Statistics for clinicians
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Transcript of Statistics for clinicians
Statistics for clinicians
Biostatistics course by Kevin E. Kip, Ph.D., FAHAProfessor and Executive Director, Research CenterUniversity of South Florida, College of NursingProfessor, College of Public HealthDepartment of Epidemiology and BiostatisticsAssociate Member, Byrd Alzheimer’s InstituteMorsani College of MedicineTampa, FL, USA
1
SECTION 4.1SECTION 4.1
Module OverviewModule Overviewand Introductionand Introduction
Hypothesis testing for 2 or more independent groups and non-parametric methods.
Module 4 Learning Outcomes:
1. Calculate and interpret 2 sample hypotheses:a) 2 sample – continuous outcomeb) >2 samples – continuous outcomec) 2 sample dichotomous outcomed) >2 samples dichotomous outcome
2. Specify 2-sample hypotheses and conduct formal testing using SPSS
3. Differentiate between parametric and non-parametric tests
4. Identify properties of non-parametric tests
Module 4 Learning Outcomes:
5. Calculate and interpret non-parametric tests:a) 2 independent samples – Wilcoxon Rank Sum
Testb) Matched samples – Wilcoxon Signed Rank Testc) >2 independent samples – Kruskal Wallis Test
6. Conduct and interpret non-parametric analyses using SPSS.
Assigned Reading:
Textbook: Essentials of Biostatistics in Public Health
Chapter 7Sections 7.5, 7.7 to 7.9Pages 138-141 and 144-162
Chapter 10
General Steps for Hypothesis Testing:
1)Set up the hypothesis and determine the level of
statistical significance (including 1 versus 2-sided
hypothesis).
2)Select the appropriate test statistic
3)Set up the decision rule
4)Compute the test statistic
5)Conclusion (interpretation)
Hypothesis Testing Calculations:
1)Two Sample – Independent Groups
a) Continuous outcome (student t test)b) Dichotomous outcome (risk difference or
risk ratio—chi-square test)
2)More than 2 Samples – Independent Groups
a) Continuous outcome (analysis of variance-ANOVA)
b) Categorical Outcome (chi-square test)
Framework of Hypothesis Testing
Goal is to compare sample parameter estimates (e.g. mean,
proportion, etc.) between 2 or more independent groups.
The groups can be defined from a clinical trial, such as treatment
versus placebo, or an observational study, such as men versus
women, or exposed versus not exposed.
With 2 groups, one group serves as the “comparison” or “control”
group representing the null value.
Groups do not need to be of the same size.
With more than 2 groups, can compare whether any groups differ
(e.g. means) or whether groups differ in an ordered manner.
SECTION 4.3SECTION 4.3
Two-sample: independent Two-sample: independent groups – continuous outcome groups – continuous outcome
1. Two-Sample: Independent Groups-Continuous Outcome
Parameter: Difference in population means: μ1 – μ2
H0: μ1 – μ2 = 0; μ1 = μ2
H1: μ1 > μ2;μ1 < μ2;μ1 = μ2;
Test statistics:
n1 > 30 and n2 > 30
n1 < 30 or n2 < 30
Critical value of zin Table 1C
Critical value of tin table 2
d.f. = n1 + n2 - 2
1. Two-Sample: Independent Groups-Continuous Outcome
Example: From the Framingham Heart Study (offspring), compare mean systolic blood pressure between men and women.
n X sMen 1623 128.2 17.5Women 1911 126.5 20.1
1) Set up the hypothesis and determine the level of statistical significance (including 1 versus 2-sided hypothesis).
