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Exercises #1-#3 utilize a data set provided by Afifi, Clark and May (2004). The data are a study of depression and was a longitudinal study. The purpose of the study was to obtain estimates of the prevalence and incidence of depression and to explore its risk factors. The study variables were of several types – demographics, life events, stressors, physical health, health services utilization, medication use, lifestyle, and social support.

Consider the following three variables.

Variable

Codings

Label in STATA

drink

1 = yes 2 = no

Regular Drinker

sex

1 = male 2 = female

cases

0 = Normal 1 = Case of Depression

Depressed is cesd > 16

Complete the following table by supplying the cell frequencies.

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Sex Regular Drinker

Female

Male

Total

Yes No

139

95

234

44

16

Total

183

111

60 294

. tabulate drink sex

What are the odds that a woman is a regular drinker? 139 / 44 = 3.2 What are the odds that a man is a regular drinker? 95 / 16 = 5.9

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What is the odds ratio? That is, compared to a man, what is the relative odds (odds ratio) that a woman is a regular drinker?

OR = [odds for woman] / [odds for man] = 3.2/5.9 = 0.54 2. Repeat the tabulation that you produced for problem #1 two times, one for persons who are depressed and the other for persons who are not depressed. . * Use command SORT to sort the data by case status (depressed or not depressed) . sort case . * Use the command BY in front of the command TABULATE . by case: tabulate drink sex

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Among Persons Who are Depressed Sex Regular Drinker

Female

Male

Total

Yes

33

8

41

No

7

2

9

Total

40

10

50

OR (Relative odds, compared to a man, that a woman is a regular drinker): OR = [(33)(2)] / [(7)(8) ] = 1.18

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Among Persons Who are NOT Depressed Sex Regular Drinker

Female

Male

Total

Yes

106

87

193

No

37

14

51

Total

143

101

244

OR (Relative odds, compared to a man, that a woman is a regular drinker): OR = [(106)(14)] / [(37)(87)] = 0.46 3. Fit a logistic regression model using these variables. Use DRINK as the dependent variable and CASES and SEX as independent variables. Also include as an independent variable the appropriate interaction term. .* Some variable creation commands .* Create 0/1 indicators of drinker and female gender

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. generate drink01= . (294 missing values generated) . replace drink01=1 if drink==1.

(234 real changes made) . replace drink01=0 if drink==2 (60 real changes made) . label define drinkf 0 "0=nondrinker" 1 "1=drinker" . label values drink01 drinkf . generate female=.

(294 missing values generated) . replace female=0 if sex==1 (111 real changes made) . replace female=1 if sex==2 (183 real changes made) . label define sexf 0 "0=male" 1 "1=female" . label values female sexf .* Create a new variable called FEM_CASE that is the interaction of FEMALE and CASES . generate fem_case=female*cases statisticsassignmenthelp.com

. * Use the command LOGISTIC if you want output to include ODDS RATIOS . * Use the command LOGIT if you want the output to include BETAs and SEs . * LOGISTIC OUTCOME PREDICTOR PREDICTOR etc.. . logit drink01 cases female fem_case Logistic regression

Number of obs = 294 LR chi2(3) = 5.62

Log likelihood = -145.95772

Pseudo R2 = 0.0189

------------------------------------------------------------------------------ ------------------drink01 | Coef. Std. Err. z P>|z| [95% Conf. Interval] -------------------- +--------------------------------------------------------- ----------------cases | -.4405564 .8413815 -0.52 0.601 -2.089634 1.208521 female |

-.7743296 .3455196

-2.24 0.025 -1.451536 -.0971237

fem_case | .9386327 .9578851 0.98 0.327 -.9387877 2.816053 _cons | 1.826851 .2879632 6.34 0.000 1.262453 2.391248 ------------------------------------------------------------------------------ -------------------

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Fitted Model: logit [ pr (drinker=yes) ] = 1.8269 - 0.4406 CASES - 0.7743 FEMALE [ ] [ ] + 0.9386 FEM_CASE [ ] where CASES =1 if depressed; 0 otherwise FEMALE = 1 if female; 0 otherwise FEM_CASE = (CASES) * (FEMALE) Is the interaction term in your model significant? No. How does your answer to problem #3 compare to your answer to problem #2? Comment.

The answers match.

Among Depressed : OR = 1.18 Among NON-depressed : OR = 0.46 logit [ pr (drinker=yes) ] = 1.8269 - 0.4406 CASES - 0.7743 FEMALE [ ] [ ] + 0.9386 FEM_CASE [ ]

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Among Depressed “1” = Female CASES FEMALE FEM_CASE

“0” = Male

1 1 1

1 0 0

Logit [ female ] = 1.8269 – 0.4406 – 0.7743 + 0.9386 = 1.5506 Logit [male] = 1.8269 – 0.4406 = 1.3863 logit [ female ] - logit [ male ] = 1.5506 - 1.3863 = + 0.1643 OR [women compared to men ] = exp { logit [ p1 ] - logit [ p0 ] } = exp { + 0.1643 } = 1.1786 statisticsassignmenthelp.com

Among NON Depressed “1” = Female “0” = Male CASES FEMALE FEM_CASE

0 1 0

0 0 0

logit [ female ] = 1.8269 – 0.7743 = 1.0526 logit [male] = 1.8269 logit [ female ] - logit [ male ] = 1.0526 - 1.8269 = -0.7743 OR [women compared to men ] = exp { logit [ p1 ] - logit [ p0 ] } = exp { -0.7743 } = 0.4610

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4. Source: Kleinbaum, Kupper, Miller, and Nizam. Applied Regression Analysis and Other Multivariable Methods, Third Edition. Pacific Grove: Duxbury Press, 1998. p 683 (problem 2).

A five year follow-up study on 600 disease free subjects was carried out to assess the effect of 0/1 exposure E on the development (or not) of a certain disease. The variables AGE (continuous) and obesity status (OBS), the latter a 0/1 variable were determined at the start of the follow-up and were to be considered as control variables in analyzing the data.

(A) State the logit form of a logistic regression model that assesses the effect of the 0/1 exposure variable E controlling for the confounding effects of AGE and OBS and the interaction effects of AGE with E and OBS with E.

Solution:

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I used the following notation: π = Probability [ disease ] AGEE = AGE * E. This is a created variable that is the interaction of AGE with E OBSE = OBS * E Similarly, this is the interaction of OBS with E.

(B) Given the model you have for part “A”, give a formula for the odds ratio for the exposure-disease relationship that controls for the confounding and interactive effects of AGE and OBS. Solution: The solution here follows the ideas on pp 9-11 in Lecture Notes 5, Logistic Regression.

Value of Predictor for Person who is Predictor E AGE OBS

Exposed 1 AGE1 OBS1

Not Exposed 0 AGE0 OBS0

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AGEE OBSE

AGE1 OBS 1

0 0

Then OR = exp { logit[π for exposed person] - logit[π for NON exposed person] }

(C) Now use the formula that you have for part “B” to write an expression for the estimated odds ratio for the exposure-disease relationship that considers both confounding and interaction when AGE=40 and OBS=1.

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Solution:

Predictor E

Value of Predictor for Person who is Exposed Not Exposed 1

AGE OBS

40 1

40 1

AGEE

40

0 OBSE

1

0

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0