# 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 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:

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.

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

Residual = 10 – 16.766 = - 6.766

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

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

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.

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

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