Why I get p value = NA in Shapiro Wilk test after a BoxCox transformation?












0















this is my first question in this forum which I find very helpful! I have got a problem that I don't know how to interpret or solve it:



I've got measurements of a biochemical parameter (named 'droms') which doesn't reach normality. So, I did a BoxCox transformation following this code: 



sum(is.na(dataframe$droms))

[1] 0

boxcoxnc(dataframe$droms, method='sw', lambda =seq(-5,5,0.01),plot=TRUE, alpha=0.05, verbose=TRUE)



Error in boxcoxnc(data_turbidez$droms, method = "sw", lambda = seq(-5,  : 

Data must include positive values. Specify shifting parameter, lambda2


So, I changed the code to this new one:



 boxcoxnc(dataframe$droms, method='sw', lambda2=2, lambda =seq(-5,5,0.01),plot=TRUE, alpha=0.05, verbose=TRUE)



lambda.hat : -0.53 

statistic  : 0.9895904 

p.value    : 0.1819954 



Result     : Transformed data are normal.


And I proceeded to apply the transformation: 



dataframe$droms_normalized <- (dataframe$droms^-0.53-1)/-0.53


But when I try to check again the normality looking at the histogram, I don't get the same histogram given by the boxcox transformation and I get this output from the Shapiro Wilk Test:



W = NaN, p-value = NA


Could you help me to identify what I am doing wrong in the transformation?



Thank you so much!






r statistics normalization







share|improve this question























  • It's a pity that you don't provide reproducible sample data. Provided I understood correctly, you're estimating the parameter lambda of a two parameter Box-Cox transformation (with a non-zero shift parameter lambda2). So you need to account for the shift parameter when calculating dataframe$droms_normalized. Try ((dataframe$droms + 2)^-0.53-1)/-0.53.

    – Maurits Evers
    Jan 2 at 12:38








  • 1





    Thank you so much!

    – Iraida
    Jan 2 at 12:43
















0















this is my first question in this forum which I find very helpful! I have got a problem that I don't know how to interpret or solve it:



I've got measurements of a biochemical parameter (named 'droms') which doesn't reach normality. So, I did a BoxCox transformation following this code: 



sum(is.na(dataframe$droms))

[1] 0

boxcoxnc(dataframe$droms, method='sw', lambda =seq(-5,5,0.01),plot=TRUE, alpha=0.05, verbose=TRUE)



Error in boxcoxnc(data_turbidez$droms, method = "sw", lambda = seq(-5,  : 

Data must include positive values. Specify shifting parameter, lambda2


So, I changed the code to this new one:



 boxcoxnc(dataframe$droms, method='sw', lambda2=2, lambda =seq(-5,5,0.01),plot=TRUE, alpha=0.05, verbose=TRUE)



lambda.hat : -0.53 

statistic  : 0.9895904 

p.value    : 0.1819954 



Result     : Transformed data are normal.


And I proceeded to apply the transformation: 



dataframe$droms_normalized <- (dataframe$droms^-0.53-1)/-0.53


But when I try to check again the normality looking at the histogram, I don't get the same histogram given by the boxcox transformation and I get this output from the Shapiro Wilk Test:



W = NaN, p-value = NA


Could you help me to identify what I am doing wrong in the transformation?



Thank you so much!






r statistics normalization







share|improve this question























  • It's a pity that you don't provide reproducible sample data. Provided I understood correctly, you're estimating the parameter lambda of a two parameter Box-Cox transformation (with a non-zero shift parameter lambda2). So you need to account for the shift parameter when calculating dataframe$droms_normalized. Try ((dataframe$droms + 2)^-0.53-1)/-0.53.

    – Maurits Evers
    Jan 2 at 12:38








  • 1





    Thank you so much!

