using excel>data >data analysis >regression
we have
| Simple Linear Regression Analysis |
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| Regression Statistics |
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| Multiple R |
0.9521 |
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| R Square |
0.9065 |
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| Adjusted R Square |
0.8878 |
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| Standard Error |
21.0780 |
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| Observations |
7 |
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| ANOVA |
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df |
SS |
MS |
F |
Significance F |
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| Regression |
1 |
21547.3807 |
21547.3807 |
48.4994 |
0.0009 |
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| Residual |
5 |
2221.4078 |
444.2816 |
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| Total |
6 |
23768.7886 |
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Coefficients |
Standard Error |
t Stat |
P-value |
Lower 95% |
Upper 95% |
| Intercept |
-38.7124 |
18.0116 |
-2.1493 |
0.0843 |
-85.0126 |
7.5878 |
| Weight (tons) |
66.2700 |
9.5159 |
6.9641 |
0.0009 |
41.8087 |
90.7314 |
Ans 1 )
2) For a least squares regression line, the sum of the residuals
is b) always zero
3) In the regression equation, y = 2.164 + 1.3657x, n = 6, the
mean of x is 8.667, SSxx = 89.333 and Se = 3.44. A 95% confidence
interval for the average of y when x = 8 is 13.08+/- 2.78*3.44 =
(3.56, 22.62)
option a is true
a) (3.56, 22.62)
