```Computer, Minitab, and other printouts
%   = −42.734 + 1.70()
r2 = 67.8%
67.8% of the variation in percent body fat can be explained by the
variation in waist size.
2 =
.678 = .823
= .823
The correlation of .823 tells us that there is a
fairly strong and positive relationship between
percent body fat and waist size.
Residual =  −  = observed – predicted
%   = −42.734 + 1.70()
%   = −42.734 + 1.70 35 = 16.766
Residual = 10 – 16.766 = - 6.766
Variable
C1
Total
Count
16
Variable
C1
Median
0.4100
Mean
0.3281
SE Mean
0.0480
Q3
0.4650
Mean = 0.3281
Median = 0.4100
StDev
0.1920
Maximum
0.7000
Variance
0.0369
CoefVar
58.51
Minimum
0.0300
Q1
0.1300
Variable
C1
Total
Count
16
Variable
C1
Median
0.4100
Mean
0.3281
Q3
0.4650
SE Mean
0.0480
StDev
0.1920
Variance
0.0369
Maximum
0.7000
Standard Deviation = 0.1920
IQR = Q3 – Q1 = 0.4650 – 0.1300 = 0.3350
CoefVar
58.51
Minimum
0.0300
Q1
0.1300
Variable
C1
Total
Count
16
Variable
C1
Median
0.4100
Mean
0.3281
SE Mean
0.0480
Q3
0.4650
StDev
0.1920
Variance
0.0369
CoefVar
58.51
Minimum
0.0300
Q1
0.1300
Maximum
0.7000
Use the IQR = 0.3350.
IQR X 1.5 = 0.3350 X 1.5 = 0.5025
Q3 + 0.5025 = 0.9675
Q1 – 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.
Y-intercept
Standard error
t - statistic
Slope
Correlation
p - value
```