Question

If the proportion of algae in a body of water exceeds 70%, then aquatic plant life will die. This in turn kills the fish. A researcher wants to know if a certain pond has enough algae to kill fish in it. In a sample of 100 lbs of water, he finds 85 lbs of algae. The researcher is interested in an alpha of .05.

1. How would you calculate an appropriate confidence interval (Using two sided, lower, or upper bounds)? How can it be used to test the hypothesis?

2. Find the power of the test when p = .76 and alpha = .05

3. How can you increase the power of the test? State every method and justify why each works.

Thank you!

Answer 1

Null hypothesis :  proportion of algae in a body of water is 70% or less, P_0 <= 0.7

Alternate hypothesis :  proportion of algae in a body of water exceeds 70% , P_0 > 0.7

alpha = 0.05

sample size, n = 100

sample proportion, p = 85/100 = 0.85

sample variance = p*(1 - p)/n = 0.7*0.3 / 100 = 0.0021

sample standard deviation, sd = 0.0458

1. 95% confidence interval

z critical = 1.654 ( for one tailed )

P = p +/- sd*z critical = 0.85 +/- 0.0458*1.654 = 0.85 +/- 0.0757 = ( 0.7743, 0.9257 )

2. Power of test = P[ Reject H0 | H1 is true ]

p = 0.76

Reject H0 , when Z > Z critical

( p_critical - 0.7 ) / 0.0458 > 2.33

or, p_critical - 0.7 > 0.1067

or, p_critical > 0.7 + 0.1067

or, p_critical > 0.8067

when , H1 is true p = 0.76

P[ Reject H0 | H1 is true ] = P[ Z > Z critical | H1 is true ] = P[ Z > ( p_critical - 0.76 )/sd ] = P[ Z > ( 0.8067 - 0.76 )/0.0458 ] = P[ Z > 1.0196 ] = 1 - P[ Z < 1.0196 ] = 1 - 0.846 = 0.154

Power of test = 0.154

3. to increase power of test :

1. increase sample size, that will reduce the variance of sample and hence, increase the power

2. increase level of significance, that will reduce p_critical and hence increase the power