Statistic in Health Care Management Week 4
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chapter7.pdf
Chapter 7
Hypothesis Testing Procedures
Learning Objectives
• Define null and research hypothesis, test
statistic, level of significance and decision rule
• Distinguish between Type I and Type II errors
and discuss the implications of each
• Explain the difference between one- and two-
sided tests of hypothesis
Learning Objectives
• Estimate and interpret p-values
• Explain the relationship between confidence interval
estimates and p-values in drawing inferences
• Perform analysis of variance by hand
• Appropriately interpret the results of analysis of
variance tests
• Distinguish between one and two factor analysis of
variance tests
Learning Objectives
• Perform chi-square tests by hand
• Appropriately interpret the results of chi-square tests
• Identify the appropriate hypothesis testing procedures
based on type of outcome variable and number of
samples
Hypothesis Testing
• Research hypothesis is generated about
unknown population parameter
• Sample data are analyzed and determined to
support or refute the research hypothesis
Hypothesis Testing Procedures
Step 1
Null hypothesis (H0):
No difference, no change
Research hypothesis (H1):
What investigator
believes to be true
Hypothesis Testing Procedures
Step 2
Collect sample data and determine whether sample
data support research hypothesis or not.
For example, in test for m, evaluate . X
Hypothesis Testing Procedures
Step 3
• Set up decision rule to decide when to believe null
versus research hypothesis
• Depends on level of significance, a = P(Reject H0|H0 is true)
Hypothesis Testing Procedures
Steps 4 and 5
• Summarize sample information in test statistic (e.g.,
Z value)
• Draw conclusion by comparing test statistic to
decision rule. Provide final assessment as to whether
H1 is likely true given the observed data.
P-values
• P-values represent the exact significance of the data
• Estimate p-values when rejecting H0 to summarize significance of the data (can approximate with statistical tables, can get exact value with statistical computing package)
• P-value is the smallest a where we still reject H0
Hypothesis Testing Procedures
1. Set up null and research hypotheses, select a
2. Select test statistic
2. Set up decision rule
3. Compute test statistic
4. Draw conclusion & summarize significance
Errors in Hypothesis Tests
Hypothesis Testing for m
• Continuous outcome
• 1 Sample
H0: m=m0
H1: m>m0, m<m0, m≠m0
Test Statistic
n>30 (Find critical
value in Table 1C,
n<30 Table 2, df=n-1)
ns/
μ-X Z
0 =
ns/
μ-X t
0 =
Example 7.2.
Hypothesis Testing for m
The National Center for Health Statistics (NCHS)
reports the mean total cholesterol for adults is 203. Is
the mean total cholesterol in Framingham Heart
Study participants significantly different?
In 3310 participants the mean is 200.3 with a standard
deviation of 36.8.
Example 7.2.
Hypothesis Testing for m
1. H0: m=203
H1: m≠203 a=0.05
2. Test statistic
3. Decision rule
Reject H0 if z > 1.96 or if z < -1.96
ns/
μ-X Z
0 =
Example 7.2.
Hypothesis Testing for m
4. Compute test statistic
5. Conclusion. Reject H0 because -4.22 <-1.96. We have
statistically significant evidence at a=0.05 to show that
the mean total cholesterol is different in the Framingham
Heart Study participants.
22.4 3310/8.36
2033.200
ns/
μ-X Z 0 =
==
Example 7.2.
Hypothesis Testing for m
Significance of the findings. Z = -4.22.
Table 1C. Critical Values for Two-Sided Tests
a Z
0.20 1.282
0.10 1.645
0.05 1.960
0.010 2.576
0.001 3.291
0.0001 3.819 p<0.0001.
New Scenario
• Outcome is dichotomous (p=population proportion)
– Result of surgery (success, failure)
– Cancer remission (yes/no)
• One study sample
• Data
– On each participant, measure outcome (yes/no)
– n, x=# positive responses,
n
x p̂ =
Hypothesis Testing for p
• Dichotomous outcome
• 1 Sample
H0: p=p0 H1: p>p0, p<p0, p≠p0
Test Statistic
(Find critical value in Table 1C)
5)]pn(1,min[np 00
n
)p-(1p
p-p̂ Z
00
0 =
Example 7.4.
Hypothesis Testing for p
The NCHS reports that the prevalence of cigarette smoking among adults in 2002 is 21.1%. Is the prevalence of smoking lower among participants in the Framingham Heart Study?