H0: μ1 = μ2
H1: μ1 = μ2 (two-sided hypothesis) α = 0.05
1. Two-Sample: Independent Groups-Continuous Outcome
Example: From the Framingham Heart Study (offspring), compare mean systolic blood pressure between men and women.
n X sMen 1623 128.2 17.5Women 1911 126.5 20.1
2)Select the appropriate test statistic:n1 > 30 and n2 > 30, so use z
3)Set up the decision rule:Reject H0 if z < 1.96 or z > 1.96
1. Two-Sample: Independent Groups-Continuous Outcome
Example: From the Framingham Heart Study (offspring), compare mean systolic blood pressure between men and women.
n X sMen 1623 128.2 17.5Women 1911 126.5 20.1
4) Compute the test statistic:
= sqrt(359.12) = 19.0
5) Conclusion: Reject H0 because 2.66 > 1.96
1. Two-Sample: Independent Groups-Continuous Outcome (Practice)
Example: From the Heart SCORE Study, compare mean total cholesterol levels between men and women. (α = 0.05)
n X sMen 165 198.88 38.416Women 337 222.23 42.023
1) Set up the hypothesis and determine the level of statistical significance (including 1 versus 2-sided hypothesis).
H0: _________________H1: _________________
2)Select the appropriate test statistic:n1 > 30 and n2 > 30, so use z __________________
3)Set up the decision rule:_________________________________________
1. Two-Sample: Independent Groups-Continuous Outcome (Practice)
Example: From the Heart SCORE Study, compare mean total cholesterol levels between men and women. (α = 0.05)
n X sMen 165 198.88 38.416Women 337 222.23 42.023
1) Set up the hypothesis and determine the level of statistical significance (including 1 versus 2-sided hypothesis).
H0: μ1 = μ2
H1: μ1 = μ2 (two-sided hypothesis)
2)Select the appropriate test statistic:n1 > 30 and n2 > 30, so use z
3)Set up the decision rule:Reject H0 if z < 1.96 or z > 1.96
1. Two-Sample: Independent Groups-Continuous Outcome (Practice)
Example: From the Heart SCORE Study, compare mean total cholesterol levels between men and women. (α = 0.05)
n X sMen 165 198.88 38.416Women 337 222.23 42.023
4)Compute the test statistic
5) Conclusion: _________________________________
1. Two-Sample: Independent Groups-Continuous Outcome (Practice)
Example: From the Heart SCORE Study, compare mean total cholesterol levels between men and women. (α = 0.05)
n X sMen 165 198.88 38.416Women 337 222.23 42.023
4)Compute the test statistic
165 + 337 - 2
(165–1)(38.416)2 + (337–1)(42.023)2
Sp = = 40.875
198.88 – 222.23 -23.35z = ----------------------- = ------- = -6.01 40.875 1/165 + 1/337 3.884
5) Conclusion: Reject H0: abs(-6.01) > 1.96
1. Two-Sample: Independent Groups-Continuous Outcome (Practice)
Example: From the Heart SCORE Study, compare mean total cholesterol levels between men and women. (α = 0.05)
SPSS
AnalyzeCompare Means
Independent Samples T-TestTest Variable: Total cholesterolGroup Variable: Gender (defined as 1,2)
Options: 95% C.I.
SECTION 4.4SECTION 4.4
Two-sample: independent Two-sample: independent groups – dichotomous groups – dichotomous outcome outcome
2. Two-Sample: Independent Groups-Dichotomous Outcome
Parameter: Risk Difference (RD)(p1 – p2) or Risk Ratio (RR)(p1 / p2)
H0: RD: p1 = p2; or p1 – p2 = 0; RR: p1 / p2 = 1.0 H1: RD: p1 = p2; or p1 – p2 = 0; RR: p1 / p2 = 1.0
Test statistics: Critical valueof z in
Table 1C
min[n1p1, n1(1 – p1)] > 5
min[n2p2, n2(1 – p2)] > 5
Note: p = proportion of successes (outcomes)
2. Two-Sample: Independent Groups-Dichotomous Outcome
Example: From the Framingham Heart Study (offspring), compare the prevalence of CVD between smokers and non-smokers.
