A sample of 26women is obtained, and their heights (in inches) and pulse rates (in beats per minute) are measured. The linear correlation coefficient is 0.208and the equation of the regression line is ModifyingAbove y with caret equals 17.8 plus 0.940 x,where x represents height. The mean of the 26heights is 63.5in and the mean of the 26pulse rates is 77.9beats per minute. Find the best predicted pulse rate of a woman who is 71in tall.
Question
Answer:
The pulse rate is 84.54bpm
Step-by-step explanation:
Given
[tex]\^{} y = 17.8 + 0.940x[/tex]
Required
Find [tex]\^{}y[/tex] when [tex]x = 71[/tex]
Substitute 71 for x in: [tex]\^{} y = 17.8 + 0.940x[/tex]
[tex]\^{} y = 17.8 + 0.940 * 71[/tex]
[tex]\^{} y = 17.8 + 66.74[/tex]
[tex]\^{} y = 84.54[/tex]
The pulse rate is 84.54bpm