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Week Three Journal

Course Reflection

Provide reflection of this week that contains 1-2 paragraph which address at least one of the following topics:

· Learning (i.e., information learned from course and/or content)

· Likes (i.e., liked most about the course and/or content)

· Dislikes (i.e., liked least about the course and/or content)

· Suggestions (i.e., suggestions for improvement about the course, content and/or assignments)

Chapter 9

Multivariable Methods

Learning Objectives (1 of 2)

Define and provide examples of dependent and independent variables in a study of a public health problem

Explain the principle of statistical adjustment to a lay audience

Organize data for regression analysis

Learning Objectives (2 of 2)

Define and provide an example of confounding

Define and provide an example of effect modification

Interpret coefficients in multiple linear and multiple logistic regression analysis

Definitions

Confounding—the distortion of the effect of a risk factor on an outcome

Effect modification—a different relationship between the risk factor and an outcome depending on the level of another variable

Confounding

A confounder is related to the risk factor and also to the outcome.

Assessing confounding

Formal tests of hypothesis

Clinically meaningful associations

We wish to assess the association between obesity and incident cardiovascular disease.

Example 9.1.

Confounding (1 of 2)

Example 9.1.

Confounding (2 of 2)

Is age a confounder?

A clinical trial is run to assess the efficacy of a new drug to increase HDL cholesterol.

H0: m1 = m2 versus H1:m1 ≠ m2

Z = –1.13 is not statistically significant.

Example 9.2.

Effect Modification (1 of 2)

Is there effect modification by gender?

Example 9.2.

Effect Modification (2 of 2)

Effect Modification

Cochran-Mantel-Haenszel Method

Technique to estimate association between risk factor and outcome accounting for confounding

Data are organized into stratum and associations are estimated in each stratum and combined.

Correlation and Simple Linear

Regression Analysis

Two continuous variables

Y = dependent, outcome variable

X = independent, predictor variable

Relationship between age and SBP, number of hours of exercise and percent body fat, caffeine consumption and blood sugar level.

Correlation and Simple

Linear Regression

Correlation—nature and strength of linear association between variables

Regression—equation that best describes relationship between variables

Scatter Diagram

Y 10 14 15 17 19 21 25 28 30 35 39 5 8 3 4 9 12 16 10 17 18 21 X

Y

Correlation Coefficient

Population correlation r

Sample correlation r, –1 ≤ r ≤ +1

Sign indicates nature of relationship (positive or direct, negative, or inverse).

Magnitude indicates strength.

Direct Relationship Between

X and Y, r = 0.6

Y 10 14 15 17 19 21 25 28 30 35 39 5 8 3 4 9 12 16 10 17 18 21 X

Y

Inverse Relationship Between

X and Y, r = –0.6

Y 40 37 38 32 28 25 17 19 12 15 10 5 8 3 4 9 12 16 10 17 18 21 X

Y

Sample Correlation Coefficient

Example (1 of 6)

Suppose we are interested in the relationship between body mass index (BMI; computed as the ratio of weight in kilograms to height in meters squared) and systolic blood pressure in males 50 years of age.

Example (2 of 6)

A random sample of 10 males 50 years of age is selected and their weights, heights, and systolic blood pressures are measured.

Their weights and heights are transformed into body mass index scores (see next slide).

In this analysis, the independent (or predictor) variable is BMI and the dependent (or response) variable is systolic blood pressure.

