Section 7.4 Estimating  1 –  2 and p 1 – p 2

2 - create confidence intervals for the difference between population parameters -So we need a sample from each population -samples may be independent or dependent -Independent, if data from one popu. is unrelated to the data in the other (this section is all ind.) -Dependent, if data from one popu. can be paired with data from the other (measurements from the same objects twice – before/after, pretest/postest) Comparing two populations

3 Sampling Distribution for  1 –  2 Diagram:

4 Procedure: Normal distributions or sample size for BOTH > 30 CI for  1 –  2 (  1 and  2 known)

5 Ex: 1 A period before a fire, a random sample of 167 reported average catch was 5.2 trout per day. Assume that the st. dev. for this period is yrs after the fire, a random sample of 125 reported 5 average catch per day was 6.8 trout. Assume the st.dev for this period was 2.3. Compute a 95% confidence interval for  1 –  2, the difference of population means.

6 Interpretation - interval contains only negative values, we can be 95% sure that  1 –  2 < 0 Or 95% sure that  1 <  2 ****we are 95% sure that the average catch before the fire was less than the average catch after the fire.****

7 2-Sample Mean Z Interval in calc: Stat  TESTS  #9: 2-SampleZInt (give stats or data) Go back and try Ex: 1 with this

8 CI for  1 –  2, When  1 and  2 Are unknown Procedure: ***d.f. of t is the smaller of the sample sizes.

9 Ex: 2 Rapid brain waves (wakefulness) are in the range of 16 to 25 hertz. Slow brain waves (sleep) are in the range of 4 to 8 hertz. Claim: Alcohol is a poor sleep aid. average brain wave distribution is normal. Group 1 (x 1 values): n 1 = 15 (with alcohol) Average brain wave activity from 4 to 6 A.M x 1  and s 1  1.86

10 Group 2 (x 2 values): n 2 = 14 (no alcohol) Average brain wave activity from 4 to 6 A.M x 2  6.59 and s 2  1.91 Compute a 90% confidence interval for  1 –  2, the difference of population means. cont’d

11 Interpretation contains only positive values - This means that  1 –  2 > 0. Thus, we are 90% confident  1 >  2 We are 90% confident avg. brain wave activity from 4 to 6 A.M. for wine-drinkers is more than avg. brain wave activity for non-drinkers. cont’d

12 2-Sample Mean T Interval Stat  TESTS  #0: 2-SampleTInt (give stats or data) (“No” for Pooled option) Go back and try Ex: 2 with this (“Yes” for Pooled only when the standard deviations of the two populations is said to be equal or approximately equal)

13 Estimating the Difference of Proportions p 1 – p 2

14 CI for p 1 – p 2 Procedure:

15 Ex: 3 Based on a study of over 650 people in the Zurich Sleep Laboratory, it was found that about one-third of all dream reports contain feelings of fear, anxiety, or aggression. If a person is in a good mood when going to sleep, the proportion of “bad” dreams (fear, anxiety, aggression) might be reduced. Suppose that two groups of subjects were randomly chosen for a sleep study.

16 Group I: before sleeping, watched a comedy movie, there were 175 dreams recorded, of which 49 were bad dreams. Group II: before sleeping, no movie, there were 180 dreams recorded, of which 63 were bad dreams. Compute a 95% confidence interval for p 1 – p 2, the difference in percentage of bad dreams. cont’d

17 Interpretation Interval contains both positive/negative values. we cannot say that p 1 – p 2 0). Thus, we cannot conclude that p 1 p at a 95% confidence, we cannot say that the proportion of bad dreams was reduced by watching a comedy movie.

18 2-Sample Proportion Z Interval in calc. Stat  TESTS  #B: 2-PropZInt  enter 2 proportions Now try Ex: 3 with this

19 Interpreting CI for Differences 1.If the CI only contains negative values, we conclude 2.If the CI only contains positive values, we conclude 3.If the CI contains both positive and negative values, we conclude neither population mean or population is larger. (then we could increase n or decrease confidence to see if the conclusion changes.)