Hypothesis Testingand

Sample Size Calculation

Po Chyou, Ph. D.

Director, BBC

Hypothesis Testingon

• Population mean(s)• Population median(s)Population proportion(s)

• Population variance(s)• Population correlation(s)Association based on

contingency table(s)

• Coefficients based on regression model

• Odds ratio• Relative risk• Trend analysis• Survival distribution(s) /

curve(s)• Goodness of fit

Hypothesis Testing1. Definition of a Hypothesis

An assumption made for the sake of argument

2. Establishing Hypothesis

Null hypothesis - H0 Alternative hypothesis - Ha

3. Testing Hypotheses

Is H0 true or not?

Hypothesis Testing

4.Type I and Type II Errors

Type I error: we reject H0 but H0 is true

α = Pr(reject H0 / H0 is true) = Pr(Type I error) = Level of significance in hypothesis testing

Type II error: we accept H0 but H0 is false = Pr(accept H0 / H0 is false) = Pr(Type II error)

Hypothesis Testing5. Steps of Hypothesis Testing

- Step 1 Formulate the null hypothesis H0 in statistical terms

- Step 2 Formulate the alternative hypothesis Ha in statistical terms

- Step 3 Set the level of significance α and the sample size n

- Step 4 Select the appropriate statistic and the rejection region R

- Step 5 Collect the data and calculate the statistic

Hypothesis Testing

5. Steps of Hypothesis Testing (continued)

- Step 6 If the calculated statistic falls in the rejection region R, reject H0 in favor of

Ha; if the calculated statistic falls outside R, do not reject H0

α/2 α/2

Z0

Z α/2

De

ns

ity

–Z α/2

Reject H0 Do not reject H0 Reject H0

Hypothesis Testing

A random sample of 400 persons included 240 smokers and 160 non-smokers. Of the smokers, 192 had CHD, while only 32 non-smokers had CHD.

Could a health insurance company claim the proportion of smokers having CHD differs from the proportion of non-smokers having CHD?

6. An Example

CHD No CHDSmokers x1 n1 - x1 n1

Non-Smokers x2 n2 - x2 n2

n = n1 + n2

CHD No CHDSmokers 192 48 240Non-Smokers 32 128 160

400

Hypothesis TestingExample (continued)

Let P1 = the true proportion of smokers having CHD

P2= the true proportion of non-smokers having CHD

- Step 1 H0 : P1 = P2

- Step 2 Ha : P1 P2

- Step 3 α = .05, n = 400

Hypothesis TestingExample (continued)

- Step 4 statistic = = P1 - P2

where P1 = x1 , P2 = x2 and P = x1 + x2

n1 n2 n1 + n2

P(1-P) (1/n1 + 1/n2)

α/2 = .025 α/2 = .025

Z0

= -1.96 = 1.96

Z.025

De

ns

ity

–Z .025

Reject H0 Do not reject H0 Reject H0

CHD No CHDSmokers x1 n1 - x1 n1

Non-Smokers x2 n2 - x2 n2

n = n1 + n2

CHD No CHDSmokers 192 48 240Non-Smokers 32 128 160

400

Hypothesis TestingExample (continued)

P1 = x1

n1

= 192 = .80 240

P2 = x2

n2

= 32 = .20 160

P = x1 + x2

n1 + n2

= 192 + 32 = 224 = 0.56 240 + 160 400

= P1 - P2

P(1-P) (1/n1 + 1/n2)

= .80 - .20 = .60 = 11.84 > 1.96

(.56) (1-.56) (1/240 + 1/160) .05066

- Step 5

Hypothesis TestingExample (continued)

- Step 6

Reject H0 and conclude that smokers had

significantly higher proportion of CHD than

that of non-smokers.

[P-value < .0000001]

Hypothesis Testing7. Contingency Table Analysis

The Chi-square distribution (2)

α=.05

0 =3.841

De

ns

ity

Do not reject H0

Reject H0

.05, 1

2

2

Hypothesis TestingEquation for chi-square for a contingency table

For i = 1, 2 and j =1, 2

2= (O11 - E11)2 + (O12 - E12)2 + (O21 - E21)2 + (O22 - E22)2

E11 E12

E21 E22

2 = (Oij - Eij )2

i, j Eij

Hypothesis TestingEquation for chi-square for a contingency table (cont.)

