Download presentation

Presentation is loading. Please wait.

Published byEthan Draine Modified over 2 years ago

1
Computer, Minitab, and other printouts

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

Similar presentations

Presentation is loading. Please wait....

OK

DESCRIBING DISTRIBUTION NUMERICALLY

DESCRIBING DISTRIBUTION NUMERICALLY

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google