Question

A claim has been made that only 5% of men in the U.S. play golf. If I want to be 90% confident, and have 90% statistical power, what sample size would I need to disprove this claim if the true percentage of men playing golf is 8%?

Answer 1

Null hypothesis: Proportion of men play goal, P_0 = 5% = 0.05

Alternate hypothesis: Proportion of men play goal,P_1 = 8% = 0.08

Criteria : reject H0 if p < p_critical

90% confidence means ( 100 - 90 )% = 10% alpha level.

Since, the value of null hypothesis is less than alternate hypothesis, the test is one tailed.

power of test = 90% = 0.9

alpha = 10% = 0.1

Under H0

Variance = V_0 = P_0*( 1 - P_0 )/n = 0.05*(1-0.05)/n = 0.05*0.95/n = 0.0475/n

standard deviation, S_0 = sqrt(V_0) = 0.2179/sqrt(n)

Under H1

Variance = V_1 = P_1*( 1 - P_1 )/n = 0.08*(1-0.08)/n = 0.08*0.92/n = 0.0736/n

standard deviation, S_1 = sqrt(V_1) = 0.2713/sqrt(n)

alpha = P[ Reject H0 when it is true ] = P[ p < p_critical | H0 ] = P[ ( p - 0.05 ) / S_0 < ( p_critical - 0.05 ) / S_0 ] = P[ Z < ( p_critical - 0.05 ) / S_0 ] = 0.1

P[ Z < Z_alpha ] = 0.1

therefore, Z_alpha = 1.28 ( 10% alpha for one tailed )

Comparing

( p_critical - 0.05 ) / S_0 = 1.28

p_critical - 0.05 = 1.28*S_0 ........................................(1)

Power of test = P[ Reject H0 when it is false ] = P[ p < p_critical | H1 ] = P[ ( p - 0.08 ) / S_1 < ( p_critical - 0.08 ) / S_1 ] = P[ Z < ( p_critical - 0.08 ) / S_1 ] = 0.9

P[ Z < Z_beta ] = 0.9

therefore, Z_beta = -1.28 ( 90% power for one tailed )

Comparing

( p_critical - 0.08 ) / S_1 = -1.28

p_critical - 0.08 = S_1*(-1.28) ..........................................(2)

from (1)

p_critical = 1.28*S_0 +0.05

substitute in (2)

1.28*S_0 +0.05 - 0.08 = S_1*(-1.28)

1.28*S_0 + 1.28*S_1 = 0.03

1.28*( 0.2179/sqrt(n) + 0.2713/sqrt(n) ) = 0.03

1.28*( 0.4892/sqrt(n) ) = 0.03

0.6262 = 0.03*sqrt(n)

sqrt(n) = 0.6262/0.03 = 20.8725

n = 20.8725^2 = 435.6 = 436 ( approx )