In 3536 participants, 482 reported smoking.
Example 7.2.
Hypothesis Testing for p
1. H0: p=0.211
H1: p<0.211 a=0.05
2. Test statistic
3. Decision rule
Reject H0 if z < -1.645
n
)p-(1p
p-p̂ Z
00
0 =
Example 7.2.
Hypothesis Testing for p
4. Compute test statistic
5. Conclusion. Reject H0 because -10.93 < -1.645. We have
statistically significant evidence at a=0.05 to show that the
prevalence of smoking is lower among the Framingham Heart
Study participants. (p<0.0001)
93.10
3536
)211.01(211.0
211.0136.0
n
)p-(1p
p-p̂ Z
00
0 =
==
Hypothesis Testing for Categorical and
Ordinal Outcomes*
• Categorical or ordinal outcome
• 1 Sample
H0: p1=p10, p2=p20,…,pk=pk0 H1: H0 is false
Test Statistic
(Find critical value in Table 3,
df=k-1)
* c2 goodness-of-fit test
= E
E)-(O χ
2 2
Chi-Square Tests
c2 tests are based on the agreement between
expected (under H0) and observed (sample)
frequencies.
Test statistic
E
)E - (O Σ = χ
2 2
Chi-Square Distribution
If H0 is true c 2 will be close to 0, if H0 is false, c
2 will
be large
Reject H0 if c 2 > Critical Value from Table 3
Example 7.6.
c2 goodness-of-fit test
A university survey reveals that 60% of students get
no regular exercise, 25% exercise sporadically and
15% exercise regularly. The university institutes a
health promotion campaign and re-evaluates exercise
one year later.
None Sporadic Regular
Number of students 255 125 90
Example 7.6.
c2 goodness-of-fit test
1. H0: p1=0.60, p2=0.25,p3=0.15
H1: H0 is false a=0.05
2. Test statistic
3. Decision rule df=k-1=3-1=2
Reject H0 if c 2 > 5.99
= E
E)-(O χ
2 2
Example 7.6.
c2 goodness-of-fit test
4. Compute test statistic
None Sporadic Regular Total
No. students (O) 255 125 90 470
Expected (E) 282 117.5 70.5 470
(O-E)2/E 2.59 0.48 5.39
c2 = 8.46
= E
E)-(O χ
2 2
Example 7.6.
c2 goodness-of-fit test
5. Conclusion. Reject H0 because 8.46 > 5.99. We have
statistically significant evidence at a=0.05 to show that
the distribution of exercise is not 60%, 25%, 15%.
Using Table 3, the p-value is p<0.005.
New Scenario
• Outcome is continuous
– SBP, Weight, cholesterol
• Two independent study samples
• Data
– On each participant, identify group and measure
outcome
– )s(ors,X,n),s(ors,X,n 2
2
2221
2
111
Two Independent Samples
RCT: Set of Subjects Who Meet
Study Eligibility Criteria
Randomize
Treatment 1 Treatment 2
Mean Trt 1 Mean Trt 2
Two Independent Samples
Cohort Study - Set of Subjects Who
Meet Study Inclusion Criteria
Group 1 Group 2
Mean Group 1 Mean Group 2
Hypothesis Testing for (m1m2)
• Continuous outcome
• 2 Independent Sample
H0: m1=m2 (m1m2 = 0)
H1: m1>m2, m1<m2, m1≠m2
Hypothesis Testing for (m1m2)
• Continuous outcome
• 2 Independent Sample
H0: m1=m2
H1: m1>m2, m1<m2, m1≠m2
Test Statistic
n1>30 and (Find critical value
n2> 30 in Table 1C,
n1<30 or Table 2,
df=n1+n2-2)
n2<30
21
21
n
1
n
1 Sp
X - X Z
=
21
21
n
1
n
1 Sp
X - X t
=
Pooled Estimate of Common Standard
Deviation, Sp
• Previous formulas assume equal variances
(s1 2=s2
2)
• If 0.5 < s1 2/s2
2 < 2, assumption is reasonable
2nn
1)s(n1)s(n Sp
21
2
22
2
11
=
Example 7.9.
Hypothesis Testing for (m1m2)
A clinical trial is run to assess the effectiveness of a new drug in lowering cholesterol. Patients are randomized to receive the new drug or placebo and total cholesterol is measured after 6 weeks on the assigned treatment.
Is there evidence of a statistically significant reduction in cholesterol for patients on the new drug?
