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Statistics for clinicians • Biostatistics course by Kevin E. Kip, Ph.D., FAHA Professor and Executive Director, Research Center University of South Florida, College of Nursing Professor, College of Public Health Department of Epidemiology and Biostatistics Associate Member, Byrd Alzheimer’s Institute Morsani College of Medicine Tampa, FL, USA 1 SECTION 3.8 SECTION 3.8 Guidelines and primary Guidelines and primary steps involved in steps involved in hypothesis testing, hypothesis testing, including “null” and the including “null” and the “alternative” hypothesis “alternative” hypothesis
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Statistics for clinicians• Biostatistics course by Kevin E. Kip, Ph.D., FAHA
Professor 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
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SECTION 3.8SECTION 3.8
Guidelines and primary steps Guidelines and primary steps involved in hypothesis involved in hypothesis testing, including “null” and testing, including “null” and the “alternative” hypothesisthe “alternative” hypothesis
Hypothesis Formulation
Scientific Method(not unique to health sciences)
--- Formulate a hypothesis
--- Test the hypothesis
Basic Strategy of Analytical Epidemiology
1. Identify variables you are interested in:• Exposure• Outcome
2. Formulate a hypothesis
3. Compare the experience of two groups of subjects with respect to the exposure and outcome
Basic Strategy of Analytical Epidemiology
Hypothesis Testing
• Two competing hypotheses--- “Null” hypothesis (no association) –
typically, but not always assumed--- “Alternative” hypothesis (postulates there
is an association)
• Hypothesis testing is based on probability theory and the Central Limit Theorem
Direct risk estimate(e.g. best estimate of risk of diseaseassociated with the exposure).
Hypothesis Formulation
Biostatistician:
H0: There is no association between theexposure and disease of interest
H1: There is an association between theexposure and disease of interest(beyond what might be expected from random error alone)
Hypothesis Formulation
Epidemiologist:
What is the best estimate of the risk of disease in those who are exposed compared to those who are unexposed (i.e. exposed are at XX times higher risk of disease).
This moves away from the simple dichotomy of yes or no for an exposure/disease association – to the estimated magnitude of effect irrespective of whether it differs from the null hypothesis.
The degree to which the rate of diseasein persons with a specific exposure iseither higher or lower than the rate ofdisease among those without thatexposure.
Hypothesis Formulation
Ways to Express Hypotheses:
1. Suggest possible events…
The incidence of tuberculosis will increase in the next decade.
Hypothesis Formulation
Ways to Express Hypotheses:
2. Suggest relationship between specificexposure and health-related event…
A high cholesterol intake is associatedwith the development (risk) of coronaryheart disease.
Hypothesis Formulation
Ways to Express Hypotheses:
3. Suggest cause-effect relationship….
Cigarette smoking is a cause of lung cancer
Hypothesis Formulation
Ways to Express Hypotheses:4. “One-sided” vs. “Two-sided”
One-sided example:Helicobacter pylori infection is associatedwith increased risk of stomach ulcer
Two-sided example:Weight-lifting is associated with risk oflower back injury
Hypothesis Formulation
Guidelines for Developing Hypotheses:
1. State the exposure to be measured as specifically as possible.
2. State the health outcome as specifically as possible.
Strive to explain the smallest amountof ignorance
Hypothesis Formulation
Example Hypotheses:
POOREating junk food is associated with the development of cancer.
GOODThe human papilloma virus (HPV) subtype 16 is associated with the development of cervical cancer.
SECTION 3.9SECTION 3.9
Parameters used in Parameters used in hypothesis testinghypothesis testing
Level of significance: A fixed value of the probability of rejectingthe null hypothesis (in favor of the alternative) when the nullhypothesis is actually true (i.e. type I or alpha (α) error rate)
Common levels of significance:0.10, 0.05, 0.01, 0.001
α = 0.05: The probability of incorrectly rejecting the null hypothesisin favor of the alternative is 5% when the null hypothesis is true.
P-value: A calculated probability of obtaining a test statistic at least as extreme as the one that was actually observed, assuming that thenull hypothesis is true.
A low p-value means that it is unlikely that the null hypothesis is actually true, and the alternative hypothesis should be considered (e.g. “reject” the null hypothesis).
