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Hypothesis Testing and Sample Size Calculation

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Hypothesis Testing and Sample Size Calculation

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Hypothesis TestingandSample Size Calculation

Po Chyou, Ph. D.

Director, BBC

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

1.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 H0true or not?

4.Type I and Type II Errors

Type I error: we reject H0but H0is true

α= Pr(reject H0 / H0 is true) = Pr(Type I error)

= Level of significance in hypothesis testing

Type II error: we accept H0but H0is false

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

5.Steps of Hypothesis Testing

- Step 1Formulate the null hypothesis H0 in statistical terms

- Step 2Formulate the alternative hypothesis Ha in statistical terms

- Step 3Set the level of significance αand the sample size n

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

- Step 5Collect the data and calculate the statistic

5.Steps of Hypothesis Testing (continued)

- Step 6If 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

6. An Example

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?

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

- Step 4statistic = = P1 - P2

where P1 = x1 ,P2 = x2 and P= x1 + x2 n1 n2n1 + n2

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

- Step 5

P1= x1

= 192 = .80

240

n1

P2= x2

n2

= 32 = .20

160

P= x1 + x2

n1 + n2

= 192 + 32 = 224 = 0.56

240 + 160 400

=P1 - P2

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

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

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

- Step 6

Reject H0 and conclude that smokers had significantly higher proportion of CHD than that of non-smokers.

[P-value < .0000001]

7. Contingency Table Analysis

The Chi-square distribution (2)

Equation for chi-square for a contingency table

2 = (Oij - Eij )2

i, j

Eij

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

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

E11 E12E21 E22

Equation for chi-square for a contingency table (cont.)

E11= n1m1

E12= n1 - n1m1 = n1m2

n

n

n

E21= n2m1

E22= n2 - n2m1 = n2m2

n

n

n

- 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 4statistic =

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

E11 E12 E21 E22

- Step 5

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)

Expectation Counts

- Step 5 (continued)

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

E11E12E21 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

- Step 6

Reject H0 and conclude that there is an association between smoker status and CHD.

[P-value < .0000001]

Definition of Power

Recall :

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

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

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%.

Assumptions

Summary Table 1

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.

Assumptions

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 1

“Statistical Power Analysis for the Behavioral Sciences”

Jacob Cohen

Academic Press, 1977

- 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