How to estimate the 95%CI of Absolute Difference Between Two Independent Proportions

[Qiguo Lian]

I have two variable (0= negative, 1=positive) generated from two different Guidelines on HP, one is old published in 2021, and the other is updated in last year.

I want to estimated the Absolute Difference of Two Independent Proportions with the 95%CI as the Tables in JAMA Network Open（https://jamanetwork.com/journals/jam...cle/2687386）. How can I do it efficiently?



I tried the commands:

proportion hp2024, cformat(%9.4f)
scalar prop_2024 = r(table)["b",2]
scalar se2024 = r(table)["se", 2]

proportion hp2001, cformat(%9.4f)
scalar prop_2001 = r(table)["b",2]
scalar se2001 = r(table)["se", 2]

scalar se=sqrt(se2024^2 +se2001^2 )
scalar z = invnormal(0.975)
scalar p=prop_2001-prop_2024
scalar ci_lower = p - z * se
scalar ci_upper = p + z * se
format p ci_lower ci_upper %9.4f
list p ci_lower ci_lower

I want if there are other user written commands to do it more efficient and flexible. Like:

command_name hp2024 hp2001
command_name hp2024 hp2001 if sex==1
svy: command_name hp2024 hp2001

The dataset is listed below:

* Example generated by -dataex-. For more info, type help dataex
clear
input byte (hp2024 hp2001 sex)
1 1 2
1 1 1
1 1 1
1 1 2
1 1 1
1 1 2
0 0 2
0 0 1
1 1 2
0 1 2
1 1 1
0 0 2
0 0 2
1 1 1
1 1 2
1 1 1
0 0 1
0 1 2
0 0 1
0 0 2
0 0 2
1 1 1
1 1 1
1 1 1
1 1 1
1 1 1
1 1 1
1 1 2
0 1 1
0 0 2
0 0 2
1 1 2
1 1 2
0 0 2
0 0 1
0 0 1
1 1 2
1 1 2
1 1 2
0 0 2
0 0 2
1 1 1
0 0 2
1 1 1
1 1 1
0 0 1
0 0 2
1 1 1
1 1 1
1 1 2
1 1 1
0 0 2
0 0 2
0 0 2
1 1 2
1 1 2
1 1 2
0 1 1
0 0 2
0 0 1
0 1 1
0 0 1
0 0 1
1 1 1
1 1 1
0 0 2
0 0 2
0 0 1
1 1 1
1 1 2
0 0 2
0 0 2
0 1 2
1 1 1
0 0 2
0 0 2
1 1 1
0 1 1
1 1 1
0 0 1
0 1 2
0 0 1
1 1 1
0 1 1
1 1 1
1 1 2
0 1 1
0 0 2
0 0 2
0 1 2
0 0 2
0 1 1
0 1 1
0 1 1
0 0 2
0 0 2
0 0 2
1 1 2
1 1 1
0 1 2
end

[Felix Bittmann]

prtest hp2024 = hp2001

However, I am not sure the command allows for the svy prefix.

[Qiguo Lian]

prtest hp2024 = hp2001

However, I am not sure the command allows for the svy prefix.

Thank you for remind me the prtest command, which is useful too. It seems that svy does not allows with prtest.

[Tiago Pereira]

It is unclear the design of your study. I have assumed that the proportions are risks. You can also use GLM if the assumption of independence is OK.

  clear
    set obs 10000
    gene time = runiform()>0.50
    *! Risk of 60% for time 1
    gene status = runiform()>0.40 if time ==1
    *! Risk of 40% for time 0
    replace status = runiform()>0.60 if time ==0
    glm status time, family(binomial) link(identity) vce(robust)
    tab status time, col

[Qiguo Lian]

It is unclear the design of your study. I have assumed that the proportions are risks. You can also use GLM if the assumption of independence is OK.

 clear
set obs 10000
gene time = runiform()>0.50
*! Risk of 60% for time 1
gene status = runiform()>0.40 if time ==1
*! Risk of 40% for time 0
replace status = runiform()>0.60 if time ==0
glm status time, family(binomial) link(identity) vce(robust)
tab status time, col

Thanks. The design is estimating the prevalence of a disease using two cutoffs (one was released in 2011 and the updated was published in 2024).

[Joseph Coveney]

I have two variable (0= negative, 1=positive) generated from two different Guidelines on HP, one is old published in 2021, and the other is updated in last year.

I want to estimated the Absolute Difference of Two Independent Proportions with the 95%CI as the Tables in JAMA Network Open

Those data in that article were paired, not independent. Yours look paired, too. Are you sure that a method for use with independent observations is what you're looking for?