1.0H0
0.750.500.250.0 2.0 3.0 4.0 5.0
H1 H1
Null hypothesis
Alternative hypothesis
Alternative hypothesis
RelativeRisk
RelativeRisk
A B C D E
PA < PB < PC > PD > PE
Refer to the figure below.Assume that 3 studies (A, B, and C) are conducted all with the same sample size. Which of the following is most likely true? Study B will have a lower p-value than Study CStudy C will have the lowest p-valueNone of the studies will be statistically significantNone of the above
1.0H0
0.750.500.250.0 2.0 3.0 4.0 5.0
H1 H1
Null hypothesis
Alternative hypothesis
Alternative hypothesis
EstimateFrom
Study A
EstimateFrom
Study B
EstimateFrom
Study C
RelativeRisk
RelativeRisk
Refer to the figure below.Assume that 3 studies (A, B, and C) are conducted all with the same sample size. Which of the following is most likely true? Study B will have a lower p-value than Study CStudy C will have the lowest p-valueNone of the studies will be statistically significantNone of the above
1.0H0
0.750.500.250.0 2.0 3.0 4.0 5.0
H1 H1
Null hypothesis
Alternative hypothesis
Alternative hypothesis
EstimateFrom
Study A
EstimateFrom
Study B
EstimateFrom
Study C
RelativeRisk
RelativeRisk
Interpreting Results
The p-value is NOT theindex of causality
It is an arbitrary quantitywith no direct relationship to biology
SECTION 3.10SECTION 3.10
Type I and Type II error and Type I and Type II error and factors that impact statistical factors that impact statistical powerpower
Interpreting Results
DECISION
H0 True
(No assoc.)
H1 True
(Yes assoc.)
Do not reject H0
(not stat. sig.)
Correct decision
Type II
(beta error)
Reject H0
(stat. sig.)
Type I
(alpha error)
Correct decision
Four possible outcomes of any epidemiologic study:
Interpreting Results
When evaluating the incidence of diseasebetween the exposed and non-exposed groups,we need guidelines to help determine whether
there is a true difference between the twogroups, or perhaps just random variation
from the study sample.
Interpreting Results
DECISION H0 True H1 True
Do not reject H0
(not stat. sig.)
Reject H0
(stat. sig.)
Type I
(alpha error)
“Conventional” Guidelines:• Set the fixed alpha level (Type I error) to 0.05This means, if the null hypothesis is true, theprobability of incorrectly rejecting it is 5% of less.The “p-value” is a measure of the compatibility ofthe data and the null hypothesis.
Although it appears that the incidence of disease may behigher in the exposed than in the non-exposed (RR=1.5),the p-value of 0.30 exceeds the fixed alpha level of 0.05.This means that the observed data are relativelycompatible with the null hypothesis. Thus, we do notreject H0 in favor of H1 (alternative hypothesis).
Interpreting Results
DECISION H0 True H1 True
Do not reject H0
(not stat. sig.)
Type II
(beta error)
Reject H0
(stat. sig.)
Conventional Guidelines:• Set the fixed beta level (Type II error) to 0.20This means, if the null hypothesis is false, theprobability of not rejecting it is 20% of less.The “power” of a study is (1 – beta). This meanshaving 80% probability to reject H0 when H1 is true.
Interpreting ResultsExample:
With the above sample size of 400, and if the alternativehypothesis is true, we need to expect a RR of about2.1 (power = 82%) or higher to be able to reject the null hypothesis in favor of the alternative hypothesis.
N Incid.
E- 200 0.10 0.10 0.10
E+ 200 0.18 0.21 0.24
RR 1.8 2.1 2.4
Power 58% 82% 95%
Interpreting ResultsFactors that affect the power of a study:
1. The fixed alpha level (the lower the level, thethe lower the power).
2. The total and within group sample sizes (thesmaller the sample size, the lower the power --unbalanced groups have lower power thanbalanced groups).
3. The anticipated effect size (the higher the expected/observed effect size, the higher the power).
Interpreting Results
Trade-offs between fixed alpha and beta levels:
Reducing the fixed alpha level (e.g. to < 0.01)is considered “conservative.” This reduces thelikelihood of a type I error (erroneouslyrejecting the null hypothesis), but at theexpense of increasing the probability of atype II error if the alternative hypothesis is true.
Interpreting Results
Trade-offs between fixed alpha and beta levels:
Increasing the fixed alpha level (e.g. to < 0.10)reduces the probability of a type II error(failing to reject H0 when H1 is true), but at theexpense of increasing the probability of atype I error if the null hypothesis is true.
Interpreting Results
Expos-ure N Incid.
