Hypothesis testing for non-independent data

Question

Researchers collected the following data concerning comparability of diagnoses of schizophrenia obtained from primary-care physician report as compared with proxy report (from spouses). Data was collected from 953 people. The researchers found that schizophrenia was identified as present on 115 physician reports and 124 proxy reports. Both physician and proxy informants identified 34 people as positive for schizophrenia.

How do I compare the percentage of subjects identified as schizophrenic by the two reports? And how do I test if there is a significant agreement between the two reports. Can someone please help me on how to approach this problem?

Answer 1

Clue, to get you started: My interpretation of your question is that you need to finish filling in the frequencies in the following $2 \times 2$ table. (This can be done by simple arithmetic from the counts provided.) Then use the completed table to do a chi-squared test of independence.

                   Physician
                ---------------
                Yes          No         Total 
---------------------------------------------
       Yes       34                      124
Spouse
       No
--------------------------------------------
       Total    115                      953

Addendum, Results from Minitab statistical software are shown below. Do you know how to find Expected counts, Contributions, and Pearson chi-sq statistic? Null hypothesis that Physicians and Spouses make judgments about schizophrenia in independent ways is rejected. That is, both physicians and spouses must be looking at some of the same symptoms. Briefly put, under independence, we would 'expect' only about 15 agreements on 'Yes', but we observe 34 agreements (more than twice as many as expected).

Chi-Square Test for Association: Spouse, Worksheet columns 

Rows: Spouse   Columns: Worksheet columns

       PhyYes  PhyNo  All

Yes        34     90  124
         15.0  109.0
       24.219  3.324

No         81    748  829
        100.0  729.0
        3.623  0.497

All       115    838  953

Cell Contents:      Count
                    Expected count
                    Contribution to Chi-square

Pearson Chi-Square = 31.662, DF = 1, P-Value = 0.000

Note; Formally, a standard test comparing the physicians' estimate $115/953$ with the spouses' estimate $124/953$ is shown below. It shows no significant difference. However, this is supposed to be a comparison of two independent proportions, which we do not seem to have here.

Test and CI for Two Proportions 

Sample    X    N  Sample p
1       124  935  0.132620
2       115  935  0.122995

Difference = p (1) - p (2)
Estimate for difference:  0.00962567
95% CI for difference:  (-0.0206363, 0.0398876)
Test for difference = 0 (vs ≠ 0):  Z = 0.62  P-Value = 0.533