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kjetil b halvorsen ♦
kjetil b halvorsen ♦
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If the null hypothesis is that the distributions in the two groups are equal, the alternative that they are different, you can estimate the weibull parameters under the two distributions, and compute a likelihood ratio test. Look at fit GLM for weibull family and its answers.

In your case, you would create variables diameter containing the measured diameters for both groups, and an indicator variable group (factor) with values G11 and G22. Then you could fitThe R glm function do not have a model likeweibull family, but the gamlss package do have (multiple ones, with different parameterization). Setting up and simulating some data:

mod1set.seed(7*11*13)
diameter <- glmc(diameter ~ grouprweibull(100, family1,1),rweibull(100,2,1.2))
group =<- weibullc(link='log'rep(1,100), datarep(2,100))
simdata =<- data.frame(diameter=diameter, group=groupgroup=as.factor(group))    

library(MASS)
summarylibrary(gamlss)

Then fitting the two nested models, and calculating the likelihood ratio test:

mod1 <- gamlss(diameter ~ 1, family=WEI(), data=simdata)
anova...
mod2 <- gamlss(diameter ~ group, diameter ~ group, family=WEI(), data=simdata)
...
LR.test(mod1, mod2)
 Likelihood Ratio Test for nested GAMLSS models. 
 (No check whether the models are nested is performed). 
 
       Null model: deviance= 424.7234 with  2 deg. of freedom 
 Alternative model: deviance= 373.4748 with  4 deg. of freedom 
 
 LRT = 51.24854 with 2 deg. of freedom and p-value= 7.439049e-12

and you could answer from that outputso the null hypothesis of equal distribution in the two groups is clearly rejected.

Finally we show some plots of the data: enter image description here

If the null hypothesis is that the distributions in the two groups are equal, the alternative that they are different, you can estimate the weibull parameters under the two distributions, and compute a likelihood ratio test. Look at fit GLM for weibull family and its answers.

In your case, you would create variables diameter containing the measured diameters for both groups, and an indicator variable group (factor) with values G1 and G2. Then you could fit a model like

mod1 <- glm(diameter ~ group, family = weibull(link='log'), data = data.frame(diameter=diameter, group=group) )
summary(mod1)
anova(mod1)

and you could answer from that output.

If the null hypothesis is that the distributions in the two groups are equal, the alternative that they are different, you can estimate the weibull parameters under the two distributions, and compute a likelihood ratio test. Look at fit GLM for weibull family and its answers.

In your case, you would create variables diameter containing the measured diameters for both groups, and an indicator variable group (factor) with values 1 and 2. The R glm function do not have a weibull family, but the gamlss package do have (multiple ones, with different parameterization). Setting up and simulating some data:

set.seed(7*11*13)
diameter <- c(rweibull(100,1,1),rweibull(100,2,1.2))
group <- c(rep(1,100),rep(2,100))
simdata <- data.frame(diameter=diameter, group=as.factor(group))    

library(MASS)
library(gamlss)

Then fitting the two nested models, and calculating the likelihood ratio test:

mod1 <- gamlss(diameter ~ 1, family=WEI(), data=simdata)
...
mod2 <- gamlss(diameter ~ group, diameter ~ group, family=WEI(), data=simdata)
...
LR.test(mod1, mod2)
 Likelihood Ratio Test for nested GAMLSS models. 
 (No check whether the models are nested is performed). 
 
       Null model: deviance= 424.7234 with  2 deg. of freedom 
 Alternative model: deviance= 373.4748 with  4 deg. of freedom 
 
 LRT = 51.24854 with 2 deg. of freedom and p-value= 7.439049e-12

so the null hypothesis of equal distribution in the two groups is clearly rejected.

Finally we show some plots of the data: enter image description here

Source Link
kjetil b halvorsen ♦
kjetil b halvorsen ♦
  • 82.8k
  • 32
  • 201
  • 663

If the null hypothesis is that the distributions in the two groups are equal, the alternative that they are different, you can estimate the weibull parameters under the two distributions, and compute a likelihood ratio test. Look at fit GLM for weibull family and its answers.

In your case, you would create variables diameter containing the measured diameters for both groups, and an indicator variable group (factor) with values G1 and G2. Then you could fit a model like

mod1 <- glm(diameter ~ group, family = weibull(link='log'), data = data.frame(diameter=diameter, group=group) )
summary(mod1)
anova(mod1)

and you could answer from that output.