The POWER of a hypothesis test Section 10.4.2 Lesson 10.4.2

Starter 10.4.2 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

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.

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.

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?

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?

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

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

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%

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

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.

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.

Homework Read p 574 Do problems 69-71