bs704 class 7 hypothesis testing procedures

BS704 Class 7 Hypothesis Testing Procedures

HW Set #6

Chapter 7Problems 4, 7, 8, 20 and 25R Problem Set 6 (on Blackboard)Due October 26

Please complete Quiz 8 Before Oct 26

Objectives

Define null and research hypothesis, test statistic, level of significance and decision rule

Understand Type I and Type II errors Differentiate 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 Tests About m

Example: A large, national study was conducted in 2007 and found that the mean systolic blood pressure for males aged 50 was 130. In 2008, an investigator hypothesizes systolic blood pressures have increased.

To test this hypothesis we set up two competing hypotheses

Null H0: m = 130Research H1: m > 130

Hypothesis Tests About m

To test the hypotheses a random sample is selected from the population of interest.

Suppose a sample of n=108 males age 50 in 2008 is selected and their systolic blood pressures are analyzed. Of primary interest is the mean systolic blood pressure in the sample ( ). X

If the sample mean is 130, which is more likely true?

Yes N

o

0%0%

1. H0

2. H1

If the sample mean is 150, which is more likely true?

Yes N

o

0%0%

1. H0

2. H1

If the sample mean is 135, which is more likely true?

Yes N

o

0%0%

1. H0

2. H1

Hypothesis Tests About m

We must determine a critical value such that if our sample mean is less than the critical value we will conclude that H0 is true (i.e., m=130), and if our sample mean is greater than the critical value we will conclude that H1 is true (i.e., m > 130).

Hypothesis Tests About m

Instead of determining critical values for the sample mean - which would be specific to each application (since depends on the unit of measurement), - appeal to the Central Limit Theorem.

Hypothesis Tests About m

For large n - is approximately normally distributed. Assuming that H0 is true (or "under the null hypothesis") we can standardize , producing a Z score (test statistic).

If 130 then Z 0 H0 probably true.

If >130 then Z > 0 H1 probably true.

What value of Z is considered “large” ?

In Example suppose that s=15 and n=108.

X

XX

Hypothesis Tests About m

We must select a level of significance, denoted a, which is defined as the probability of rejecting H0 when H0 is true.

The level of significance is generally in the range of 0.01 to 0.10.Once a level of significance is selected, a decision rule is formulated.

Hypothesis Tests About m

Decision rule: 

Reject H0 if Z > 1.645

Do Not Reject H0 if Z < 1.645

Once the decision rule is in place, we compute the value of the test statistic.

Suppose in example, X=135.

3.46 =

108

15130 - 135

=

n

sμ - X

= Z 0

Hypothesis Tests About m

The final step - draw a conclusion.

The test statistic falls in the rejection region - we reject H0 (3.46 > 1.645). We have significant evidence, a = 0.05, to show that the mean systolic blood pressure for males aged 50 in 2008 has increased from 130.

Hypothesis Testing Procedures

1. Set up null and research hypotheses, select a

2. Select test statistic3. Set up decision rule4. Compute test statistic5. Draw conclusion & summarize

significance (p-value)

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 for m

Continuous outcome 1 Sample

H0: =m m0

H1: >m m0, m<m0, m≠m0

Test Statisticn>30 (Find critical

value in Table 1C,

n<30 Table 2)

ns/

μ-X Z 0

ns/

μ-X t 0

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.

Hypothesis Testing for m

1. H0: =203m

H1: m≠203 =0.05a

2. Test statistic

3. Decision rule Reject H0 if z > 1.96 or if z < -1.96

ns/

μ-X Z 0

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.

22.43310/8.36

2033.200

ns/

μ-X Z 0

Hypothesis Testing for m

Significance of the findings. Z = -4.22.

Table 1C. Critical Values for Two-Sided Tests a Z0.20 1.2820.10 1.6450.05 1.9600.010 2.5760.001 3.2910.0001 3.819 p<0.0001.

Interpreting P-Values

If p < a then reject H0

Errors in Hypothesis Tests

Conclusion of Statistical TestDo Not Reject H0 Reject

H0

H0 true Correct Type I error

H0 false Type II error Correct

Practice Example – Is social networking a health risk?

Hypertexting (>120 text messages per day) has been associated with health risks (Frank et al, Nov 2010). In 2010, the mean number of texts per day was 55. Is texting increasing in 2012? A sample of 75 teens report sending a mean of 61 texts per day (SD = 15). Is there evidence of an increase in texting in 2012?

Practice Example – Is social networking a health risk?

1. H0: = 55m

H1: m > 55 =0.05a

2. Test statistic 4. Test Statistic

3. Decision rule 5. Reject H0.

Reject H0 if z > 1.645

ns/

μ-X Z 0 5.3

7515/

55-61 Z

In an upper tailed test with a=0.05. If Z=-2.5 would you reject H0: m=50?

Yes N

o

0%0%

1. Yes2. No

We run a test and do not reject H0. Which is most likely…

We

mad

e th

e co

rrec

t...

We

com

mitt

ed a

Typ

e ...

We

com

mitt

ed a

Typ

e ..

1 o

r 2

1 o

r 3

0% 0% 0%0%0%

1. We made the correct decision

2. We committed a Type I error

3. We committed a Type II error

4. 1 or 25. 1 or 3

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

xp̂

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

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 (482/3536=0.136).

