513
Spring BIOST/EPI Quarter 2011
Dr. McKnight
HOMEWORK 3 KEY
See Appendix I for STATA commands and Appendix II for output.
1. In a matched case-control study of risk factors for benign breast disease (BBD),
investigators were interested in whether a woman had given birth affected her risk of the
disease. Cases were women diagnosed with fibrocystic benign breast disease; for each case,
an age-matched control was identified from women admitted for general surgery, orthopedic
or otolaryngologic services to the same hospital as the case. The numbers of cases and
matched controls who had and had not ever given birth are given in the table below.
(a) The first part of this question asks you to analyze these data adjusting for the matching.
i. An estimate of the OR for the association between having had 1 birth and BBD. 1.8
ii. A 95% Confidence Interval for this OR. (0.74, 4.6)
iii. A P-value for the two-sided test of H0 : OR = 1. 0.230
iv. We observed little evidence that prior birth was associated with benign breast disease in
our age-matched case-control data (OR=1.8, 95% CI: 0.74-4.6; p=0.230).
(b) Repeat all four parts of (a) above, using an analysis that ignores the fact that controls
were matched to cases, and does not adjust for matching variables.
Cases
Controls
Total
1 Birth
25
18
43
No Births
25
32
57
Total
50
50
100
i. An estimate of the OR for the association between having had 1 birth and BBD. 1.8
ii. A 95% Confidence Interval for this OR. (0.74, 4.3)
iii. A P-value for the two-sided test of H0 : OR = 1. 0.157
iv. Without accounting for age, women who had given birth were 1.8 times as likely to
develop benign breast disease than women who never gave birth (95% CI: 0.74-4.3;
p=0.157).
(c) Odds ratio estimates from (a) and (b) are equal, but the CI in (a) is slightly wider. Note that
accounting for the matching in case-control studies can decrease or increase the standard error of
the estimated log OR; here it slightly increases it.
1
2. In the same study that produced the ejection fraction assessments discussed in lecture,
coronary arteriography was used to determine how many of the three major arterial systems
of the heart were significantly narrowed. The table below gives the numbers of subjects
classified by the clinical site and a quality-control site as having significant narrowing
(stenosis) in at least one of the three systems.
(a) Provide an estimate and 95% confidence interval for an appropriate measure of how
well the clinical and quality control sites determinations agreed.
Kappa estimate and 95% CI: 0.75 (0.68, 0.81)
(b) Explain why you chose the measure of agreement you did. I chose kappa, because it accounts
for agreement as expected by chance under independence.
(c) Our data show that substantial agreement exists between the clinical site and the quality
control site regarding stenosis classification (kappa=0.75; CI: 0.68-0.81).
3. In fact, the data in problem 2 are collapsed from a larger table, given below, for the amount
of narrowing and number of major arterial systems significantly impacted. Both clinical
and quality control sites classified coronary arteriographs as showing no narrowing, some
narrowing or significant narrowing in one, two or all three major arterial systems of the
heart. The results of clinical site and quality-control site classifications are given in the
table below.
(a) Provide an estimate and 95% confidence interval for an appropriate measure of how
well the clinical and quality control sites determinations agreed.
Weighted kappa estimate and 95% CI: 0.69 (0.65, 0.74).
(b) Explain what measure of agreement you chose, and why you made the choices you
did. I chose weighted kappa, because the weights will allow us to penalize disagreement
according to how off the classifications are. My weighting scheme (see matrix in Appendix
II, p. 7) penalizes disagreement based on the distance between the ordered categories, i.e., the
clinical site reporting None and the quality control site reporting Some is not as bad as the
clinical site reporting None and the quality control site reporting Three.
(c) Our data show substantial agreement between the clinical site and the quality control site on
the number of arterial systems narrowed (weighted kappa=0.69; CI: 0.65-0.74).
2
4. This question is about data from a study of patients with phenylketonuria (PKU). The PKU
cases were identified as infants and placed on dietary therapy during the first three months
of life. Between the ages of 4 and 6 their IQ was measured using the Stanford-Binet test,
and then the IQ of their normal sibling closest in age was also measured using the
Stanford-Binet test. The data for 15 PKU-sibling pairs are given below. The investigators
were interested in whether adoption of dietary therapy before three months of age prevented
PKU-associated mental retardation.