No CVD CVD TotalSmoker 663 81 744 p1 = 81/744 = 0.1089 Non-smoker 2757 298 3055 p2 = 298/3055 = 0.0975
(RD)(p1 – p2 = 0.0114); Risk Ratio (RR)(p1 / p2 = 1.12)
1) Set up the hypothesis and determine the level of statistical significance (including 1 versus 2-sided hypothesis).
H0: p1 = p2
H1: p1 = p2 (two-sided hypothesis) α = 0.05
2. Two-Sample: Independent Groups-Dichotomous Outcome
Example: From the Framingham Heart Study (offspring), compare the prevalence of CVD between smokers and non-smokers.
No CVD CVD TotalSmoker 663 81 744 p1 = 81/744 = 0.1089 Non-smoker 2757 298 3055 p2 = 298/3055 = 0.0975
2) Select the appropriate test statistic: min[n1p1, n1(1 – p1)] > 5min[n2p2, n2(1 – p2)] > 5 --- use z
3) Set up the decision rule:Reject H0 if z < 1.96 or z > 1.96
2. Two-Sample: Independent Groups-Dichotomous Outcome
Example: From the Framingham Heart Study (offspring), compare the prevalence of CVD between smokers and non-smokers.
No CVD CVD TotalSmoker 663 81 744 p1 = 81/744 = 0.1089 Non-smoker 2757 298 3055 p2 = 298/3055 = 0.0975
4) Compute the test statistic:
81 + 298 379p = ---------------- = -------- = 0.0988 744 + 3055 3799
0.1089 – 0.0975z = -------------------------------------------- = 0.927 0.0988(1 – 0.0988)(1/744 + 1/3055)
5) Conclusion: Do not reject H0: -1.96 < 0.927 < 1.96
2. Two-Sample: Independent Groups-Dichotomous Outcome (Practice)
Example: From the Heart SCORE Study, compare the prevalence of diabetes by level of weekly exercise.
Exercise No diabetes Diabetes Total< 3 times/wk 177 18 195 p1 = _____________ > 3 times/wk 278 23 301 p2 = _____________
(RD) (p1 – p2 = _______); Risk Ratio (RR) (p1 / p2 = _______)
1) Set up the hypothesis and determine the level of statistical significance (including 1 versus 2-sided hypothesis).
H0: _____________________________H1: _____________________________ α = 0.05
2. Two-Sample: Independent Groups-Dichotomous Outcome (Practice)
Example: From the Heart SCORE Study, compare the prevalence of diabetes by level of weekly exercise.
Exercise No diabetes Diabetes Total< 3 times/wk 177 18 195 p1 = 18/195 = 0.0923 > 3 times/wk 278 23 301 p2 = 23/301 = 0.0764
(RD) (p1 – p2 = 0.0159); Risk Ratio (RR) (p1 / p2 = 1.21)
1) Set up the hypothesis and determine the level of statistical significance (including 1 versus 2-sided hypothesis).
H0: p1 = p2
H1: p1 = p2 (two-sided hypothesis) α = 0.05
2. Two-Sample: Independent Groups-Dichotomous Outcome (Practice)
Example: From the Heart SCORE Study, compare the prevalence of diabetes by level of weekly exercise.
Exercise No diabetes Diabetes Total<3 times/wk 177 18 195 p1 = _______ >3 times/wk 278 23 301 p2 = _______
2) Select the appropriate test statistic: min[n1p1, n1(1 – p1)] > 5min[n2p2, n2(1 – p2)] > 5 --- use z
3) Set up the decision rule:Reject H0 if: ________________________
2. Two-Sample: Independent Groups-Dichotomous Outcome (Practice)
Example: From the Heart SCORE Study, compare the prevalence of diabetes by level of weekly exercise.