Example (3 of 6)

Data

X = BMI Y = SBP

18.4 120

20.1 110

22.4 120

25.9 135

26.5 140

28.9 115

30.1 150

32.9 165

33.0 160

34.7 180

X = BMI (X – ) (X – )2

18.4 –8.89 79.032

20.1 –7.19 51.696

22.4 –4.89 23.912

25.9 –1.39 1.932

26.5 –0.79 0.624

28.9 1.61 2.592

30.1 2.81 7.896

32.9 5.61 31.472

33.0 5.71 32.604

34.7 7.41 54.908

272.9 286.669

= 27.29

Example (4 of 6)

Y = SBP (Y – ) (Y – )2

120 –19.5 380.25

110 –29.5 870.25

120 –19.5 380.25

135 –4.5 20.25

140 0.5 0.25

115 –24.5 600.25

150 10.5 110.25

165 25.5 650.25

160 20.5 420.25

180 40.5 1640.2

1395 5072.50

= 139.5

Example (5 of 6)

(X – ) (Y – ) (X – )(Y –)

–8.89 –19.5 173.355

–7.19 –29.5 212.105

–4.89 –19.5 95.355

–1.39 –4.5 6.255

–0.79 0.50 –0.395

1.61 –24.5 –39.455

2.81 10.5 29.505

5.61 25.5 143.055

5.71 20.5 117.055

7.41 40.5 300.105

1036.95

cov (X,Y) = 1036.95/9

= 115.22

Example (6 of 6)

Sample Correlation Coefficient

Example (1 of 2)

Suppose in the same study we also measure the number of hours of vigorous exercise per week.

Is there a relationship between the number of hours of exercise and SBP in males 50 years of age?

Example (2 of 2)

Data

X = Hours of ExerciseY = SBP

4 120

10 110

2 120

3 135

3 140

5 115

1 115

2 165

2 160

0 180

Sample Correlation Coefficient

Simple Linear Regression

Y = Dependent, outcome variable

X = Independent, predictor variable

= b0 + b1 x

b0 is the Y-intercept, b1 is the slope

Simple Linear Regression Assumptions

Linear relationship between X and Y

Independence of errors

Homoscedasticity (constant variance) of the errors

Normality of errors

Least Squares Estimates of

Regression Parameters

Regression Analysis: BMI and SBP

Using Regression Equation

What is expected SBP for a male with BMI = 20?

Compare two males whose BMIs differ by 2 units. How do SBPs compare?

Person with higher BMI will have SBP that is 2(3.61) = 7.22 units higher.

Regression Analysis: Exercise and SBP

Example 9.6.

Linear Regression Analysis

Clinical trial to assess the efficacy of a new drug to increase HDL cholesterol

where 1 = new drug and 0 = placebo

Multiple Linear Regression

Y = continuous outcome variable

X1, X2, …, Xp = set of independent or predictor variables

Multiple Regression Analysis (1 of 2)

Model is conditional, parameter estimates are conditioned on other variables in model.

Perform overall test of regression.

If significant, examine individual predictors.

Relative importance of predictors by p-values (or standardized coefficients)

Multiple Regression Analysis (2 of 2)

Predictors can be continuous, indicator variables (0/1), or a set of dummy variables.

Dummy variables (for categorical predictors)

Race: white, black, Hispanic

Black (1 if black, 0 otherwise)

Hispanic (1 if Hispanic, 0 otherwise)

Example 9.7.

Multiple Linear Regression Analysis

Outcome = infant birth weight, grams

Independent Regression

Variable Coefficient t p-value

Intercept –3850.92 –11.56 0.0001

Male gender 174.79 6.06 0.0001

Gestational age, weeks 179.89 22.35 0.0001

Mother’s age, years 1.38 0.47 0.6361

Black race –138.46 –1.93 0.0535

Hispanic race –13.07 –0.37 0.7103

Other race –68.67 –1.05 0.2918

Simple Logistic Regression Analysis

Outcome is dichotomous (1 = event, 0 = non-event) and p = P(event).

Outcome is modeled as log odds.

Multiple Logistic Regression Analysis

Outcome is dichotomous (1 = event, 0 = non-event) and p = P(event).

Outcome is modeled as log odds.

Example 9.8.

Logistic Regression Analysis

Study to assess the relationship between obesity and incident CVD

Estimation of Regression Coefficients

Model parameters are estimated using maximum likelihood techniques.

b1 is the log odds ratio.

exp(b1) is the odds ratio estimate from a logistic regression model.

