In: Statistics and Probability
tudies have shown that people who suffer sudden cardiac arrest (SCA) have a better chance of survival if a defibrillator is administered very soon after cardiac arrest. How is survival rate related to the time between when cardiac arrest occurs and when the defibrillator shock is delivered? This question is addressed in the paper “Improving Survival from Sudden Cardiac Arrest: The Role of Home Defibrillators” (by J.K. Stross, University of Michigan, February 2002). The accompanying data give y = survival rate (percent) and x = mean call-to-shock time (minutes) for a cardiac rehabilitative center (where cardiac arrests occurred while victims were hospitalized and so the call-to-shock time tended to be short) and for four communities of different sizes.
Mean call-to-shock time,x |
2 |
6 |
7 |
9 |
12 |
Survival Rate, y |
90 |
45 |
30 |
5 |
2 |
Do the following by hand and on Minitab.
Construct a scatter plot.
Calculate the Pearson correlation coefficient.
Determine equation of least squares line that can be used for predicting a value of y based on a value of x.
Compute SSE = ( y yˆ)2 for the least squares line.
Why do we call the least squares line the “best fitting line”?
tudies have shown that people who suffer sudden cardiac arrest (SCA) have a better chance of survival if a defibrillator is administered very soon after cardiac arrest. How is survival rate related to the time between when cardiac arrest occurs and when the defibrillator shock is delivered? This question is addressed in the paper “Improving Survival from Sudden Cardiac Arrest: The Role of Home Defibrillators” (by J.K. Stross, University of Michigan, February 2002). The accompanying data give y = survival rate (percent) and x = mean call-to-shock time (minutes) for a cardiac rehabilitative center (where cardiac arrests occurred while victims were hospitalized and so the call-to-shock time tended to be short) and for four communities of different sizes.
Mean call-to-shock time,x |
2 |
6 |
7 |
9 |
12 |
Survival Rate, y |
90 |
45 |
30 |
5 |
2 |
Do the following by hand and on Minitab.
Construct a scatter plot.
Calculate the Pearson correlation coefficient.
Determine equation of least squares line that can be used for predicting a value of y based on a value of x.
Compute SSE = ( y yˆ)2 for the least squares line.
Why do we call the least squares line the “best fitting line”?
tudies have shown that people who suffer sudden cardiac arrest (SCA) have a better chance of survival if a defibrillator is administered very soon after cardiac arrest. How is survival rate related to the time between when cardiac arrest occurs and when the defibrillator shock is delivered? This question is addressed in the paper “Improving Survival from Sudden Cardiac Arrest: The Role of Home Defibrillators” (by J.K. Stross, University of Michigan, February 2002). The accompanying data give y = survival rate (percent) and x = mean call-to-shock time (minutes) for a cardiac rehabilitative center (where cardiac arrests occurred while victims were hospitalized and so the call-to-shock time tended to be short) and for four communities of different sizes.
Mean call-to-shock time,x |
2 |
6 |
7 |
9 |
12 |
Survival Rate, y |
90 |
45 |
30 |
5 |
2 |
Do the following by hand and on Minitab.
Construct a scatter plot.
Calculate the Pearson correlation coefficient.
Determine equation of least squares line that can be used for predicting a value of y based on a value of x.
Compute SSE = ( y yˆ)2 for the least squares line.
Why do we call the least squares line the “best fitting line”?
tudies have shown that people who suffer sudden cardiac arrest (SCA) have a better chance of survival if a defibrillator is administered very soon after cardiac arrest. How is survival rate related to the time between when cardiac arrest occurs and when the defibrillator shock is delivered? This question is addressed in the paper “Improving Survival from Sudden Cardiac Arrest: The Role of Home Defibrillators” (by J.K. Stross, University of Michigan, February 2002). The accompanying data give y = survival rate (percent) and x = mean call-to-shock time (minutes) for a cardiac rehabilitative center (where cardiac arrests occurred while victims were hospitalized and so the call-to-shock time tended to be short) and for four communities of different sizes.
Mean call-to-shock time,x |
2 |
6 |
7 |
9 |
12 |
Survival Rate, y |
90 |
45 |
30 |
5 |
2 |
Do the following by hand and on Minitab.
Construct a scatter plot.
Calculate the Pearson correlation coefficient.
Determine equation of least squares line that can be used for predicting a value of y based on a value of x.
Compute SSE = ( y yˆ)2 for the least squares line.
Why do we call the least squares line the “best fitting line”?
Calculate r2 using the following formula: r 2
( y y)2 ( y yˆ)2
Interpret the r2 value.
Using your equation in part c, draw the least squares line on the scatterplot you constructed in part a.
Use your prediction equation to predict SCA survival rate for a community with a mean call-to-shock time of 5 min.
Sol:
1).
Scatter Plot: Take Time variable on X - axis, Survival Rate on Y - axis. Plot all (xi,yi) points on XY plane
2).
Time x | Survival Rate y | X^2 | Y^2 | XY | |
2 | 90 | 4 | 8100 | 180 | |
6 | 45 | 36 | 2025 | 270 | |
7 | 30 | 49 | 900 | 210 | |
9 | 5 | 81 | 25 | 45 | |
12 | 2 | 144 | 4 | 24 | |
Total: | 36 | 172 | 314 | 11054 | 729 |
X and Y are negative correlation
(4)
Time x | Survival Rate y | Ycap | (Y-Ycap)^2 |
2 | 90 | 82.73723 | 52.74788 |
6 | 45 | 45.55474 | 0.307741 |
7 | 30 | 36.25912 | 39.17663 |
9 | 5 | 17.66788 | 160.4753 |
12 | 2 | -10.219 | 149.3034 |
36 | 172 | 402.0109 |
3)
4).
SSE = 402.0109
Time x | Survival Rate y | Ycap | (Y-Ycap)^2 |
2 | 90 | 82.73723 | 52.74788 |
6 | 45 | 45.55474 | 0.307741 |
7 | 30 | 36.25912 | 39.17663 |
9 | 5 | 17.66788 | 160.4753 |
12 | 2 | -10.219 | 149.3034 |
36 | 172 | 402.0109 |
5)
The least square line is called "Best fitting line" since sum of square of resitudal (SSE) is minimum compare to any other lines.
According to least square principle, sum of the squares of the deviation of actual values from their estiamated value is minimum.
Least square line is derived through the above least square principle.
6).
Correlation Determination
r2 =(-0.96008)2 = 0.92175 (very high or large)
Usually, the higher the value of r2 , the better is the regression model. This is so because if r2 is larger, a greater portion of the total errors is explained by the included indepdent variable, and a smaller portion of errors is attributed to other variables and randomness
7)
8) .
Given X = 5
Y = -9.2956(5) + 101.3285
Y = 54.8505
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