Exercise No diabetes Diabetes Total<3 times/wk 177 18 195 p1 = 18/195 = 0.0923 >3 times/wk 278 23 301 p2 = 23/301 = 0.0764
2) Select the appropriate test statistic: min[n1p1, n1(1 – p1)] > 5min[n2p2, n2(1 – p2)] > 5 --- use z
3) Set up the decision rule:Reject H0 if z < 1.96 or z > 1.96
2. Two-Sample: Independent Groups-Dichotomous Outcome (Practice)
Example: From the Heart SCORE Study, compare the prevalence of diabetes by level of weekly exercise.
Exercise No diabetes Diabetes Total<3 times/wk 177 18 195 p1 = _______ >3 times/wk 278 23 301 p2 = _______
4) Compute the test statistic:
p = ______________ = __________
z = __________________________________
5) Conclusion: ________________________________
2. Two-Sample: Independent Groups-Dichotomous Outcome (Practice)
Example: From the Heart SCORE Study, compare the prevalence of diabetes by level of weekly exercise.
Exercise No diabetes Diabetes Total<3 times/wk 177 18 195 p1 = 18/195 = 0.0923 >3 times/wk 278 23 301 p2 = 23/301 = 0.0764
4) Compute the test statistic:
18 + 23 41p = ---------------- = -------- = 0.0827 195 + 301 496
0.0923 – 0.0764z = ------------------------------------------- = 0.628 0.0827(1 – 0.0827)(1/195 + 1/301)
5) Conclusion: Do not reject H0: -1.96 < 0.628 < 1.96
2. Two-Sample: Independent Groups-Dichotomous Outcome (Practice)
Example: From the Heart SCORE Study, compare the prevalence of diabetes by level of weekly exercise (α = 0.05)
SPSS
AnalyzeDescriptive Statistics
CrosstabsRow Variable: Exercise >times/weekColumn Variable: History of diabetes
Statistics – Chi-squareCells – Observed, Expected
Note: Pearson chi-square test in SPSS includes Yates correction.
SECTION 4.5SECTION 4.5
More than two-samples: More than two-samples: independent groups – independent groups – continuous outcome continuous outcome
3. More Than Two Independent Groups-Continuous Outcome
Parameter: Difference in means for more than 2 groups (ANOVA)
H0: μ1 = μ2 = …μk
H1: Means are not all equal Test statistic: F value
Find critical value in Table 4(df1 = k – 1; df2 = N – k)
Where nj = sample size in the jth group (e.g. j = 1,2,3……)
Xj = mean in the jth group
X = overall mean
k = number of independent groups (k > 2)
N = total number of observations in analysis
ANOVA Assumptions: Outcome follows a normal distribution (all groups)Variances approximately equal among groups
3. More Than Two Independent Groups-Continuous Outcome
“Between-group” variabilityF = ---------------------------------------- “Residual or error” variability (“within-group” variability) (i.e. variability in the outcome); (null hypothesis is that all groups are random samples)
F statistic assesses whether differences among the means (the numerator) are larger than expected by chance.