Example 7.9.
Hypothesis Testing for (m1m2)
Sample SizeMean Std Dev
New Drug 15 195.9 28.7
Placebo 15 227.4 30.3
Example 7.2.
Hypothesis Testing for (m1m2)
1. H0: m1=m2
H1: m1<m2 a=0.05
2. Test statistic
3. Decision rule, df=n1+n2-2 = 28
Reject H0 if t < -1.701
21
21
n
1
n
1 Sp
X - X t
=
Assess Equality of Variances
• Ratio of sample variances: 28.72/30.32 = 0.90
2nn
1)s(n1)s(n Sp
21
2
22
2
11
=
5.2989.870 21515
1)30.3(151)28.7(15 Sp
22
==
=
Example 7.2.
Hypothesis Testing for (m1m2)
4. Compute test statistic
5. Conclusion. Reject H0 because -2.92 <
-1.701. We have statistically significant evidence at a=0.05 to show that the mean cholesterol level is lower in patients on treatment as compared to placebo. (p<0.005)
92.2
15
1
15
1 5.29
4.2279.195
n
1
n
1 Sp
X - X t
21
21 =
=
=
New Scenario
• Outcome is continuous
– SBP, Weight, cholesterol
• Two matched study samples
• Data
– On each participant, measure outcome under each
experimental condition
– Compute differences (D=X1-X2)
– dd s,Xn,
Two Dependent/Matched Samples
Subject ID Measure 1 Measure 2
1 55 70
2 42 60
.
.
Measures taken serially in time or under different
experimental conditions
Crossover Trial
Treatment Treatment
Eligible R
Participants
Placebo Placebo
Each participant measured on Treatment and placebo
Hypothesis Testing for md
• Continuous outcome
• 2 Matched/Paired Sample
H0: md=0
H1: md>0, md<0, md≠0
Test Statistic
n>30 (Find critical value
in Table 1C,
n<30 Table 2, df=n-1)
ns
μ - X Z
d
dd =
ns
μ - X t
d
dd =
Example 7.10.
Hypothesis Testing for md
Is there a statistically significant difference in mean systolic blood pressures (SBPs) measured at exams 6 and 7 (approximately 4 years apart) in the Framingham Offspring Study?
Among n=15 randomly selected participants, the mean difference was -5.3 units and the standard deviation was 12.8 units. Differences were computed by subtracting the exam 6 value from the exam 7 value.
Example 7.10.
Hypothesis Testing for md
1. H0: md=0
H1: md≠0 a=0.05
2. Test statistic
3. Decision rule, df=n-1=14
Reject H0 if t > 2.145 or if z < -2.145
ns
μ - X t
d
dd =
Example 7.2.
Hypothesis Testing for md
4. Compute test statistic
5. Conclusion. Do not reject H0 because -2.145 < -1.60 <
2.145. We do not have statistically significant evidence
at a=0.05 to show that there is a difference in systolic
blood pressures over time.
60.1 15/8.12
03.5
ns
μ - X t
d
dd =
==
New Scenario
• Outcome is dichotomous
– Result of surgery (success, failure)
– Cancer remission (yes/no)
• Two independent study samples
• Data
– On each participant, identify group and measure
outcome (yes/no)
– 2211
p̂,n,p̂,n
Hypothesis Testing for (p1-p2)
• Dichotomous outcome
• 2 Independent Sample
H0: p1=p2
H1: p1>p2, p1<p2, p1≠p2
Test Statistic
(Find critical value
in Table 1C)
5)]p̂(1n,p̂n),p̂(1n,p̂min[n 22221111
=
21
21
n
1
n
1 )p̂-(1p̂
p̂-p̂ Z
Example 7.12.
Hypothesis Testing for (p1-p2)
Is the prevalence of CVD different in smokers as compared to
nonsmokers in the Framingham Offspring Study?
Free of CVD
History of CVD
Total
Nonsmoker 2757 298 3055
Current smoker 663 81 744
Total 3420 379 3799
Example 7.2.
Hypothesis Testing for (p1-p2)
1. H0: p1=p2
H1: p1≠p2 a=0.05
2. Test statistic
3. Decision rule
Reject H0 if Z < -1.96 or if Z > 1.96
=
21
21
n
1
n
1 )p̂-(1p̂
p̂-p̂ Z
Example 7.2.
