How to do a paired t-test using two different weights
I am doing an analysis on the well-being of children placed in foster care, and my dataset is based on a survey answered by a sample of all children placed in foster care in 2014 and 2018. About 50 percent of all children placed in foster care are 17 years old, but because this was not acoounted for in the process of collecting the data, I need to calculate a set of weights according to this age problem.
However, only about a third of the children who answered the survey in 2014 also answers the survey in 2018. That means that 2 thirds of the sample in 2018 are new respondents, and weights have therefore been calculated in both 2014 and 2018 and are not the equal to each other.
I am now going to compare the group of respondents in 2014 to the respondents in 2018 to find out if there has been a statistically significant change in the well being over the course of these four years.
My manager has asked me to do a paired t-test, because we have made the assumption the two samples contain same group of children (although this is not exactly true).
I am working in SAS and normally I have no problems doing a paired t-test, but because I have to use the aforementioned weights, I am a bit lost. My problem is that I have no idea how to state, that the variable from 2014 needs to be weighed with the weight from 2014, and that the variable from 2018 needs to be weighed with the weight from 2018 when I am doing the t-test.
I have posted an example of my code below:
Data have;
input id age_2014 age_2018 w_2014 w_2018 q_2014 q_2018;
lines;
1 11 15 1 1.3 0 1
2 11 15 1 1.3 0 1
3 11 . 1 . 1 0
4 13 17 2 2.1 1 0
5 13 17 2 2.1 0 1
6 . 17 . 2.1 1 0
7 13 17 2 2.1 1 0
8 . 15 . 1.3 1 0
9 11 . 1 . 1 0
10 . 17 . 2.1 0 1
;
run;
Proc ttest data=have H0=0 sides=2 alpha=0.05;
weight w_2014 w_2018;
paired q_2014*q_2018;
run;
My main problem is that I don't know where to state the two weights. At this point I get an error because I can't write two weights in the weight-statement.
Maybe it is actually wrong to assume that the two sample consists of the same people, and therefore it is not correct to use a paired-test?
sas weight t-test
asked Nov 20 at 15:19
BeeBee
335
add a comment |
I am doing an analysis on the well-being of children placed in foster care, and my dataset is based on a survey answered by a sample of all children placed in foster care in 2014 and 2018. About 50 percent of all children placed in foster care are 17 years old, but because this was not acoounted for in the process of collecting the data, I need to calculate a set of weights according to this age problem.
However, only about a third of the children who answered the survey in 2014 also answers the survey in 2018. That means that 2 thirds of the sample in 2018 are new respondents, and weights have therefore been calculated in both 2014 and 2018 and are not the equal to each other.
I am now going to compare the group of respondents in 2014 to the respondents in 2018 to find out if there has been a statistically significant change in the well being over the course of these four years.
My manager has asked me to do a paired t-test, because we have made the assumption the two samples contain same group of children (although this is not exactly true).
I am working in SAS and normally I have no problems doing a paired t-test, but because I have to use the aforementioned weights, I am a bit lost. My problem is that I have no idea how to state, that the variable from 2014 needs to be weighed with the weight from 2014, and that the variable from 2018 needs to be weighed with the weight from 2018 when I am doing the t-test.
I have posted an example of my code below:
Data have;
input id age_2014 age_2018 w_2014 w_2018 q_2014 q_2018;
lines;
1 11 15 1 1.3 0 1
2 11 15 1 1.3 0 1
3 11 . 1 . 1 0
4 13 17 2 2.1 1 0
5 13 17 2 2.1 0 1
6 . 17 . 2.1 1 0
7 13 17 2 2.1 1 0
8 . 15 . 1.3 1 0
9 11 . 1 . 1 0
10 . 17 . 2.1 0 1
;
run;
Proc ttest data=have H0=0 sides=2 alpha=0.05;
weight w_2014 w_2018;
paired q_2014*q_2018;
run;
My main problem is that I don't know where to state the two weights. At this point I get an error because I can't write two weights in the weight-statement.
Maybe it is actually wrong to assume that the two sample consists of the same people, and therefore it is not correct to use a paired-test?
sas weight t-test
asked Nov 20 at 15:19
BeeBee
335
1
It only makes sense to talk about a paired test in relation to cases where you can actually calculate the difference - you are testing whether the difference is zero. You can discard cases where this isn't possible, or impute missing values. You can assign an overall weight to each difference, but a 'weighted difference' is not a meaningful concept for individual rows.
– user667489
Nov 20 at 16:49
add a comment |
I am doing an analysis on the well-being of children placed in foster care, and my dataset is based on a survey answered by a sample of all children placed in foster care in 2014 and 2018. About 50 percent of all children placed in foster care are 17 years old, but because this was not acoounted for in the process of collecting the data, I need to calculate a set of weights according to this age problem.
However, only about a third of the children who answered the survey in 2014 also answers the survey in 2018. That means that 2 thirds of the sample in 2018 are new respondents, and weights have therefore been calculated in both 2014 and 2018 and are not the equal to each other.
I am now going to compare the group of respondents in 2014 to the respondents in 2018 to find out if there has been a statistically significant change in the well being over the course of these four years.
My manager has asked me to do a paired t-test, because we have made the assumption the two samples contain same group of children (although this is not exactly true).
