In: Statistics and Probability
1. Data on ICU hospitalization (X) and number of death (Y) for 5 days due to COVID-19 is given.
Date | In ICU (X) | Death (Y) |
04/25 | 50 | 21 |
04/26 | 80 | 35 |
04/27 | 87 | 31 |
04/28 | 92 | 45 |
04/29 | 118 | 50 |
Total | 427 | 182 |
Find estimate βˆ1β^1
2. Data on ICU hospitalization (X) and number of death (Y) for 5 days due to COVID-19 is given.
Date | In ICU (X) | Death (Y) |
04/25 | 50 | 21 |
04/26 | 80 | 35 |
04/27 | 87 | 31 |
04/28 | 92 | 45 |
04/29 | 118 | 50 |
Total | 427 | 182 |
Find estimate βˆ0β^0
3. Data on ICU hospitalization (X) and number of death (Y) for 5 days due to COVID-19 is given.
Date | In ICU (X) | Death (Y) |
04/25 | 50 | 21 |
04/26 | 80 | 35 |
04/27 | 87 | 31 |
04/28 | 92 | 45 |
04/29 | 118 | 50 |
Total | 427 | 182 |
Use the fitted regression model to estimate the number of death when number of patients in ICU is 500 rounded to closest number.
Solution:
Regression equation can be written as
Y = β^0 +β^1X
Here Y is the dependent variable
X is the independent variable
β^0 is the intercept of the regression line
β^1 is the slope of the regression line
The slope of the regression line can be calculated as
Slope = ((n*Xi*Yi)
- (Xi
*
Yi))/((n*Xi^2)
- (Xi)^2))
X | Y | X^2 | Y^2 | XY |
50 | 21 | 2500 | 441 | 1050 |
80 | 35 | 6400 | 1225 | 2800 |
87 | 31 | 7569 | 961 | 2697 |
92 | 45 | 8464 | 2025 | 4140 |
118 | 50 | 13924 | 2500 | 5900 |
427 | 182 | 38857 | 7152 | 16587 |
Slope = ((5*16587) - (427*182))/((5*38857)-(427*427)) = 5221/11956
= 0.4367
Solution(b)
The intercept of the regression line can be calculated as
Intercept = (Yi
- Slope*Xi)/n
= (182 - 0.4367*182)/5 = -4.64/5 = -0.8929
Solution(c)
Regression equation can be written as
Y = -0.8929 + 0.4367*X
If X = 500
Than Y = -0.8929 + 0.4367*X = -0.8929 + 0.4367*500 = 217.45 or
218