1. The POWER of a hypothesis test
Section
Lesson

2. Starter
The weight of the eggs produced by a certain breed of hen is N(65 g, 5 g). Consider a carton of 12 eggs as an SRS from the population of all eggs. What is the probability that the weight of a carton falls between 750 g and 825 g?
If 750 ≤ w ≤ 825, then 62.5 ≤  ≤ 68.75
normalcdf(62.5, 68.75, 65, 1.443) = .954

3. Today’s Objectives
Students will be able to define what is meant by the “power” of a hypothesis test
Students will be able to calculate the power of a test against a given alternative
California Standard 18.0
Students determine the P- value for a statistic for a simple random sample from a normal distribution.

4. So what is power?
Power is the probability that a level α significance test will properly reject Ho when a given alternative value is true.
In other words, power is the probability that you get it right! (When Ho false)
To calculate power, find 1 minus the probability of a Type II error for the given alternative hypothesis.

5. Example
Recall the agronomist testing sugar content in plants to see if the sugar had increased.
σ=8 n = 15
Ho: μ = 140
Ha: μ > 140
Part 1: For which sample means will you fail to reject the null at α = .05?

6. Example Recall the agronomist testing sugar content in plants
Ho: μ = 140
Ha: μ > 140
Part 1: For which sample means will you fail to reject the null at α = .05?

7. Example Recall the agronomist testing sugar content in plants
Ho: μ = 140
Ha: μ > 140
Part 1: For which sample means will you fail to reject the null at α = .05?
OR: Stat:Tests:Zinterval (use C=.90) Ë<143.4

8. Example continued Part 2: Now consider a particular alternative mean
What is the probability that Ë<143.4 if the true mean is 145?
normalcdf(-999,143.4,145,2.066)22% = P(II)
Now calculate the power
1 – 22% = 78%
So if we assume a mean of 140 when in fact the mean is 145, there is a 78% probability that we will draw the correct conclusion by properly rejecting Ho

9. Example concluded
Now re-do part 2 twice, first assuming the true mean is 147 and then 150
normalcdf(-999,143.4,147, 2.066)4%
So power against µa = 147 is 96%
normalcdf(-999,143.4,150,2.066)0.0007
So power against µa = 150 is virtually 100%

10. Power Summary
So power is the probability we draw the right conclusion when the null is not true
The farther from the null a particular alternative is, the greater the power
How could we change our sampling strategy to increase power against a close alternative?
Increase sample size

11. Why does increasing sample size help?
Increasing sample size decreases the spread of the sampling distribution.
So for a given alternative mean, there is less chance that a particular sample mean will be that far (or farther) from the assumed mean.
Also: Increased sample size reduces confidence interval width (why?), so the “acceptance region” is reduced.

12. Today’s Objectives
Students will be able to define what is meant by the “power” of a hypothesis test
Students will be able to calculate the power of a test against a given alternative
California Standard 18.0
Students determine the P- value for a statistic for a simple random sample from a normal distribution.

13. Homework
Read p 574
Do problems 69-71