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Computer, Minitab, and other printouts. r 2 = 67.8% 67.8% of the variation in percent body fat can be explained by the variation in waist size.

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Presentation on theme: "Computer, Minitab, and other printouts. r 2 = 67.8% 67.8% of the variation in percent body fat can be explained by the variation in waist size."— Presentation transcript:

1 Computer, Minitab, and other printouts

2

3 r 2 = 67.8% 67.8% of the variation in percent body fat can be explained by the variation in waist size.

4 The correlation of.823 tells us that there is a fairly strong and positive relationship between percent body fat and waist size.

5 Residual = 10 – 16.766 = - 6.766

6 Total Variable Count Mean SE Mean StDev Variance CoefVar Minimum Q1 C1 16 0.3281 0.0480 0.1920 0.0369 58.51 0.0300 0.1300 Variable MedianQ3 Maximum C1 0.4100 0.4650 0.7000 Mean = 0.3281 Median = 0.4100

7 Total Variable Count Mean SE Mean StDev Variance CoefVar Minimum Q1 C1 16 0.3281 0.0480 0.1920 0.0369 58.51 0.0300 0.1300 Variable MedianQ3 Maximum C1 0.4100 0.4650 0.7000 Standard Deviation = 0.1920 IQR = Q 3 – Q 1 = 0.4650 – 0.1300 = 0.3350

8 Total Variable Count Mean SE Mean StDev Variance CoefVar Minimum Q1 C1 16 0.3281 0.0480 0.1920 0.0369 58.51 0.0300 0.1300 Variable MedianQ3 Maximum C1 0.4100 0.4650 0.7000 Use the IQR = 0.3350. IQR X 1.5 = 0.3350 X 1.5 = 0.5025 Q 3 + 0.5025 = 0.9675Q 1 – 0.5025 = - 0.3725 Compare to Minimum and Maximum values. If either are outside of this range, there are outliers. If not, there aren’t any. Because Min = 0.0300 and Max = 0.7000, there are no outliers.

9 t - statistic p - value Standard error Y-intercept Slope Correlation


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