Risk ratio
P-
value
Power
*
RR
**
None 1000 0.10 1.0 --- --- ---
Low 500 0.15 1.5 0.006 77% 1.52
Medium 250 0.15 1.5 0.02 60% 1.64
High 100 0.15 1.5 0.12 27% 2.08
* Power with given sample size and risk ratio ( = 0.05)** Risk ratio needed for 80% power with given sample size
SECTION 3.11SECTION 3.11
Calculate and interpret sample Calculate and interpret sample hypotheses – one sample hypotheses – one sample continuous outcomecontinuous outcome
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)
1. Hypothesis Testing – One Sample Continuous Outcome
Compare a “historical control” mean (µ0) from a population to a “sample” mean.
Example: Average annual health care expenses per person in the year 2005 (n=100, X = $3,190, s = $890) are lower than “historical” control costs in the year 2002 ($3,302)
1) Set up the hypothesis and determine the level of statistical significance (including 1 versus 2-sided hypothesis).
1. Hypothesis Testing – One Sample Continuous Outcome
Example: Average annual health care expenses per person in the year 2005 (n=100, X = $3,190, s = $890) are lower than “historical” control costs in the year 2002 ($3,302)
2) Select the appropriate test statistic.
If (n < 30), then use tIf (n > 30), then use z
1. Hypothesis Testing – One Sample Continuous Outcome
Example: Average annual health care expenses per person in the year 2005 (n=100, X = $3,190, s = $890) are lower than “historical” control costs in the year 2002 ($3,302)
3) Set up the decision rule (look up z value – Table 1c).
Reject H0 if z < -1.645
4) Compute the test statistic:
3190 - 3302z = --------------- = -1.26 890 / 100
5) Conclusion:-1.26 > -1.645 (critical value): Do not reject H0
Note: we cannot confirm the null hypothesis because perhaps the sample size was too small for a conclusive result (i.e. low power)
1. One Sample Continuous Outcome (Practice)
Compare historical “control” mean (µ0) from a population to a “sample” mean.
Example: Total cholesterol levels in the Framingham Heart Study in the year 2002 (n=3,310, X = 200.3, s = 36.8) are different than the national average in 2002 (203.0)
1) Set up the hypothesis and determine the level of statistical significance (including 1 versus 2-sided hypothesis).
Compare historical “control” mean (µ0) from a population to a “sample” mean.
Example: Total cholesterol levels in the Framingham Heart Study in the year 2002 (n=3,310, X = 200.3, s = 36.8) are different than the national average in 2002 (203.0)
1) Set up the hypothesis and determine the level of statistical significance (including 1 versus 2-sided hypothesis).
Example: Total cholesterol levels in the Framingham Heart Study in the year 2002 (n=3,310, X = 200.3, s = 36.8) are different than the national average in 2002 (203.0)
2) Select the appropriate test statistic.
If (n < 30), then use t If (n > 30), then use z
3) Set up the decision rule (look up ??? value – Table 1c).
Reject H0 if ______________________________
4) Compute the test statistic:
5) Conclusion:
1. One Sample Continuous Outcome (Practice)
Example: Total cholesterol levels in the Framingham Heart Study in the year 2002 (n=3,310, X = 200.3, s = 36.8) are different than the national average in 2002 (203.0)
2) Select the appropriate test statistic.
If (n < 30), then use t If (n > 30), then use z
3) Set up the decision rule (look up z value – Table 1c).
5) Conclusion: Reject H0 because -4.22 < -1.96What about statistical versus clinical significance?
SECTION 3.12SECTION 3.12
Calculate and interpret sample Calculate and interpret sample hypotheses – one sample hypotheses – one sample dichotomous outcomedichotomous outcome
2. Hypothesis Testing – One Sample Dichotomous Outcome
Compare a “historical control” proportion (p) from a population to a “sample” proportion.
Note: The example below assumes a “large” sample defined as:np0 > 5 and n(1-p0) > 5
If not, then “exact” methods must be used.
Example: The prevalence of smoking in the 2002 Framingham Heart Study (n=3,536, p = 482/3,536 = 0.1363) is lower than a national (“historical”) report in the year 2002 (p=0.211)
1) Set up the hypothesis and determine the level of statistical significance (including 1 versus 2-sided hypothesis).
