Question

If the proportion of algae in a body of water exceeds 70%, then aquatic plant life will die. This, in turn, kills all of 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.

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

Answer 1

If the proportion of algae in a body of water exceeds 70%, then aquatic plant life will die. This, in turn, kills all of the fish.

And the researcher wants to know if a certain pond has enough algae to kill fish in it.

So the Null hypothesis ( H₀ ) and the alternative hypothesis ( H_a ) is as follows:

H₀ : p = 0.70

H_a : p > 0.70

Sample size = n = 100

x = 85 = number of algae

Level of significance = = 0.05

Here we need to use one tailed proportional test.

Let's use minitab:

The command for one sample proportion z test in minitab is

Stat>>>Basic statistics>>>1-proportion...

Then click on summarized data

number of events = x = 85

Number of trials = n = 100

Click on "Perform hypothesis test

Hypothesized proportion = P = 0.70 (because under the null hypothesis P = 0.70)

then click on option

Level of confidence in percentage = c = ( 1- )*100 = = (1 - 0.05)*100 = 95.0

so put "Confidence level " = 95.0

Alternative = Greater than

then click on "Use test and interval based on normal approximation"

Then click on OK and again click on OK

So we get the following output

P-value = 0.001

Decision rule:

1) If p-value < level of significance (alpha) then we reject null hypothesis

2) If p-value > level of significance (alpha) then we fail to reject null hypothesis.

Here p value = 0.001 < 0.05 so we used first rule.

That is we reject the null hypothesis

Conclusion: At 5% level of significance there are sufficient evidence to say that the sample data indicates certain pond has enough algae to kill fish in it.

Let's find power for the given value of p = 0.76

power = P( P > 0.76) = 1 - P( P < 0.76) ....( 1 )

The approximate distribution of sample proportion is normal with mean p = 0.70 ( under null hypothesis) and standard deviation ( ) is as follows:

Let's use excel:

P( P < 0.76) = "=NORMDIST(0.76,0.7,0.0458,1)" = 0.9049

Plug this value in equation ( 1 ), we get:

power = P( P > 0.76) = 1 - 0.9049 = 0.0951  ( This is the power )