E11 = n1m1 E12

= n1 - n1m1 = n1m2

n n nE21

= n2m1 E22 = n2 - n2m1 = n2m2

n n n

O11 O12 n1

O21 O22 n2

m1 m2 n = n1 + n2 = m1 + m2

E11 E12 n1

E21 E22 n2

m1 m2

Hypothesis TestingExample : Same as before

- Step 1 H0 : there is no association between smoker status and CHD

- Step 2 Ha : there is an association between smoker status and CHD

- Step 3 = .05, n = 400

- Step 4 statistic = 2= (O11 - E11)2 + (O12 - E12)2 + (O21 - E21)2 + (O22 - E22)2

E11 E12

E21 E22

Hypothesis TestingExample (continued) : Same as before

α=.05

0 =3.841

De

ns

ity

Do not reject H0

Reject H0

.05, 1

2

2

Hypothesis TestingExample (continued) : Same as before

- Step 5 CHD No CHDSmokers O11 O12 n1

Non-Smokers O21 O22 n2

m1 m2 n

CHD No CHDSmokers 192 48 240Non-Smokers 32 128 160

224 176 400

CHD No CHDSmokers E11 E12

Non-Smokers E21 E22

Hypothesis TestingExample (continued) : Same as before

E11 = n1m1 = 240 * 224 = 134.4

n 400

E12 = n1 - n1m1 = 240 - 134.4 = 105.6

n

E21 = n2m1 = 160 * 224 = 89.6

n 400

E22 = n2 - n2m1 = 160 - 89.6 = 70.4

n

- Step 5 (continued)

CHD No CHDSmokers 134.4 105.6Non-Smokers 89.6 70.4

Expectation Counts

Hypothesis TestingExample (continued) : Same as before

- Step 5 (continued)

2= (O11 - E11)2 + (O12 - E12)2 + (O21 - E21)2 + (O22 - E22)2

E11 E12

E21 E22

= (192 - 134.4)2 + (48 - 105.6)2 + (32 - 89.6)2 + (128 - 70.4)2

134.4 105.6 89.6 70.4

= 24.68 + 31.42 + 37.03 + 47.13

= 140.26 > 3.841

Hypothesis TestingExample (continued) : Same as before

- Step 6

Reject H0 and conclude that there is an

association between smoker status and CHD.

[P-value < .0000001]

Sample Size Estimation andStatistical Power Calculation

Definition of Power

Recall :

= Pr (accept H0 / H0 is false) = Pr (Type II error)

Power = 1 - = Pr(reject H0 / H0 is false)

Sample Size Estimationfor Intervention on Tick Bites Among

CampersAssumptions

1. Given that the proportion (PCON) of tick bites among campers in the control group is constant.

2. Given that the proportion (PINT) of tick bites among campers in the intervention group is reduced by 50% compared to that of the control group after intervention has been implemented.

3. Given that a one- or two- tailed test is of interest with 80% power and a type-I error of 5%.

Sample Size Estimationfor Intervention on Tick Bites Among

Campers

Summary Table 1

Required N for each groupPCON PINT Two-tailed One-tailed

.01 .005 (50% reduction) 4500 3600

.05 .025 (50% reduction) 1170 922

.10 .050 (50% reduction) 475 374

.15 .075 (50% reduction) 305 240

Statistical Power Calculationfor Intervention on Obesity of Women in

MESAAssumptions

1. Given that the proportion (PCON) of women who are obese at baseline (i.e., the control group) is constant. There are a total of 840 women in the control group. Based on our preliminary data analysis results, approximately 50% of these 840 women at baseline are obese (BMI >= 27.3).

2. Given that the proportion (PINT) of women who are obese in the intervention group is reduced by 5% or more compared to that of the control group after intervention has been implemented. There are a total of 680 women who had been newly recruited. Based on our preliminary data analysis results, 50% of these 680 newly recruited women are obese. Assume that 60% of these women will agree to participate, we will have 200 women to be targeted for intervention.

Statistical Power Calculationfor Intervention on Obesity of Women in

MESA (continued)

3. Given that a one-tailed test is of interest with a type-I error of 5%, then the estimated statistical powers are shown in Table 1 for detecting a difference of 5% or more in the proportion of obesity between the control group and the intervention group.

Assumptions

Table 1PCON (n=840) PINT (n=680) Difference Power

.50 .45 .05 61%

.50 .44 .06 75%

.50 .43 .07 85%

.50 .42 .08 92%

.50 .41 .09 96%

.50 .40 .10 98%

Reference

“Statistical Power Analysis for the Behavioral Sciences”

Jacob Cohen

Academic Press, 1977

Take Home Message:

• You’ve got questions : Data ? STATISTICS?...

• Contact Biostatistics and consult with an experienced biostatistician

– Po Chyou, Director, Senior Biostatistician (ext. 9-4776)– Dixie Schroeder, Secretary (ext. 1-7266)

OR

• Do it at your own risk

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