A radius of a circle is perpendicular to a chord if and only if it
1 answer:
You might be interested in
Answer:
Step-by-step explanation:
The estimated regression equation to predict the daily steps taken is:
Daily step= 6350.02 - 217.90BMI + 2702.12Grade_middle + 601.44BMI * Grade_middle
where Grade_middle=1 if the student is in middle school, and Grade_middle=0 if the student is in high school
b) The slope of BMI is -217.90. The negative value of the slope indicates that BMI and Daily steps taken move in opposite direction.
We can say that for 1 unit increase in the BMI, the predicted value of Daily Steps taken by the student decreases by 217.90, while keeping other variables unchanged.
c) The regression equation for middle school students can be obtained by setting Grade_middle=1
Daily step= 6350.02 - 217.90BMI + 2702.12 * 1 + 601.44BMI * 1
= 6350.02 - 217.90BMI + 2702.12 + 601.44BMI
= 9052.14 + 383.54BMI
Therefore, the slope of BMI on steps taken for students in middle school is 383.54
The regression equation for high school students can be obtained by setting Grade_middle=0
Daily Steps = 6350.02 - 217.90BMI + 2702.12 x 0 +601.44BMI*0 = 6350.02 - 217.90BMI
Therefore, the slope of BMI on steps taken for students in high school is -217.90
The model says that variables, the daily steps taken and the BMI of a middle school student are positively correlated (move in the same direction) and the daily steps taken and the BMI are negatively correlated (move in opposite direction) for a high school student. The above slopes say that for 1 unit increase in the BMI for students in middle school increases the predicted daily steps taken by 383.54 and for 1 unit increase in the BMI for students in high school decreases the predicted daily steps taken by 217.90
d) The predicted value of daily steps taken for a BMI=20 and Grade_middle=1 is
Daily Steps = 6350.02 – 217.90 x 20 + 2702.12 x1 +601.44 x 20 x 1 = 16722.94
The actual value of the daily steps taken by a middle school student with a BMI of 20 is 7000
The residual is calculated as
Therefore, this student’s residual is -9722.94