x_{i} |
135 | 110 | 130 | 145 | 175 | 160 | 120 |
---|---|---|---|---|---|---|---|
y_{i} |
145 | 100 | 125 | 120 | 135 | 130 | 110 |
I need the standardized residual for this data, with my answer
rounded to 2 decimal points
I did data analysis in excel, I ran a regression using standardized
residuals, but only "observation 3= 0.49" is coming out correctly.
What am I not understanding about this concept, I followed the
instructions on excel and the video.
OVERALL FIT | ||||||
Multiple R | 0.663031 | |||||
R Square | 0.43961 | |||||
Adjusted R Square | 0.327532 | |||||
Standard Error | 12.46227 | |||||
Observations | 7 | |||||
ANOVA | Alpha | 0.05 | ||||
df | SS | MS | F | p-value | sig | |
Regression | 1 | 609.1736 | 609.1736 | 3.922355 | 0.104512 | no |
Residual | 5 | 776.5407 | 155.3081 | |||
Total | 6 | 1385.714 | ||||
coeff | std err | t stat | p-value | lower | upper | |
Intercept | 61.5407 | 31.67305 | 1.942999 | 0.109649 | -19.8775 | 142.9589 |
xi | 0.445349 | 0.224868 | 1.980494 | 0.104512 | -0.13269 | 1.023389 |
standardized residuals= residual/std error
Obs | xi | yi | Pred Y | Residual | SResidual | |
1 | 135 | 145 | 121.6628 | 23.34 | 1.87 | |
2 | 110 | 100 | 110.5291 | -10.53 | -0.84 | |
3 | 130 | 125 | 119.436 | 5.56 | 0.45 | |
4 | 145 | 120 | 126.1163 | -6.12 | -0.49 | |
5 | 175 | 135 | 139.4767 | -4.48 | -0.36 | |
6 | 160 | 130 | 132.7965 | -2.80 | -0.22 | |
7 | 120 | 110 | 114.9826 | -4.98 | -0.40 |
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