In: Statistics and Probability
Exposure | Change | Exposure | Change |
---|---|---|---|
0.89 | − 6.1 | 0.33 | 1.8 |
0.05 | 0.0 | 0.11 | 1.9 |
0.35 | 0.2 | 0.36 | 2.2 |
0.37 | − 1.2 | 0.08 | 2.3 |
0.11 | 6.3 | 0.10 | − 3.0 |
0.19 | − 4.6 | 0.06 | 8.4 |
0.10 | 1.1 | 0.34 | − 2.7 |
0.45 | 1.3 | 0.12 | 3.0 |
0.23 | 6.9 | 0.14 | − 2.2 |
0.46 | 1.7 | 0.21 | − 1.4 |
0.10 | − 3.6 | 0.20 | 0.7 |
Scatter plot is given as
There is weak negative relationship exists
Because value of r=-0.30 which is negative
Let us perform the regression
Regressing subconcussion exposure on white matter change with below mentioned steps in excel
Data->Data Analysis->Regression and choose these as X and Y variables respectively>Ok
SUMMARY OUTPUT | ||||||||
Regression Statistics | ||||||||
Multiple R | 0.368868063 | |||||||
R Square | 0.136063648 | |||||||
Adjusted R Square | 0.09286683 | |||||||
Standard Error | 3.467466463 | |||||||
Observations | 22 | |||||||
ANOVA | ||||||||
df | SS | MS | F | Significance F | ||||
Regression | 1 | 37.87170833 | 37.87171 | 3.149854 | 0.091157025 | |||
Residual | 20 | 240.4664735 | 12.02332 | |||||
Total | 21 | 278.3381818 | ||||||
Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | Lower 95.0% | Upper 95.0% | |
Intercept | 2.272015257 | 1.201555964 | 1.890894 | 0.073215 | -0.234386557 | 4.778417071 | -0.234386557 | 4.778417071 |
X Variable 1 | -6.912959936 | 3.895102237 | -1.77478 | 0.091157 | -15.0380008 | 1.212080931 | -15.0380008 | 1.212080931 |
Here X=exposure & Y=change considered as variable
So least squares regression line will be
Change (Y) = -6.913 *Exposure(X) + 2.272
Regression std error= 3.467
The low p-value of 0.0912 of the Exposure coefficient, indicates that we can say with a confidence level of 90% (which has p-value of 0.1) that the Exposure is a significant predictor of white matter Change.
Hence, the null hypothesis of no relationship between exposure and white matter change can be rejected in favor of the alternative hypothesis at a confidence level of 10%.
Let us remove the outlier 0.89 then regression output will be
SUMMARY OUTPUT | ||||||||
Regression Statistics | ||||||||
Multiple R | 0.104318418 | |||||||
R Square | 0.010882332 | |||||||
Adjusted R Square | -0.041176492 | |||||||
Standard Error | 3.471079833 | |||||||
Observations | 21 | |||||||
ANOVA | ||||||||
df | SS | MS | F | Significance F | ||||
Regression | 1 | 2.518586244 | 2.518586 | 0.209039 | 0.652706126 | |||
Residual | 19 | 228.919509 | 12.0484 | |||||
Total | 20 | 231.4380952 | ||||||
Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | Lower 95.0% | Upper 95.0% | |
Intercept | 1.475869209 | 1.451936943 | 1.016483 | 0.322169 | -1.563069732 | 4.514808149 | -1.563069732 | 4.514808149 |
X | -2.66664874 | 5.832463125 | -0.45721 | 0.652706 | -14.87413433 | 9.54083685 | -14.87413433 | 9.54083685 |
Standard error is same
But the p value is 0.65
Exposure is not good predictor of Change in white matter.
Also, the Exposure coefficient has come down from -6.91 to -2.67,
which indicates that the Change in white matter will not actually change as fast
when we will eliminate the outlier data points.
the Outlier will tend to increase the correlation between the two variables.