Answers using the corrected data set are in red.
(a) Choose suitable measure for the association between PKU/Sibling status and IQ, and
in one sentence, explain why you chose this measure.
I chose to compute the mean difference between the IQ of PKU cases and that of their
normal siblings. If dietary therapy before three months of age were effective, then the
average difference between sibling IQs should be small, i.e., not significantly different from
0.
(b) Provide an estimate of this measure and a 95% confidence interval for it.
Estimated mean difference and CI: 1.9 (-15.0, 18.7) points
Estimated mean difference and CI: -4.1 (-12.9, 4.7) points
(c) State a suitable null and alternative hypothesis to be tested using these data.
Let d = IQcase IQsib.
H0: d= 0
HA: d 0
(d) Test this hypothesis and report test statistic, reference (null) distribution, and P-value.
Paired t-test test statistic: 0.238 [-1.01]
Null distribution: t distribution with 14 degrees of freedom
2-sided 0.815 p-value: [0.331]
(e) The estimated mean IQ of PKU cases measured at 4 to 6 years was 1.9 points higher [4.1
points lower] than their normal siblings (95% CI: -15.0-18.7; p = 0.815) [(95% CI: -12.94.7; p = 0.331)], and we conclude that adoption of dietary therapy at infancy was successful
at preventing a high level of PKU-associated mental retardation.
3
****************************Appendix I: STATA commands***************************
*** Useful commands:
* mcci <Ecase-Ectrl> <Ecase-Uctrl> <Ucase-Ectrl> <Ucase-Uctrl>
* cci <exposed cases> <unexposed cases> <exposed controls> <unexposed controls>
* return
* bitesti <N> <n> p (matched pairs hypothesis test)
* kapwgt (for weighted ratings)
* kap <var1> <var2> [freq=<count>], tab wgt(<mywgt>)
* ttest <pair1> = <pair2>, level()
** Q1: matched case-control study of childbirth and risk of benign beast disease
* reference: Slides 163, 166-167
* exposure is 'ever given birth'
mcci 9 16 9 16
cci 25 25 18 32
** Q2: significant stenosis as classified by the clinical site and a quality-control
site
* reference: Slide 176
clear
set obs 4
input clin qc count
0 0 70
1 0 29
0 1 12
1 1 759
kap clin qc [freq=count], tab
di 0.7474 - invnormal(0.975)*0.0337
di 0.7474 + invnormal(0.975)*0.0337
**Q3: finer classification of stenosis
* reference: Slides 183-86
clear
set obs 25
input qc clin count
1 1 13
128
131
140
150
216
2 2 43
2 3 19
244
255
311
329
3 3 155
3 4 54
3 5 24
410
422
4 3 18
4 4 162
4 5 68
510
520
5 3 11
5 4 27
5 5 240
4
* check table
tab qc clin [freq=count]
kapwgt mywgt 1 \ 0.75 1 \ 0.5 0.75 1 \ 0.25 0.5 0.75 1 \ 0 0.25 0.5 0.75 1
kap clin qc [freq=count], tab wgt(mywgt)
di 0.6919 - invnormal(0.975)*0.0233
di 0.6919 + invnormal(0.975)*0.0233
**Q4: infant dietary therapy and PKU-associated mental retardation
clear
set obs 15
input IQcase IQsib
89 77
98 110
116 94
67 91
128 122
81 94
96 121
116 114
110 88
90 91
76 99
71 93
100 104
108 12 * 102
74 82
ttest IQcase=IQsib, level(95)
5
*****************************Appendix II: STATA output***************************
1.
| Controls
|
Cases
|
Exposed
Unexposed |
Total
-----------------+------------------------+-----------Exposed |
9
16 |
25
Unexposed |
9
16 |
25
-----------------+------------------------+-----------Total |
18
32 |
50
McNemar's chi2(1) =
1.96
Prob > chi2 = 0.1615
Exact McNemar significance probability
= 0.2295
Proportion with factor
Cases
.5
Controls
.36
--------difference
.14
ratio
1.388889
rel. diff.