Hypothesis Testing for p

1. H0: p=0.211

H1: p<0.211 =0.05a

2. Test statistic

3. Decision rule Reject H0 if z < -1.645

n

)p-(1p

p-p̂ Z

00

0

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

Practice Example – Is social networking a health risk?

Hypertexting (>120 text messages per day) has been associated with health risks (Frank et al, Nov 2010). In 2010, 19% of teens were hypertexting. Is hypertexting increasing in 2012? In a sample of 75 teens, 16 report sending more than 120 texts per day. Is hyper-texting increasing in 2012?

Practice Example – Is social networking a health risk?

1. H0: p=0.19

H1: p > 0.19 =0.05a

2. Test statistic 4. Test Statistic

3. Decision rule 5. Do not reject H0.

Reject H0 if z > 1.645

n

)p-(1p

p-p̂ Z

00

0 44.0

750.19)-0.19(1

0.19-0.21 Z

Sample Data: n=75, x=16

=0.21

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 22

22212

111

Two Independent Samples

RCT: Set of Subjects Who Meet Study Eligibility Criteria

Randomize

Treatment 1 Treatment 2Mean Trt 1 Mean Trt 2

Two Independent Samples

Cohort Study - Set of Subjects Who Meet Study Inclusion Criteria

Group 1 Group 2Mean Group 1 Mean Group 2

Hypothesis Testing for (m1-m2)

Continuous outcome 2 Independent Sample

H0: m1=m2 (m1-m2 = 0)

H1: m1>m2, m1<m2, m1≠m2

An RCT is planned to show the efficacy of a new drug vs. placebo to lower total cholesterol.

H0:

mP=m

N H1:

m...

H0:

mP=m

N H1:

...

H0:

mP=m

N H1:

...

0% 0%0%

What are the hypotheses?

1. H0: mP=mN H1: mP>mN

2. H0: mP=mN H1: mP<mN

3. H0: mP=mN H1: mP≠mN

Hypothesis Testing for (m1-m2)

Test Statisticn1>30 and (Find critical

valuen2> 30 in Table 1C,

n1<30 or Table 2)

n2<30

21

21

n

1

n

1Sp

X - XZ

21

21

n

1

n

1Sp

X - Xt

Pooled Estimate of Common Standard Deviation, Sp

Previous formulas assume equal variances (s1

2=s22)

If 0.5 < s12/s2

2 < 2, assumption is reasonable

2nn

1)s(n1)s(nSp

21

222

211

Hypothesis Testing for (m1-m2)

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?

Hypothesis Testing for (m1-m2)

Sample Size Mean Std DevNew Drug 15 195.9 28.7Placebo 15 217.4 30.3

Hypothesis Testing for (m1-m2)

1. H0: m1=m2

H1: m1<m2 =0.05a

2. Test statistic

3. Decision rule, df=n1+n2-2 = 28

Reject H0 if t < -1.701

21

21

n

1

n

1Sp

X - Xt

Assess Equality of Variances Ratio of sample variances: 28.72/30.32 =

0.90

2nn

1)s(n1)s(nSp

21

222

211

5.2989.87021515

1)30.3(151)28.7(15Sp

22

Hypothesis Testing for (m1-m2)

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

151

151

5.29

4.2279.195

n1

n1

Sp

X - Xt

21

21

A two sided test for the equality of means produces p=0.20. Reject H0?

Yes N

o

0%0%

1. Yes2. No3. Maybe

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 21 55 702 42 60..

Measures taken serially in time or under different experimental conditions

Crossover Trial

Treatment Treatment

Eligible RParticipants

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 Statisticn>30 (Find critical

value in Table 1C,

n<30 Table 2)

ns

μ - XZ

d

dd

ns

μ - Xt

d

dd

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.

Hypothesis Testing for md

1. H0: md=0

H1: md≠0 =0.05a

2. Test statistic

3. Decision rule, df=n-1=14 Reject H0 if t > 2.145 or if t < -2.145

ns

μ - Xt

d

dd

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.

60.115/8.12

03.5

ns

μ - Xt

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)

21

21

n1

n1

)p̂-(1p̂

p̂-p̂ Z

5)]p̂(1n,p̂n),p̂(1n,p̂min[n 22221111

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

Hypothesis Testing for (p1-p2)

1. H0: p1=p2

H1: p1≠p2 =0.05a

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

Hypothesis Testing for (p1-p2)

4. Compute test statistic

21

21

n

1

n

1)p̂-(1p̂

p̂-p̂ Z 0.0975

3055

298p̂ 0.1089,

744

81p̂ 21

0.09883055744

29881p̂

927.0

30551

7441

0.0988)-0.0988(1

0.0975-0.1089 Z

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.

Study of Single verses Weekly Antenatal Corticosteroids

What do p-values mean in Table 1? What do p-values mean in Table 3?

Study of Single verses Weekly Antenatal Corticosteroids

Did randomization work?

Primary outcome

Is trial a success?

Practice ProblemRun Test for Primary Outcome

Composite Morbidity

Group Sample Size # EventsWeekly 256 56Single 246 66

Solution

1. H0: p1=p2

H1: p1≠p2 =0.05a

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

Solution

4. Compute test statistic

5. Do not Reject H0 since -1.96<-1.294<1.96.

21

21

n

1

n

1)p̂-(1p̂

p̂-p̂ Z 0.268

246

66p̂ 0.219,

256

56p̂ 21

0.243246256

6656p̂

294.1

2461

2561

0.243)-0.243(1

0.268-0.219 Z