How to calculate probability of being uninsured

[#1]

Dear all,

Fairly new to JMP, so my knowledge is pretty limited.

I have a study to find the probability of being uninsured by education background.

I have data set with consists of the insurance status (yes/none) and education profile (no high school degree, high school degree, college degree, advanced degree).

I need help to perform in JMP or SAS with no high school degree as the reference. I wish the result as the table which attached below.

Any and all help is appreciated.

 

[#2]

Hi @henggar313 

  It sounds like what you want to do is a Logistic Fit in JMP.  Your insured column should be nominal and your education should be numeric. That way, you can get a column function that predicts the insured outcome based on education level. Each row should be an individual with their insurance status and degree listed. I made a mock table (attached). After you run the logistic fit, you can save the formula to the data table and use that to predict future events.

  This is at least the easiest way to go about it at first. There are other ways that can be more complicated, but it doesn't sound like you need those options.

  If this isn't on the right track of what you're after, can you explain more?

Hope this helps!,

DS 

[#3]

Hi @SDF1 

thank you for the answer. your hint was very clear.

But I need the number of p value in each degree education to being uninsured person. 
I hope you can give another way.
thank you forward. 

[#4]

Hi @henggar313 ,

  When you say p value, do you mean a significance test p-value? like p < 0.00276?

  Or, do you mean probability of the outcome "yes" or "no"? The two columns Prob[No] and Prob[Yes] give the probability of that outcome given the Logist function (1/(1+e^-x)), where "x" is the linear function found from the Logistic fit of the data.

  Otherwise, I'm not sure I follow what it is you need.

DS

[#5]

I don't think (s)he is looking for an inferential result.  I am guessing that insured is a dichotomous variable and education is nominal in the categories suggested.  So, if you do a Fit Y (insured) by X (education) the resulting table row % are the probability of being insured for each educational level.

[#6]

Hi @dale_lehman ,

  So you're thinking they're looking more for a contingency table? That also makes sense. That's just changing the data type in my example table. Something more like this:

Is this what you're after @henggar313 ?

DS