    – Iraida
    Jan 2 at 12:43














0












0








0








this is my first question in this forum which I find very helpful! I have got a problem that I don't know how to interpret or solve it:



I've got measurements of a biochemical parameter (named 'droms') which doesn't reach normality. So, I did a BoxCox transformation following this code: 



sum(is.na(dataframe$droms))

[1] 0

boxcoxnc(dataframe$droms, method='sw', lambda =seq(-5,5,0.01),plot=TRUE, alpha=0.05, verbose=TRUE)



Error in boxcoxnc(data_turbidez$droms, method = "sw", lambda = seq(-5,  : 

Data must include positive values. Specify shifting parameter, lambda2


So, I changed the code to this new one:



 boxcoxnc(dataframe$droms, method='sw', lambda2=2, lambda =seq(-5,5,0.01),plot=TRUE, alpha=0.05, verbose=TRUE)



lambda.hat : -0.53 

statistic  : 0.9895904 

p.value    : 0.1819954 



Result     : Transformed data are normal.


And I proceeded to apply the transformation: 



dataframe$droms_normalized <- (dataframe$droms^-0.53-1)/-0.53


But when I try to check again the normality looking at the histogram, I don't get the same histogram given by the boxcox transformation and I get this output from the Shapiro Wilk Test:



W = NaN, p-value = NA


Could you help me to identify what I am doing wrong in the transformation?



Thank you so much!






r statistics normalization







share|improve this question














this is my first question in this forum which I find very helpful! I have got a problem that I don't know how to interpret or solve it:



I've got measurements of a biochemical parameter (named 'droms') which doesn't reach normality. So, I did a BoxCox transformation following this code: 



sum(is.na(dataframe$droms))

[1] 0

boxcoxnc(dataframe$droms, method='sw', lambda =seq(-5,5,0.01),plot=TRUE, alpha=0.05, verbose=TRUE)



Error in boxcoxnc(data_turbidez$droms, method = "sw", lambda = seq(-5,  : 

Data must include positive values. Specify shifting parameter, lambda2


So, I changed the code to this new one:



 boxcoxnc(dataframe$droms, method='sw', lambda2=2, lambda =seq(-5,5,0.01),plot=TRUE, alpha=0.05, verbose=TRUE)



lambda.hat : -0.53 

statistic  : 0.9895904 

p.value    : 0.1819954 



Result     : Transformed data are normal.


And I proceeded to apply the transformation: 



dataframe$droms_normalized <- (dataframe$droms^-0.53-1)/-0.53


But when I try to check again the normality looking at the histogram, I don't get the same histogram given by the boxcox transformation and I get this output from the Shapiro Wilk Test:



W = NaN, p-value = NA


Could you help me to identify what I am doing wrong in the transformation?



Thank you so much!






r statistics normalization




r statistics normalization






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Jan 2 at 12:25









Iraida Iraida

12




12













  • It's a pity that you don't provide reproducible sample data. Provided I understood correctly, you're estimating the parameter lambda of a two parameter Box-Cox transformation (with a non-zero shift parameter lambda2). So you need to account for the shift parameter when calculating dataframe$droms_normalized. Try ((dataframe$droms + 2)^-0.53-1)/-0.53.

    – Maurits Evers
    Jan 2 at 12:38








  • 1





    Thank you so much!

    – Iraida
    Jan 2 at 12:43



















  • It's a pity that you don't provide reproducible sample data. Provided I understood correctly, you're estimating the parameter lambda of a two parameter Box-Cox transformation (with a non-zero shift parameter lambda2). So you need to account for the shift parameter when calculating dataframe$droms_normalized. Try ((dataframe$droms + 2)^-0.53-1)/-0.53.

    – Maurits Evers
    Jan 2 at 12:38








  • 1





    Thank you so much!

    – Iraida
    Jan 2 at 12:43

















It's a pity that you don't provide reproducible sample data. Provided I understood correctly, you're estimating the parameter lambda of a two parameter Box-Cox transformation (with a non-zero shift parameter lambda2). So you need to account for the shift parameter when calculating dataframe$droms_normalized. Try ((dataframe$droms + 2)^-0.53-1)/-0.53.

– Maurits Evers
Jan 2 at 12:38







It's a pity that you don't provide reproducible sample data. Provided I understood correctly, you're estimating the parameter lambda of a two parameter Box-Cox transformation (with a non-zero shift parameter lambda2). So you need to account for the shift parameter when calculating dataframe$droms_normalized. Try ((dataframe$droms + 2)^-0.53-1)/-0.53.

– Maurits Evers
Jan 2 at 12:38






1




1





Thank you so much!

– Iraida
Jan 2 at 12:43





Thank you so much!

– Iraida
Jan 2 at 12:43












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