Hypothesis testing for non-independent data
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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?
statistics statistical-inference
asked Nov 20 at 0:33
Lady
978
|
show 8 more comments
up vote
0
down vote
favorite
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?
statistics statistical-inference
asked Nov 20 at 0:33
Lady
978
Well, you could assume that the physician reports were normative, analyze the distribution that would yield, and see whether the proxy results were probable.
– lulu
Nov 20 at 0:44
@lulu, what do you mean by normative?
– Lady
Nov 20 at 0:45
Actually, I don't understand the question. What does the $953$ have to do with anything? And what is the difference between "schizophrenia identified as present" and "positive for schizophrenia"?
– lulu
Nov 20 at 0:46
I was assuming (probably incorrectly) that you had two samples, sample $A$ had $115$ members and $34$ positives, sample $B$ had $124$ members and also had $34$ positives. But in that case I am ignoring the $953$ entirely.
– lulu
Nov 20 at 0:48
@lulu, 953 is the grand total number of people sampled. The intersection between the physician report and the proxy report is 34. Meaning 34 were sampled from physician and proxy report.
– Lady
Nov 20 at 0:50
|
show 8 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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?
statistics statistical-inference
asked Nov 20 at 0:33
Lady
978
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?
statistics statistical-inference
statistics statistical-inference
asked Nov 20 at 0:33
Lady
978
asked Nov 20 at 0:33
Lady
978
asked Nov 20 at 0:33
Lady
978
asked Nov 20 at 0:33
Lady
978
asked Nov 20 at 0:33
Lady
978
978
Well, you could assume that the physician reports were normative, analyze the distribution that would yield, and see whether the proxy results were probable.
– lulu
Nov 20 at 0:44
@lulu, what do you mean by normative?
– Lady
Nov 20 at 0:45
Actually, I don't understand the question. What does the $953$ have to do with anything? And what is the difference between "schizophrenia identified as present" and "positive for schizophrenia"?
– lulu
Nov 20 at 0:46
I was assuming (probably incorrectly) that you had two samples, sample $A$ had $115$ members and $34$ positives, sample $B$ had $124$ members and also had $34$ positives. But in that case I am ignoring the $953$ entirely.
– lulu
Nov 20 at 0:48
@lulu, 953 is the grand total number of people sampled. The intersection between the physician report and the proxy report is 34. Meaning 34 were sampled from physician and proxy report.
– Lady
Nov 20 at 0:50
|
show 8 more comments
Well, you could assume that the physician reports were normative, analyze the distribution that would yield, and see whether the proxy results were probable.
– lulu
Nov 20 at 0:44
@lulu, what do you mean by normative?
– Lady
Nov 20 at 0:45
Actually, I don't understand the question. What does the $953$ have to do with anything? And what is the difference between "schizophrenia identified as present" and "positive for schizophrenia"?
– lulu
Nov 20 at 0:46
I was assuming (probably incorrectly) that you had two samples, sample $A$ had $115$ members and $34$ positives, sample $B$ had $124$ members and also had $34$ positives. But in that case I am ignoring the $953$ entirely.
– lulu
Nov 20 at 0:48
@lulu, 953 is the grand total number of people sampled. The intersection between the physician report and the proxy report is 34. Meaning 34 were sampled from physician and proxy report.
– Lady
Nov 20 at 0:50
Well, you could assume that the physician reports were normative, analyze the distribution that would yield, and see whether the proxy results were probable.
– lulu
Nov 20 at 0:44
Well, you could assume that the physician reports were normative, analyze the distribution that would yield, and see whether the proxy results were probable.
– lulu
Nov 20 at 0:44
@lulu, what do you mean by normative?
– Lady
Nov 20 at 0:45
@lulu, what do you mean by normative?
– Lady
Nov 20 at 0:45
Actually, I don't understand the question. What does the $953$ have to do with anything? And what is the difference between "schizophrenia identified as present" and "positive for schizophrenia"?
– lulu
Nov 20 at 0:46
Actually, I don't understand the question. What does the $953$ have to do with anything? And what is the difference between "schizophrenia identified as present" and "positive for schizophrenia"?
– lulu
Nov 20 at 0:46
I was assuming (probably incorrectly) that you had two samples, sample $A$ had $115$ members and $34$ positives, sample $B$ had $124$ members and also had $34$ positives. But in that case I am ignoring the $953$ entirely.
– lulu
Nov 20 at 0:48
I was assuming (probably incorrectly) that you had two samples, sample $A$ had $115$ members and $34$ positives, sample $B$ had $124$ members and also had $34$ positives. But in that case I am ignoring the $953$ entirely.
– lulu
Nov 20 at 0:48
@lulu, 953 is the grand total number of people sampled. The intersection between the physician report and the proxy report is 34. Meaning 34 were sampled from physician and proxy report.
– Lady
Nov 20 at 0:50
@lulu, 953 is the grand total number of people sampled. The intersection between the physician report and the proxy report is 34. Meaning 34 were sampled from physician and proxy report.
– Lady
Nov 20 at 0:50
|
show 8 more comments
1 Answer
1
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oldest
votes
up vote
0
down vote
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
answered 2 days ago
BruceET
34.7k71440
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
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
answered 2 days ago
BruceET
34.7k71440
add a comment |
up vote
0
down vote
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
answered 2 days ago
BruceET
34.7k71440
add a comment |
up vote
0
down vote
up vote
0
down vote
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
answered 2 days ago
BruceET
34.7k71440
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
answered 2 days ago
BruceET
34.7k71440
edited 2 days ago
answered 2 days ago
BruceET
34.7k71440
answered 2 days ago
BruceET
34.7k71440
answered 2 days ago
BruceET
34.7k71440
34.7k71440
add a comment |
add a comment |
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Well, you could assume that the physician reports were normative, analyze the distribution that would yield, and see whether the proxy results were probable.
– lulu
Nov 20 at 0:44
@lulu, what do you mean by normative?
– Lady
Nov 20 at 0:45
Actually, I don't understand the question. What does the $953$ have to do with anything? And what is the difference between "schizophrenia identified as present" and "positive for schizophrenia"?
– lulu
Nov 20 at 0:46
I was assuming (probably incorrectly) that you had two samples, sample $A$ had $115$ members and $34$ positives, sample $B$ had $124$ members and also had $34$ positives. But in that case I am ignoring the $953$ entirely.
– lulu
Nov 20 at 0:48
@lulu, 953 is the grand total number of people sampled. The intersection between the physician report and the proxy report is 34. Meaning 34 were sampled from physician and proxy report.
– Lady
Nov 20 at 0:50