I am working in SAS and normally I have no problems doing a paired t-test, but because I have to use the aforementioned weights, I am a bit lost. My problem is that I have no idea how to state, that the variable from 2014 needs to be weighed with the weight from 2014, and that the variable from 2018 needs to be weighed with the weight from 2018 when I am doing the t-test.
I have posted an example of my code below:
Data have;
input id age_2014 age_2018 w_2014 w_2018 q_2014 q_2018;
lines;
1 11 15 1 1.3 0 1
2 11 15 1 1.3 0 1
3 11 . 1 . 1 0
4 13 17 2 2.1 1 0
5 13 17 2 2.1 0 1
6 . 17 . 2.1 1 0
7 13 17 2 2.1 1 0
8 . 15 . 1.3 1 0
9 11 . 1 . 1 0
10 . 17 . 2.1 0 1
;
run;
Proc ttest data=have H0=0 sides=2 alpha=0.05;
weight w_2014 w_2018;
paired q_2014*q_2018;
run;
My main problem is that I don't know where to state the two weights. At this point I get an error because I can't write two weights in the weight-statement.
Maybe it is actually wrong to assume that the two sample consists of the same people, and therefore it is not correct to use a paired-test?
sas weight t-test
asked Nov 20 at 15:19
BeeBee
335
I am doing an analysis on the well-being of children placed in foster care, and my dataset is based on a survey answered by a sample of all children placed in foster care in 2014 and 2018. About 50 percent of all children placed in foster care are 17 years old, but because this was not acoounted for in the process of collecting the data, I need to calculate a set of weights according to this age problem.
However, only about a third of the children who answered the survey in 2014 also answers the survey in 2018. That means that 2 thirds of the sample in 2018 are new respondents, and weights have therefore been calculated in both 2014 and 2018 and are not the equal to each other.
I am now going to compare the group of respondents in 2014 to the respondents in 2018 to find out if there has been a statistically significant change in the well being over the course of these four years.
My manager has asked me to do a paired t-test, because we have made the assumption the two samples contain same group of children (although this is not exactly true).
I am working in SAS and normally I have no problems doing a paired t-test, but because I have to use the aforementioned weights, I am a bit lost. My problem is that I have no idea how to state, that the variable from 2014 needs to be weighed with the weight from 2014, and that the variable from 2018 needs to be weighed with the weight from 2018 when I am doing the t-test.
I have posted an example of my code below:
Data have;
input id age_2014 age_2018 w_2014 w_2018 q_2014 q_2018;
lines;
1 11 15 1 1.3 0 1
2 11 15 1 1.3 0 1
3 11 . 1 . 1 0
4 13 17 2 2.1 1 0
5 13 17 2 2.1 0 1
6 . 17 . 2.1 1 0
7 13 17 2 2.1 1 0
8 . 15 . 1.3 1 0
9 11 . 1 . 1 0
10 . 17 . 2.1 0 1
;
run;
Proc ttest data=have H0=0 sides=2 alpha=0.05;
weight w_2014 w_2018;
paired q_2014*q_2018;
run;
My main problem is that I don't know where to state the two weights. At this point I get an error because I can't write two weights in the weight-statement.
Maybe it is actually wrong to assume that the two sample consists of the same people, and therefore it is not correct to use a paired-test?
sas weight t-test
sas weight t-test
asked Nov 20 at 15:19
BeeBee
335
asked Nov 20 at 15:19
BeeBee
335
asked Nov 20 at 15:19
BeeBee
335
asked Nov 20 at 15:19
BeeBee
335
asked Nov 20 at 15:19
BeeBee
335
335
1
It only makes sense to talk about a paired test in relation to cases where you can actually calculate the difference - you are testing whether the difference is zero. You can discard cases where this isn't possible, or impute missing values. You can assign an overall weight to each difference, but a 'weighted difference' is not a meaningful concept for individual rows.
– user667489
Nov 20 at 16:49
add a comment |
1
It only makes sense to talk about a paired test in relation to cases where you can actually calculate the difference - you are testing whether the difference is zero. You can discard cases where this isn't possible, or impute missing values. You can assign an overall weight to each difference, but a 'weighted difference' is not a meaningful concept for individual rows.
– user667489
Nov 20 at 16:49
1
1
It only makes sense to talk about a paired test in relation to cases where you can actually calculate the difference - you are testing whether the difference is zero. You can discard cases where this isn't possible, or impute missing values. You can assign an overall weight to each difference, but a 'weighted difference' is not a meaningful concept for individual rows.
– user667489
Nov 20 at 16:49
It only makes sense to talk about a paired test in relation to cases where you can actually calculate the difference - you are testing whether the difference is zero. You can discard cases where this isn't possible, or impute missing values. You can assign an overall weight to each difference, but a 'weighted difference' is not a meaningful concept for individual rows.
– user667489
Nov 20 at 16:49
add a comment |
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It only makes sense to talk about a paired test in relation to cases where you can actually calculate the difference - you are testing whether the difference is zero. You can discard cases where this isn't possible, or impute missing values. You can assign an overall weight to each difference, but a 'weighted difference' is not a meaningful concept for individual rows.
– user667489
Nov 20 at 16:49