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Concept explainers
a.
Justify whether the transformation x′=√x or x′′=ln(x) is recommended, by keeping y unchanged.
a.
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Answer to Problem 59E
The transformation x′′=ln(x)_ is recommended.
Explanation of Solution
Calculation:
Reference: Exercise 5.58: The data on success (%), y and energy of shock, x is given.
The suitable transformation can be identified by constructing
Consider the transformed variable, x′ as sqrt(x)=√x. The transformation can be obtained using software.
Data transformation sqrt(x)=√x:
Software procedure:
Step-by-step procedure to transform the data using MINITAB software is given below:
- Choose Calc > Calculator.
- Enter the column of sqrt(x) under Store result in variable.
- Enter the formula SQRT(‘x’) under Expression.
- Click OK.
The transformed variable is stored in the column sqrt(x).
Scatterplot:
Software procedure:
Step-by-step procedure to draw the scatterplot using MINITAB software is given below:
- Choose Graph > Scatterplot.
- Choose Simple, and then click OK.
- Enter the column of y under Y variables.
- Enter the column of sqrt(x) under X variables.
- Click OK.
The output obtained using MINITAB is as follows:
Consider the transformed variable, x′ as sqrt(x)=√x.
Data transformation x′′=ln(x):
Software procedure:
Step-by-step procedure to transform the data using MINITAB software is given below:
- Choose Calc > Calculator.
- Enter the column of sqrt(x) under Store result in variable.
- Enter the formula LN(‘x’) under Expression.
- Click OK.
The transformed variable is stored in the column ln(x).
Scatterplot:
Software procedure:
Step-by-step procedure to draw the scatterplot using MINITAB software is given below:
- Choose Graph > Scatterplot.
- Choose Simple, and then click OK.
- Enter the column of y under Y variables.
- Enter the column of ln(x) under X variables.
- Click OK.
The output obtained using MINITAB is as follows:
A careful observation of the scatterplot between y and x′=√x shows that the points form an extended S-shaped pattern. This does not suggest that a linear model would suitably describe the relationship between y and x′=√x.
On the other hand, the scatterplot between y and x′′=ln(x) shows a straightened plot. Although there is a slight curvature observed in this graph too, it is negligible. In comparison to the previous scatterplot, this appears to be much straighter.
Thus, the transformation x′′=ln(x)_ is recommended.
b.
Find the least-squares regression line between y and the transformation recommended in the previous part.
b.
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Answer to Problem 59E
The least-squares regression equation between y and the transformation recommended in the previous part, that is, x′′=ln(x), is ˆy=62.42+43.89ln(x)_.
Explanation of Solution
Calculation:
Regression:
Software procedure:
Step by step procedure to get regression equation using MINITAB software is given as,
- Choose Stat > Regression > Regression > Fit Regression Model.
- Under Responses, enter the column of y.
- Under Continuous predictors, enter the columns of ln(x).
- Choose Results and select Analysis of Variance, Model Summary, Coefficients, Regression Equation.
- Click OK on all dialogue boxes.
The outputs using MINITAB software is given as follows:
From the output, the least-squares regression equation between y and the transformation recommended in the previous part, that is, x′′=ln(x), is: ˆy=62.42+43.89ln(x)_.
c.
Predict the success for an energy shock 1.75 times the threshold.
Predict the success for an energy shock 0.8 times the threshold.
c.
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Answer to Problem 59E
The success for an energy shock 1.75 times the threshold is 86.98%.
The success for an energy shock 0.8 times the threshold is 52.63%.
Explanation of Solution
Calculation:
The energy of shock is given as a multiple of the threshold of defibrillation.
For an energy shock 1.75 times the threshold, x=1.75. Substitute this value in the obtained regression equation:
ˆy=62.42+43.89ln(x)=62.42+43.89ln(1.75)≈62.42+(43.89×0.5596)≈62.42+24.56=86.98.
Thus, the success for an energy shock 1.75 times the threshold is 86.98%.
For an energy shock 0.8 times the threshold, x=0.8. Substitute this value in the obtained regression equation:
ˆy=62.42+43.89ln(x)=62.42+43.89ln(0.8)≈62.42+(43.89×(−0.2231))≈62.42−9.79=52.63.
Thus, the success for an energy shock 0.8 times the threshold is 52.63%.
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Chapter 5 Solutions
Introduction To Statistics And Data Analysis
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