15)
a)
| ΣX | ΣY | Σ(x-x̅)² | Σ(y-ȳ)² | Σ(x-x̅)(y-ȳ) | |
| total sum | 24.00 | 102.00 | 10.00 | 174.0 | -37.00 |
| mean | 4.00 | 17.00 | SSxx | SSyy | SSxy |
Sample size, n = 6
here, x̅ = Σx / n= 4.000
ȳ = Σy/n = 17.000
SSxx = Σ(x-x̅)² = 10.0000
SSxy= Σ(x-x̅)(y-ȳ) = -37.0
estimated slope , ß1 = SSxy/SSxx = -37/10=
-3.7000
intercept,ß0 = y̅-ß1* x̄ = 17- (-3.7 )*4=
31.8000
Regression line is, Ŷ= 31.800 +
( -3.700 )*x
b) for every unit increase in days of training, predicted defect
per week get decrease by 370
c)
Predicted Y at X= 6 is
Ŷ= 31.80000 +
-3.70000 *6= 9.600
d)
R² = (SSxy)²/(SSx.SSy) = 0.787
Approximately 78.68% of variation in
observations of variable Y, is explained by variable x
15. United Widget Manufacturing has a problem with defective widgets. Employees make hundreds of thousands of...
24. United Widget Manufacturing has a problem with defective widgets. Employees make hundreds of thousands of them each day, and many are defective. United has instituted training for the workers, and you would like to predict the number of defects per week based on the number of days of training an employee has received. You obtain the following data: ST Employee number 1015 2023 1153 4029 1117 0012 Days of training 4 5 6 4 3 2 Defects per week...
24. United Widget Manufacturing has a problem with defective widgets. Employees make hundreds of thousands of them each day, and many are defective. United has instituted training for the workers, and you would like to predict the number of defects per week based on the number of days of training an employee has received. You obtain the following data: Employee number 1015 2023 1153 4029 1117 0012 Days of training 4 5 6 4 3 2 Defects per week (100's)...