7Bres%7D%20%7D%20%3D%20%5Cfrac%7B1000%7D%7B99.5%7D%20%3D%2010.05)
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;
, we reject the null hypothesis.