Interpretation of Regression Coefficients in Logistic Regression (1 of 2)

With a dichotomous predictor X, b1 is a log odds ratio for success for group1 versus group2.

With a continuous predictor X, b1 is a log odds ratio for success per unit change in X.

b1 = 0 No association between Y and X

b1 > 0 Probability of success increases as X increases

b1 < 0 Probability of success decreases as X increases
Interpretation of Regression Coefficients in Logistic Regression (2 of 2)
Multiple Logistic Regression Model for Hypertension (Y/N)
Predictor b p OR (95% CI for OR)
Intercept –5.4070.0001
Age 0.0520.0001 1.053 (1.044 – 1.062)
Male –0.2500.0007 0.779 (0.674 – 0.900)
BMI 0.1580.0001 1.171 (1.146 – 1.198)
Chapter 10
Nonparametric Tests
Learning Objectives (1 of 2)
Compare and contrast parametric and nonparametric tests
Identify multiple applications where nonparametric approaches are appropriate
Perform and interpret the Mann–Whitney U test
Perform and interpret the Sign test and Wilcoxon Signed Rank test
Learning Objectives (2 of 2)
Compare and contrast the Sign test and Wilcoxon Signed Rank test
Perform and interpret the Kruskal–Wallis test
Identify the appropriate nonparametric hypothesis testing procedure based on type of outcome variable and number of samples
Nonparametric Tests
Appropriate when outcome is continuous but not normally distributed
Rank scores (e.g., disease stage)
Continuous but subject to extremes
Continuous but there are limits of detection (on high or low end of scale)
General Approach
Rank data
Perform analysis on ranks
Follow same 5-step procedure for hypothesis testing
Ranking Data
Raw data
7 5 9 3 0 2
Ordered data
0 2 3 5 7 9
Ranked data
1 2 3 4 5 6
Ranking Data with Ties
Raw data
7 7 9 3 0 2
Ordered data
0 2 3 7 7 9
Ranked data
1 2 3 4.5 4.5 6
Assign mean rank to ties,
Sum of ranks = n(n + 1)/2
Tests with Two Independent Samples: Mann–Whitney U Test (1 of 2)
Continuous outcome that is not assumed to follow a normal distribution
Two independent samples
H0: Two populations are equal
H1: Two populations are not equal
Tests with Two Independent Samples: Mann–Whitney U Test (2 of 2)
Test statistic is U = min(U1, U2),
where R1 and R2 are the sums of the ranks in groups 1 and 2.
Reject H0 if U ≤ critical value in Table 5.
Example 10.1.
Mann–Whitney U Test (1 of 4)
A Phase II clinical trial is run to investigate efficacy of a new drug for asthma in children.
Outcome is number of episodes of shortness of breath over a 1-week period.
Placebo 7 5 6 4 12
Drug 3 6 4 2 1
Example 10.1.
Mann–Whitney U Test (2 of 4)
H0: The two populations are equal
H1: The two populations are not equal
a = 0.05
Test statistic is U.
Rank data in pooled sample (n = 10), and compute R1 and R2.
Example 10.1.
Mann–Whitney U Test (3 of 4)
Example 10.1.
Mann–Whitney U Test (4 of 4)
Test statistic is U = 3.
Reject H0 if U ≤ 2 (Table 5).
Do not reject H0 because 3 > 2. We do not have significant evidence to show that the two populations are not equal.

Tests with Matched Samples:

Sign Test (1 of 2)

Continuous outcome measured in matched or paired samples; differences are not assumed to follow a normal distribution.

