1. Type I and Type II Errors Section 10.4.1

2. Starter 10.4.1 An agronomist examines the cellulose content of a variety of alfalfa hay. Suppose that the cellulose content in the population has a standard deviation of 8 mg/g. A sample of 15 cuttings has mean cellulose content of 145 mg/g. Give a 90% confidence interval for the mean cellulose content in the population A previous study claimed that the mean cellulose content was 140 mg/g, but the agronomist believes that the mean is higher than 140. State H o and H a and carry out a significance test to see if the new data support this belief. Name two major assumptions that underlie the analysis.

3. Today’s Objectives Students should be able to describe the meaning of Type I and Type II errors in hypothesis testing. Students should be able to compute the probability of a Type I or a Type II error. California Standard 18.0 Students determine the P- value for a statistic for a simple random sample from a normal distribution.

4. Error Types Type I error: Incorrectly reject the null hypothesis when it is true. –Example: Convict an innocent person Type II error: Incorrectly fail to reject the null hypothesis when it is false. –Example: Acquit a guilty person H o TrueH o False Reject H o Type I ErrorCorrect “Accept” H o CorrectType II Error

5. Calculating error probabilities Type I error: We incorrectly reject H o –The significance level (alpha) is the probability of Type I error. Type II error: We should reject H o (because it is wrong!) but we incorrectly fail to do so.  Use a two step process: 1.First find the range of values that would fail to reject H o. (The “acceptance” range) 2.Then find the probability of getting a sample mean in that range if a certain given value in H a is true.

6. A more detailed look at Type II Step 1: Find the values of  that would lead you to “accept” H o Notice that this looks just like a confidence interval. Step 2: Find the probability that we get an  within that range if μ a is the true mean

7. Example The mean diameter of a certain type of ball bearing is supposed to be 2.000 cm. The diameters are N(2,.010). The customer takes an SRS of 5 bearings from a shipment and measures their diameters. He rejects the bearings if the sample mean diameter is significantly different from 2.000 at the 5% significance level. State the appropriate hypotheses  H o : μ=2H a : μ≠2 α = 5% State the probability of making a Type I error –P(Type I error) = 5%

8. Find the largest and smallest values of the sample mean that will lead to the customer accepting the shipment. (Hint: This is the same as finding a 95% confidence interval around the assumed mean of 2). –Stat:Tests:Zinterval  (1.9912, 2.0088) Now assume that the actual mean diameter of the shipment is 2.015 cm. Based on that assumption, what is the probability that the sample mean (of the 5 bearings sampled) will lie in the acceptance range you calculated above? –Normalcdf(1.9912, 2.0088, 2.015,.00447) .0828 Your result is the probability of a Type II error. Suppose the true mean of the shipment is 2.1 cm –Find the new P(Type II error) –Normalcdf(1.9913, 2.0088, 2.1,.00447) = 0 Why is the probability so low? –The new mean is MANY standard deviations away from the assumed mean under the null hypothesis. There is virtually NO CHANCE we will get it wrong.

9. Today’s Objectives Students should be able to describe the meaning of Type I and Type II errors in hypothesis testing. Students should be able to compute the probability of a Type I or a Type II error. California Standard 18.0 Students determine the P- value for a statistic for a simple random sample from a normal distribution.

10. Homework Read pages 567 - 572 Do problems 66 - 68