F statistic has 2 degrees of freedom; df1 (numerator), df2 (denominator)df1 = k – 1; df2 = N – k
Table 4 contains critical values for the F distribution
3. More Than Two Independent Groups-Continuous Outcome
Analysis of Variance (ANOVA) Table
Source of Variation
Sum of Squares (SS)
Degrees of freedom (df)
Mean Squares (MS) F
Between-groupSSB
k - 1 SSB
MSB = ---------- k – 1
MSBF = ------- MSE
Error or residual(random)
“within-group”
SSE
N - k SSE*
MSE = ---------- N – k
------
TotalSST
N - 1 ------ ------
*Textbook on page 150 has typographical error
3. More Than Two Independent Groups-Continuous Outcome
Low Calorie Low Fat Low Carb Control
8 2 3 2
9 4 5 2
6 3 4 -1
7 5 2 0
3 1 3 3
n1 = 5 n2 = 5 n3 = 5 n4 = 5
X1 = 6.6 X2 = 3.0 X3 = 3.4 X4 = 1.2
Example: Weight Loss by Treatment (in Pounds)
1) Set up the hypothesis and determine level of statistical significance
H0: µ1 = µ2 = µ3 = µ4 H1: Means are not all equal; α = 0.05
2) Select the appropriate test statistic
3. More Than Two Independent Groups-Continuous Outcome
Low Calorie Low Fat Low Carb Control
n1 = 5 n2 = 5 n3 = 5 n4 = 5
X1 = 6.6 X2 = 3.0 X3 = 3.4 X4 = 1.2
Example: Weight Loss by Treatment (in Pounds)
3) Set up the decision rule --- see critical value in Table 4df1 = k – 1 = 4 – 1 = 3df2 = N – k = 20 – 4 = 16
Reject H0 if F > 3.24
3. More Than Two Independent Groups-Continuous Outcome
Low Calorie Low Fat Low Carb Control
n1 = 5 n2 = 5 n3 = 5 n4 = 5
X1 = 6.6 X2 = 3.0 X3 = 3.4 X4 = 1.2
Example: Weight Loss by Treatment (in Pounds)
4) Compute test statistic: SSB = SSE =(ANOVA table) MSB = SSB / (k – 1) MSE = SSE / (N – k)
F = MSB / MSE
SSB = 5(6.6 – 3.6)2 + 5(3.0 – 3.6)2 + 5(3.4 – 3.6)2 + 5(1.2 – 3.6)2
= 45.0 + 1.8 + 0.2 + 28.8 = 75.8
SSE = 21.4 + 10.0 + 5.4 + 10.6 = 47.4 (see tables 7-24 to 7-28, page 151 of text)
MSB = 75.8 / (4 – 1) = 25.3
MSE = 47.4 / (20 – 4) = 3.0
F = 25.3 / 3.0 = 8.43
5) Conclusion: Reject H0; 8.43 > 3.24
3. More Than Two Independent Groups-Continuous Outcome
Low Calorie Low Fat Low Carb Control
n1 = 5 n2 = 5 n3 = 5 n4 = 5
X1 = 6.6 X2 = 3.0 X3 = 3.4 X4 = 1.2
Example: Weight Loss by Treatment (in Pounds)
ANOVA orthogonal contrasts of mean:Sometimes, rather than just comparing a difference among all means, we wish to compare specific means or whether the means increase or decrease in a monotonic (linear) manner. This can be achieved with orthogonal contrasts of the means.Sum of coefficients in each linear contrast must equal zeroIn the example above:
µ1 versus (µ2, µ3, µ4) -3 1 1 1(µ1, µ2) versus (µ3, µ4) -1 -1 1 1(µ1, µ2, µ3) versus µ4 -1 -1 -1 3linear trend -2 -1 1 2
3. More Than Two Independent Groups-Continuous Outcome (Practice)
Normal Pre-hypertensive Htn Stage I Htn Stage II
n1 = 88 n2 = 191 n3 = 139 n4 = 55
X1 = 28.42 X2 = 29.43 X3 = 30.75 X4 = 33.39
s = 5.37 s = 5.75 s = 5.89 s = 6.39
Example: Body Mass Index by Blood Pressure Classification
1) Set up the hypothesis and determine level of statistical significance
H0: __________________________________H1: ___________________________________ α = 0.05
2) Select the appropriate test statistic: __________________
3. More Than Two Independent Groups-Continuous Outcome (Practice)
Normal Pre-hypertensive Htn Stage I Htn Stage II
n1 = 88 n2 = 191 n3 = 139 n4 = 55