Hypothesis Testing for (p1-p2)
4. Compute test statistic
0.0975 3055
298 p̂ 0.1089,
744
81 p̂
21 ====
=
21
21
n
1
n
1 )p̂-(1p̂
p̂-p̂ Z
0.0988 3055744
29881 p̂ =
=
927.0
3055
1
744
1 0.0988)-0.0988(1
0.0975-0.1089 Z =
=
Example 7.2.
Hypothesis Testing for (p1-p2)
5. Conclusion. Do not reject H0 because -1.96 < 0.927 <
1.96. We do not have statistically significant evidence at
a=0.05 to show that there is a difference in prevalent
CVD between smokers and nonsmokers.
Hypothesis Testing for More than 2 Means*
• Continuous outcome
• k Independent Samples, k > 2
H0: m1=m2=m3 … =mk H1: Means are not all equal
Test Statistic
(Find critical value in Table 4) *Analysis of Variance
k)/(N)XΣΣ(X
1)/(k)XX(Σn F
2
j
2
jj
=
Test Statistic - F Statistic
• Comparison of two estimates of variability in data
• Between treatment variation, is based on the assumption
that H0 is true (i.e., population means are equal)
• Within treatment, Residual or Error variation, is
independent of H0 (i.e., we do not assume that the
population means are equal and we treat each sample
separately)
F Statistic
k)/(N)XΣΣ(X
1)/(k)XX(Σn F
2
j
2
jj
=
Difference BETWEEN each group mean and overall mean
Difference between each observation and its group mean (WITHIN group variation - ERROR)
F Statistic
F = MSB/MSE
MS = Mean Square
What values of F that indicate H0 is likely
true?
Decision Rule
Reject H0 if F > Critical Value of F with
df1=k-1 and df2=N-k
from Table 4
k= # comparison groups
N=Total sample size
ANOVA Table
Source of Sums of Mean
Variation Squares df Squares F
Between
Treatments k-1 SSB/k-1 MSB/MSE
Error N-k SSE/N-k
Total N-1
)X - X( n Σ = SSB j 2
j
)X - X( Σ Σ = SSE j 2
)X -X( Σ Σ = SST 2
Example 7.14.
ANOVA
Is there a significant difference in mean weight loss
among 4 different diet programs? (Data are pounds lost over 8 weeks)
Low-Cal 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
Example 7.14.
ANOVA
1. H0: m1=m2=m3=m4
H1: Means are not all equal a=0.05
2. Test statistic
k)/(N)XΣΣ(X
1)/(k)XX(Σn F
2
j
2
jj
=
Example 7.14.
ANOVA
3. Decision rule
df1=k-1=4-1=3
df2=N-k=20-4=16
Reject H0 if F > 3.24
Example 7.14.
ANOVA
Summary Statistics on Weight Loss by
Treatment
Low-Cal Low-Fat Low-Carb Control
N 5 5 5 5
Mean 6.6 3.0 3.4 1.2
Overall Mean = 3.6
Example 7.14.
ANOVA
)X - X( n Σ = SSB j 2
j
=5(6.6-3.6)2+5(3.0-3.6)2+5(3.4-3.6)2+5(1.2-3.6)2
= 75.8
Example 7.14.
ANOVA
Low-Cal (X-6.6) (X-6.6)2
8 1.4 2.0
9 2.4 5.8
6 -0.6 0.4
7 0.4 0.2
3 -3.6 13.0
Total 0 21.4
)X - X( Σ Σ = SSE j 2
Example 7.14.
ANOVA
Low-Fat (X-3.0) (X-3.0)2
2 -1.0 1.0
4 1.0 1.0
3 0 0
5 2.0 4.0
1 -2.0 4.0
Total 0 10.0
)X - X( Σ Σ = SSE j 2
Example 7.14.
ANOVA
Low-Carb (X-3.4) (X-3.4)2
3 -0.4 0.2
5 1.6 2.6
4 0.6 0.4
2 -1.4 2.0
3 -0.4 0.2
Total 0 5.4
)X - X( Σ Σ = SSE j 2
Example 7.14.
ANOVA
Control (X-1.2) (X-1.2)2
2 0.8 0.6
2 0.8 0.6
-1 -2.2 4.8
0 -1.2 1.4
3 1.8 3.2
Total 0 10.6
)X - X( Σ Σ = SSE j 2
Example 7.14.
ANOVA
)X - X( Σ Σ = SSE j 2
=21.4 + 10.0 + 5.4 + 10.6 = 47.4
Example 7.14.