Matched or paired samples

H0: Median difference is zero

H1: Median difference >, <, or ≠ 0
Tests with Matched Samples:
Sign Test (2 of 2)
Test statistic is the smaller of the number of positive or negative signs (of differences).
Reject H0 if the smaller of the number of positive of negative signs ≤ critical value in Table 6.
A new chemotherapy treatment is proposed for patients with breast cancer. Investigators want to assess tolerability of treatment.
Outcome is quality of life (QOL) measured on an ordinal scale (1 = poor, 2 = fair, 3 = good, 4 = very good, 5 = excellent) both before and after treatment.
Example 10.5.
Sign Test (1 of 5)
Observed data
Example 10.5.
Sign Test (2 of 5)
Difference scores
Example 10.5.
Sign Test (3 of 5)
Signs of the difference scores
NOTE: Randomly assign “+” or “–” when there are zeros.
Example 10.5.
Sign Test (4 of 5)
Example 10.5.
Sign Test (5 of 5)
Test statistic is 3.
Reject H0 if the smaller of the number of positive or negative signs ≤ 2 (Table 6).
Do not reject H0 because 3 > 2. We do not have significant evidence to show that there is a difference in QOL measured before versus after chemotherapy treatment.

Tests with Matched Samples: Wilcoxon Signed Rank Test (1 of 2)

Continuous outcome measured in matched or paired samples, differences are not assumed to follow a normal distribution.

Matched or paired samples

H0: Median difference is zero

H1: Median difference >, <, or ≠ 0
Tests with Matched Samples: Wilcoxon Signed Rank Test (2 of 2)
Test statistic is W, the smaller of W+ and W–, the sums of the positive and negative ranks of the differences scores.
Reject H0 if W ≤ critical value in Table 7.
Example 10.7.
Wilcoxon Signed Rank Test (1 of 6)
A study is run to evaluate the effectiveness of a new exercise program to reduce systolic blood pressure (SBP) in patients with pre-hypertension; n = 15 patients participate and have SBP measured before and after 6 weeks on the program.
Is there a significant difference in SBP after participating in the program?
Observed data
Example 10.7.
Wilcoxon Signed Rank Test (2 of 6)
Difference scores
Example 10.7.
Wilcoxon Signed Rank Test (3 of 6)
Ranks of the difference scores
Example 10.7.
Wilcoxon Signed Rank Test (4 of 6)
Signed ranks of the difference scores
W+ = 89 W– = 31
Example 10.7.
Wilcoxon Signed Rank Test (5 of 6)
Test statistic is W = 31.
Reject H0 if W ≤ 25 (Table 7).
Do not reject H0 because 31 > 25. We do not have significant evidence to show that the median difference in SBP is not zero.

Example 10.7.

Wilcoxon Signed Rank Test (6 of 6)

Tests with More Than Two Independent Samples: Kruskal–Wallis Test (1 of 2)

Continuous outcome that is not assumed to follow a normal distribution

k (k > 2) independent samples

H0: k population medians are equal

H1: k population medians are not all equal

Test statistic is H,

where k = number of groups, N = total sample size, nj = sample size in jth group, Rj = sum of the ranks in jth group.

Reject H0 if H ≥ critical value in Table 8.

Tests with More Than Two Independent Samples: Kruskal–Wallis Test (2 of 2)

Example 10.8.

Kruskal–Wallis Test (1 of 4)

A clinical study is run to assess differences in albumin levels in patients following 5%, 10%, and 15% protein diets.

H0: The three population medians are equal

H1: The three population medians are not equal

a = 0.05

Test statistic is H.

Rank data in pooled sample (n = 12), and compute R1, R2, and R3.

Example 10.8.

Kruskal–Wallis Test (2 of 4)

Example 10.8.

Kruskal–Wallis Test (3 of 4)

R1 = 7.5, R2 = 30.5, R3 = 40.

Test statistic is H.

Reject H0 if H ≥ 5.656 (Table 8).

Example 10.8.

Kruskal–Wallis Test (4 of 4)

Tests with More Than Two Independent Samples: Kruskal Wallis Test

Reject H0 because 7.52 > 5.656.

We have statistically significant evidence to show that there is a difference in median albumin levels among the three diets.

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