[R] predict function in regression analysis
li li
hannah.hlx at gmail.com
Tue May 5 18:53:36 CEST 2015
Hi all,
I have the following data in which there is one factor lot with six
levels and one continuous convariate time.
I want to fit an Ancova model with common slope and different intercept. So
the six lots will have seperate paralell
regression lines.I wanted to find the upper 95% confidence limit for the
mean of the each of
the regression lines. It doesnot seem straightforward to achieve this using
predict function. Can anyone give some suggestions?
Here is my data. I only show the first 3 lots. Also I show the model I
used in the end. Thanks very much!
Hanna
y lot time
[1,] 4.5 1 0
[2,] 4.5 1 3
[3,] 4.7 1 6
[4,] 6.7 1 9
[5,] 6.0 1 12
[6,] 4.4 1 15
[7,] 4.1 1 18
[8,] 5.3 1 24
[9,] 4.0 2 0
[10,] 4.2 2 3
[11,] 4.1 2 6
[12,] 6.4 2 9
[13,] 5.5 2 12
[14,] 3.5 2 15
[15,] 4.6 2 18
[16,] 4.1 2 24
[17,] 4.6 3 0
[18,] 5.0 3 3
[19,] 6.2 3 6
[20,] 5.9 3 9
[21,] 3.9 3 12
[22,] 5.3 3 15
[23,] 6.9 3 18
[24,] 5.7 3 24
> mod <- lm(y ~ lot+time)
> summary(mod)
Call:
lm(formula = y ~ lot + time)
Residuals:
Min 1Q Median 3Q Max
-1.5666 -0.3344 -0.1343 0.4479 1.8985
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 4.74373 0.36617 12.955 2.84e-14 ***
lot2 -0.47500 0.41129 -1.155 0.2567
lot3 0.41250 0.41129 1.003 0.3234
lot4 0.96109 0.47943 2.005 0.0535 .
lot5 0.98109 0.47943 2.046 0.0490 *
lot6 -0.09891 0.47943 -0.206 0.8379
time 0.02586 0.02046 1.264 0.2153
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