ANOVA
Source of Sums of Mean
Variation Squares df Squares F
Between 75.8 3 25.3 8.43
Treatments
Error 47.4 16 3.0
Total 123.2 19
Example 7.14.
ANOVA
4. Compute test statistic
F=8.43
5. Conclusion. Reject H0 because 8.43 > 3.24. We have
statistically significant evidence at a=0.05 to show that
there is a difference in mean weight loss among 4
different diet programs.
Two Factor ANOVA
• Compare means of a continuous outcome across two
grouping variables or factors
– Overall test – is there a difference in cell means
– Factor A – marginal means
– Factor B – marginal means
– Interaction – difference in means across levels of
Factor B for each level of Factor A?
Interaction
Cell Means Factor B
1 2 3
Factor A 1 45 58 70
2 65 55 38
35
40
45
50
55
60
65
70
75
1 2 3
A1
A2
No Interaction
Cell Means Factor B
1 2 3
Factor A 1 45 58 70
2 38 55 65
35
40
45
50
55
60
65
70
75
1 2 3
A1
A2
EXAMPLE 7.16
Two Factor ANOVA
• Clinical trial to compare time to pain relief of three
competing drugs for joint pain. Investigators
hypothesize that there may be a differential effect in
men versus women.
• Design – N=30 participants (15 men and 15 women)
are assigned to 3 treatments (A, B, C)
EXAMPLE 7.16
Two Factor ANOVA
• Mean times to pain relief by treatment and gender
• Is there a difference in mean times to pain relief?
Are differences due to treatment? Gender? Or both?
Men Women
A 14.8 21.4
B 17.4 23.2
C 25.4 32.4
EXAMPLE 7.16
Two Factor ANOVA Table Source Sums of Mean
Of Variation Squares df Squares F p-value
Model 967.0 5 193.4 20.7 0.0001
Treatment 651.5 2 325.7 34.8 0.0001
Gender 313.6 1 313.6 33.5 0.0001
Treatment*Gender 1.9 2 0.9 0.1 0.9054
Error 224.4 24 9.4
Total 1191.4 29
Hypothesis Testing for Categorical or Ordinal
Outcomes*
• Categorical or ordinal outcome
• 2 or More Samples
H0: The distribution of the outcome is
independent of the groups
H1: H0 is false
Test Statistic
(Find critical value in Table 3
* c2 test of independence df=(r-1)(c-1))
= E
E)-(O χ
2 2
Chi-Square Test of Independence
Outcome is categorical or ordinal (2+ levels) and there are two or more independent comparison groups (e.g., treatments).
H0: Treatment and Outcome are Independent (distributions of
outcome are the same across
treatments)
Example 7.17.
c2 Test of Independence
Is there a relationship between students’ living
arrangement and exercise status?
Exercise Status
None Sporadic Regular Total
Dormitory 32 30 28 90
On-campus Apt 74 64 42 180
Off-campus Apt 110 25 15 150
At Home 39 6 5 50
Total 255 125 90 470
Example 7.17.
c2 Test of Independence
1. H0: Living arrangement and exercise status are
independent
H1: H0 is false a=0.05
2. Test statistic
3. Decision rule df=(r-1)(c-1)=3(2)=6
Reject H0 if c 2 > 12.59
= E
E)-(O χ
2 2
Example 7.17.
c2 Test of Independence
4. Compute test statistic
O = Observed frequency
E = Expected frequency
E = (row total)*(column total)/N
= E
E)-(O χ
2 2
Example 7.17.
c2 Test of Independence
4. Compute test statistic
Table entries are Observed (Expected) frequencies
Exercise Status
None Sporadic Regular Total
Dormitory 32 30 28 90
(90*255/470=48.8) (23.9) (17.2)
On-campus Apt 74 64 42 180
(97.7) (47.9) (34.5)
Off-campus Apt 110 25 15 150
(81.4) (39.9) (28.7)
At Home 39 6 5 50
(27.1) (13.3) (9.6)
Total 255 125 90 470
Example 7.17.
c2 Test of Independence
4. Compute test statistic
5.60χ
9.6
9.6)(5 ...
17.2
17.2)(28
23.9
23.9)(30
48.8
48.8)(32 χ
2
2222 2
=
=
Example 7.17.
c2 Test of Independence
5. Conclusion. Reject H0 because 60.5 > 12.59. We have
statistically significant evidence at a=0.05 to show that
living arrangement and exercise status are not
independent. (P<0.005)