H0: p = 0.211H1: p < 0.211 (one-sided hypothesis, lower-tailed test) α = 0.05
2. Hypothesis Testing – One Sample Dichotomous Outcome
Example: The prevalence of smoking in the 2002 Framingham Heart Study (n=3,536, p = 482/3,536 = 0.1363) is lower than a national (“historical”) report in the year 2002 (p=0.211)
2) Select the appropriate test statistic.
p – p0
z = ---------------- p0(1-p0) / n
2. Hypothesis Testing – One Sample Dichotomous Outcome
Example: The prevalence of smoking in the 2002 Framingham Heart Study (n=3,536, p = 482/3,536 = 0.1363) is lower than a national (“historical”) report in the year 2002 (p=0.211)
3) Set up the decision rule (look up z value – Table 1c).
Compare historical “control” proportion (p) from a population to a “sample” mean.
Example: The prevalence of diabetes in a Tampa-based study of adults in 2002 (n=1,240, p = 108/1,240 = 0.0871) is different than a national (“historical”) report in the year 2002 (p=0.1082)
1) Set up the hypothesis and determine the level of statistical significance (including 1 versus 2-sided hypothesis).
H0: p = ______H1: p ______ (???-sided hypothesis, ???-tailed test) α = 0.05
2. One Sample Dichotomous Outcome (Practice)
Compare historical “control” proportion (p) from a population to a “sample” mean.
Example: The prevalence of diabetes in a Tampa-based study of adults in 2002 (n=1,240, p = 108/1,240 = 0.0871) is different than a national (“historical”) report in the year 2002 (p=0.1082)
1) Set up the hypothesis and determine the level of statistical significance (including 1 versus 2-sided hypothesis).
H0: p = 0.1082H1: p = 0.1082 (2-sided hypothesis, 2-tailed test) α = 0.05
2. One Sample Dichotomous Outcome (Practice)
Example: The prevalence of diabetes in a Tampa-based study of adults in 2002 (n=1,240, p = 108/1,240 = 0.0871) is different than a national (“historical”) report in the year 2002 (p=0.1082)
2) Select the appropriate test statistic (assume z – large sample)
3) Set up the decision rule (look up z value – Table 1c).
Reject H0 if _______________________
4) Compute the test statistic:
p – p0
z = ---------------- p0(1-p0) / n
5) Conclusion:
2. One Sample Dichotomous Outcome (Practice)
Example: The prevalence of diabetes in a Tampa-based study of adults in 2002 (n=1,240, p = 108/1,240 = 0.0871) is different than a national (“historical”) report in the year 2002 (p=0.1082)
2) Select the appropriate test statistic (assume z – large sample)
3) Set up the decision rule (look up z value – Table 1c).
Calculate and interpret sample Calculate and interpret sample hypotheses – one sample hypotheses – one sample categorical/ordinal outcomecategorical/ordinal outcome
3. One Sample Categorical/Ordinal Outcome
Goal is to compare the proportion of subjects in 3 or more
categories to proportions from a known distribution
Forms the basis of comparing “observed” (O) versus “expected” (E)
frequency counts
Expected frequencies (counts) are determined by multiplying the
observed sample size (n) by the proportions specified in the null
hypothesis (p10, p20,…. pk0)
The test is a comparison of distribution of responses to a χ2
distribution and “goodness of fit” test.
3. One Sample Categorical/Ordinal Outcome
Example: Compare proportions of levels of exercise among a sample of college students at USF to a national sample (historical)
No regular exercise
Sporadic exercise
Regular exercise Total
n 6,000 2,500 1,500 10,000
p 0.60 0.25 0.15 1.0
National Sample
No regular exercise
Sporadic exercise
Regular exercise Total
n 255 125 90 470
p 0.5426 0.2660 0.1915 1.0
USF
1) Set up the hypothesis and determine the level of statistical significance
H0: p1 = 0.60, p2 = 0.25, p3 = 0.15H0: Distribution of responses is 0.60, 0.25, 0.15H1: H0 is False (distribution of responses are similar) α = 0.05
3. One Sample Categorical/Ordinal Outcome
No regular exercise
Sporadic exercise
Regular exercise Total
n 6,000 2,500 1,500 10,000
p 0.60 0.25 0.15 1.0
National Sample
No regular exercise
Sporadic exercise
Regular exercise Total
n 255 125 90 470
p 0.5426 0.2660 0.1915 1.0
USF
2) Select the appropriate test statistic
3) Set up the decision ruled.f. = k(categories) – 1, so 3 categories – 1 = 2; α = 0.05
Refer to Table 3 in Appendix: critical value = 5.99
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