Question

Two drugs A and B to reduce nausea in pregnancy are tested on pregnant women to see if they reduce nausea. 15 women receive drug A & 15 women receive drug B.

Drug A, women report an average reduction in nausea about 25% of the time, with standard deviation of 5% and on drug B average reduction is 30% with a standard deviation of 8%.

(a) Test the hypothesis that the drug A is less effective than drug B. State null and alternative hypothesis, calculate the test-statistic and obtain the P-value. Clearly state your conclusions at α = .01 (Please explain how the degrees of freedom is obtained)

(b) If we assume that the true standard deviation is σ = 6% for both drugs and we assume equal sizes, how large a sample is required to detect an improvement in drug B over drug A of 5% with power 1 − β = .95 when we test at α = .01.

Answer 1

a) NULL HYPOTHESIS H0:

ALTERNATIVE HYPOTHESIS Ha:

For Drug A

n=15

sample mean= 25%

standard deviation= 5%

For Drug B

n=15

sample mean= 30%

standard deviation= 8%

Firstly we will check equal variance assumption.

F= sa^2/sb^2

F= (5)^2/(8)^2

F= 25/64

F= 0.39

P value=0.955 > 0.05

Decision: Fail to reject null hypothesis and therefore equality of population variances can be assumed.

test statistic= xbar-ybar/sp*sqrt(1/n1+1/n2)

where sp is pooled standard deviation

sp= sqrt((n1​−1)s1^2​+(n2​−1)s2^2​​/(n1​+n2​−2))

By putting all values we get

Pooled Standard Deviation: 6.6708 %

Now t= 25-30/6.67*sqrt(1/15+1/15)

Test Statistic t= -2.0527

Degrees of freedom= n1+n2-2= 15+15-2=28

The P-Value is .024761.The result is not significant because p > .01.Hence NOT significant.

Decision: Fail to reject null hypothesis H0.

Conclusion: We don't have enough evidence to conclude that Drug A is less effective than Drug B.

b) ES=

ES= 5/6

ES= 0.83

ni= (Z(1-alpha/2)+Z(1-beta)/ES)^2

ni= (2.58+1.64/0.83)^2

ni= (5.084337)^2

ni= 25.85

ni= 26

n1=n2=26