X1 = 28.42 X2 = 29.43 X3 = 30.75 X4 = 33.39
s = 5.37 s = 5.75 s = 5.89 s = 6.39
Example: Body Mass Index by Blood Pressure Classification
1) Set up the hypothesis and determine level of statistical significance
H0: µ1 = µ2 = µ3 = µ4
H1: Means are not all equal;H1: Means increase or decrease in a monotonic (linear) manner; α = 0.05
2) Select the appropriate test statistic
3. More Than Two Independent Groups-Continuous Outcome (Practice)
Normal Pre-hypertensive Htn Stage I Htn Stage II
n1 = 88 n2 = 191 n3 = 139 n4 = 55
X1 = 28.42 X2 = 29.43 X3 = 30.75 X4 = 33.39
s = 5.37 s = 5.75 s = 5.89 s = 6.39
Example: Body Mass Index by Blood Pressure Classification
3) Set up the decision rule --- see critical value in Table 4
df1 = k – 1 = ___________df2 = N – k = ___________
http://www.danielsoper.com/statcalc3/calc.aspx?id=4
Reject H0 if: _______________________
Total N = n1 + n2 + n3 + n4 = ___________________________
3. More Than Two Independent Groups-Continuous Outcome (Practice)
Normal Pre-hypertensive Htn Stage I Htn Stage II
n1 = 88 n2 = 191 n3 = 139 n4 = 55
X1 = 28.42 X2 = 29.43 X3 = 30.75 X4 = 33.39
s = 5.37 s = 5.75 s = 5.89 s = 6.39
Example: Body Mass Index by Blood Pressure Classification
3) Set up the decision rule --- see critical value in Table 4
df1 = k – 1 = 4 – 1 = 3df2 = N – k = 473 – 4 = 469
http://www.danielsoper.com/statcalc3/calc.aspx?id=4
Reject H0 if F > 2.62
Total N = n1 + n2 + n3 + n4 = 88 + 191 + 139 + 55 = 473
3. More Than Two Independent Groups-Continuous Outcome (Practice)
Normal Pre-hypertensive Htn Stage I Htn Stage II
n1 = 88 n2 = 191 n3 = 139 n4 = 55
X1 = 28.42 X2 = 29.43 X3 = 30.75 X4 = 33.39
s = 5.37 s = 5.75 s = 5.89 s = 6.39
Example: Body Mass Index by Blood Pressure Classification
Source of Variation
Sum of Squares (SS)
Degrees of freedom (df)
Mean Squares (MS) F
Between-groupSSB990.9
k – 1________
SSBMSB = ------ = _____ k – 1
MSBF = ------- MSE
Error or residual(random)
“within-group”
SSE15766.2
N – k_______
SSEMSE = ------- = _____ N – k
F = _______
TotalSST
16757.1N – 1
_______ ------ ------
4. Compute the test statistic F = ____________ N = ____
5. Conclusion: ___________________________
3. More Than Two Independent Groups-Continuous Outcome (Practice)
Normal Pre-hypertensive Htn Stage I Htn Stage II
n1 = 88 n2 = 191 n3 = 139 n4 = 55
X1 = 28.42 X2 = 29.43 X3 = 30.75 X4 = 33.39
s = 5.37 s = 5.75 s = 5.89 s = 6.39
Example: Body Mass Index by Blood Pressure Classification
Source of Variation
Sum of Squares (SS)
Degrees of freedom (df)
Mean Squares (MS) F
Between-groupSSB990.9
k – 13
SSBMSB = ------ = 330.3 k – 1
MSBF = ------- MSE
Error or residual(random)
“within-group”
SSE15766.2
N – k469
SSEMSE = ------- = 33.6 N – k
F = __9.83__
TotalSST
16757.1N – 1
472 ------ ------
4. Compute the test statistic F = MSB / MSE N = 473
5. Conclusion: Reject H0: 9.83 > 2.62 Linear trend: F=11.27
3. More Than Two Independent Groups-Continuous Outcome (Practice)
Example: Body mass index and blood pressure classification in the Heart SCORE Study (α = 0.05)
SPSS
AnalyzeCompare Means
One-Way ANOVADependent Variable: Body mass indexGroup Variable (Factor): Blood pressure class
Contrasts -2 -1 1 2 Options:
DescriptiveHomogeneity of variance testMeans plot