.21875
odds ratio
1.777778
[95% Conf. Interval]
--------------------.0721165
.3521165
.8750591
2.204437
-.0519343
.4894343
.7397548
4.564309
(exact)
Proportion
|
Exposed
Unexposed |
Total
Exposed
-----------------+------------------------+-----------------------Cases |
25
25 |
50
0.5000
Controls |
18
32 |
50
0.3600
-----------------+------------------------+-----------------------Total |
43
57 |
100
0.4300
|
|
|
Point estimate
|
[95% Conf. Interval]
|------------------------+-----------------------Odds ratio |
1.777778
|
.741719
4.283447 (exact)
Attr. frac. ex. |
.4375
|
-.3482194
.7665431 (exact)
Attr. frac. pop |
.21875
|
+------------------------------------------------chi2(1) =
2.00 Pr>chi2 = 0.1574
2.
|
qc
clin |
0
1|
Total
-----------+----------------------+---------0|
70
12 |
82
1|
29
759 |
788
-----------+----------------------+---------Total |
99
771 |
870
Expected
Agreement
Agreement
Kappa
Std. Err.
Z
Prob>Z
----------------------------------------------------------------95.29%
81.34%
0.7474
0.0337
22.17
0.0000
6
3.
|
qc
clin |
1
2
3
4
5|
Total
-----------+-------------------------------------------------------+---------1|
13
6
1
0
0|
20
2|
8
43
9
2
0|
62
3|
1
19
155
18
11 |
204
4|
0
4
54
162
27 |
247
5|
0
5
24
68
240 |
337
-----------+-------------------------------------------------------+---------Total |
22
77
243
250
278 |
870
Ratings weighted by:
1.0000
0.7500
0.7500
1.0000
0.5000
0.7500
0.2500
0.5000
0.0000
0.2500
0.5000
0.7500
1.0000
0.7500
0.5000
0.2500
0.5000
0.7500
1.0000
0.7500
0.0000
0.2500
0.5000
0.7500
1.0000
Expected
Agreement
Agreement
Kappa
Std. Err.
Z
Prob>Z
----------------------------------------------------------------91.09%
71.09%
0.6919
0.0233
29.65
0.0000
4. Paired t test
-----------------------------------------------------------------------------Variable |
Obs
Mean
Std. Err.
Std. Dev.
[95% Conf. Interval]
---------+-------------------------------------------------------------------IQcase |
15
94.66667
4.789539
18.5498
84.39413
104.9392
IQsib |
15
92.8
6.720402
26.02801
78.38617
107.2138
---------+-------------------------------------------------------------------diff |
15
1.866667
7.844845
30.38295
-14.95885
18.69219
-----------------------------------------------------------------------------mean(diff) = mean(IQcase - IQsib)
t=
0.2379
Ho: mean(diff) = 0
degrees of freedom =
14
Ha: mean(diff) < 0
Pr(T < t) = 0.5923
Ha: mean(diff) != 0
Pr(|T| > |t|) = 0.8154
Ha: mean(diff) > 0
Pr(T > t) = 0.4077
Paired t test
-----------------------------------------------------------------------------Variable |
Obs
Mean
Std. Err.
Std. Dev.
[95% Conf. Interval]
---------+-------------------------------------------------------------------IQcase |
15
94.66667
4.789539
18.5498
84.39413
104.9392
IQsib |
15
98.8
3.450604
13.36413
91.39919
106.2008
---------+-------------------------------------------------------------------diff |
15
-4.133333
4.10559
15.90088
-12.93895
4.672282
-----------------------------------------------------------------------------mean(diff) = mean(IQcase - IQsib)
t = -1.0068
Ho: mean(diff) = 0
degrees of freedom =
14
Ha: mean(diff) < 0
Pr(T < t) = 0.1656
Ha: mean(diff) != 0
Pr(|T| > |t|) = 0.3311
7
Ha: mean(diff) > 0
Pr(T > t) = 0.8344