t test with weights and no equals variances












0














When the function t.test is applied in R, the option of var.equals = TRUE or FALSE can be chosen according to whether the variances are equal or not.



Well, I'm analyzing data from a survey with weighting and I want to know if there is any equivalent function with weighting, where you can choose the option of equal or not variances.



I suppose the test of normality and equality of variances does not change, when there is weighting, right? (I apply the lillie.test and the var.test).










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  • 1




    Maybe the question is better for Crossvalidated? stats.stackexchange.com
    – otwtm
    Nov 21 '18 at 18:37










  • If you have questions about statistical tests, it's best to ask them at Cross Validated. You should know what test you want to perform before you ask how to do it in R here.
    – MrFlick
    Nov 21 '18 at 18:42










  • I suggest to start considering the svymean and svyttest functions of the survey package: stats.stackexchange.com/questions/139619/…
    – Marco Sandri
    Nov 21 '18 at 22:16












  • Ok, thank you very much :)
    – Jose Quesada
    Nov 22 '18 at 8:40
















0














When the function t.test is applied in R, the option of var.equals = TRUE or FALSE can be chosen according to whether the variances are equal or not.



Well, I'm analyzing data from a survey with weighting and I want to know if there is any equivalent function with weighting, where you can choose the option of equal or not variances.



I suppose the test of normality and equality of variances does not change, when there is weighting, right? (I apply the lillie.test and the var.test).










share|improve this question




















  • 1




    Maybe the question is better for Crossvalidated? stats.stackexchange.com
    – otwtm
    Nov 21 '18 at 18:37










  • If you have questions about statistical tests, it's best to ask them at Cross Validated. You should know what test you want to perform before you ask how to do it in R here.
    – MrFlick
    Nov 21 '18 at 18:42










  • I suggest to start considering the svymean and svyttest functions of the survey package: stats.stackexchange.com/questions/139619/…
    – Marco Sandri
    Nov 21 '18 at 22:16












  • Ok, thank you very much :)
    – Jose Quesada
    Nov 22 '18 at 8:40














0












0








0







When the function t.test is applied in R, the option of var.equals = TRUE or FALSE can be chosen according to whether the variances are equal or not.



Well, I'm analyzing data from a survey with weighting and I want to know if there is any equivalent function with weighting, where you can choose the option of equal or not variances.



I suppose the test of normality and equality of variances does not change, when there is weighting, right? (I apply the lillie.test and the var.test).










share|improve this question















When the function t.test is applied in R, the option of var.equals = TRUE or FALSE can be chosen according to whether the variances are equal or not.



Well, I'm analyzing data from a survey with weighting and I want to know if there is any equivalent function with weighting, where you can choose the option of equal or not variances.



I suppose the test of normality and equality of variances does not change, when there is weighting, right? (I apply the lillie.test and the var.test).







r






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 21 '18 at 18:34









m0nhawk

15.2k83160




15.2k83160










asked Nov 21 '18 at 18:33









Jose QuesadaJose Quesada

94




94








  • 1




    Maybe the question is better for Crossvalidated? stats.stackexchange.com
    – otwtm
    Nov 21 '18 at 18:37










  • If you have questions about statistical tests, it's best to ask them at Cross Validated. You should know what test you want to perform before you ask how to do it in R here.
    – MrFlick
    Nov 21 '18 at 18:42










  • I suggest to start considering the svymean and svyttest functions of the survey package: stats.stackexchange.com/questions/139619/…
    – Marco Sandri
    Nov 21 '18 at 22:16












  • Ok, thank you very much :)
    – Jose Quesada
    Nov 22 '18 at 8:40














  • 1




    Maybe the question is better for Crossvalidated? stats.stackexchange.com
    – otwtm
    Nov 21 '18 at 18:37










  • If you have questions about statistical tests, it's best to ask them at Cross Validated. You should know what test you want to perform before you ask how to do it in R here.
    – MrFlick
    Nov 21 '18 at 18:42










  • I suggest to start considering the svymean and svyttest functions of the survey package: stats.stackexchange.com/questions/139619/…
    – Marco Sandri
    Nov 21 '18 at 22:16












  • Ok, thank you very much :)
    – Jose Quesada
    Nov 22 '18 at 8:40








1




1




Maybe the question is better for Crossvalidated? stats.stackexchange.com
– otwtm
Nov 21 '18 at 18:37




Maybe the question is better for Crossvalidated? stats.stackexchange.com
– otwtm
Nov 21 '18 at 18:37












If you have questions about statistical tests, it's best to ask them at Cross Validated. You should know what test you want to perform before you ask how to do it in R here.
– MrFlick
Nov 21 '18 at 18:42




If you have questions about statistical tests, it's best to ask them at Cross Validated. You should know what test you want to perform before you ask how to do it in R here.
– MrFlick
Nov 21 '18 at 18:42












I suggest to start considering the svymean and svyttest functions of the survey package: stats.stackexchange.com/questions/139619/…
– Marco Sandri
Nov 21 '18 at 22:16






I suggest to start considering the svymean and svyttest functions of the survey package: stats.stackexchange.com/questions/139619/…
– Marco Sandri
Nov 21 '18 at 22:16














Ok, thank you very much :)
– Jose Quesada
Nov 22 '18 at 8:40




Ok, thank you very much :)
– Jose Quesada
Nov 22 '18 at 8:40












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