1. 8-5 Estimating mean differences

2. Comparing two populations Are two populations different? Really? Just a little? What if I wanted to compare the mean test grades of two classes. How different might they be? Independent samples are completely unrelated to each other. (Drawing two random samples such as a drug trial with placebo) Dependent samples can be paired based on correspondence. (Before and After – double measuring)

3. Two samples Given populations 1 and 2 Take samples (not necessarily of the same size) and find the mean of the samples. Do this over and over and you will see the result, stated as Theorem 8-1. The distribution of mean differences will be normal, the mean  1 –  2 will equal μ 1 – μ 2 and the standard deviation will equal

4. Two samples Given populations 1 and 2 Take samples (not necessarily of the same size) and find the mean of the samples. Do this over and over and you will see the result, stated as Theorem 8-1. 1.The distribution of mean differences will be normal 2. the mean  1 –  2 = μ 1 – μ 2 3. the standard deviation will equal

5. Two samples Given populations 1 and 2 Take samples (not necessarily of the same size) and find the mean of the samples. Do this over and over and you will see the result, stated as Theorem 8-1. 1.The distribution of mean differences will be normal 2. the mean  1 –  2 = μ 1 – μ 2 3. the standard deviation will equal Have you seen this before?

6. As before If x 1 and x 2 have normal distributions then the difference will be a normal distribution. If not, as long both n subsets are 30 or larger, then the CLT applies.

7. Confidence Interval For μ 1 – μ 2 The confidence interval will be (  1 –  2 ) – E < μ 1 – μ 2 < (  1 –  2 ) + E Where

8. Confidence Interval For μ 1 – μ 2 The confidence interval will be (  1 –  2 ) – E < μ 1 – μ 2 < (  1 –  2 ) + E Where

9. And logic dictates that If σ 1 – σ 2 are unknown, the student T distribution applies. The degrees of freedom idea still applies, and you choose the smaller of n 1 –1 and n 2 – 1. The formula looks a little different

10. Confidence Interval For μ 1 – μ 2 The confidence interval will be (  1 –  2 ) – E < μ 1 – μ 2 < (  1 –  2 ) + E Where

11. Confidence Interval For μ 1 – μ 2 The confidence interval will be (  1 –  2 ) – E < μ 1 – μ 2 < (  1 –  2 ) + E Where Warning: The calculator will give a slightly different d.f.

12. Example Independent random samples of professional football and basketball players gave the following information Sports Encyclopedia of Pro Football; Official NBA Basketball Encyclopeia)

13. Weight in lbs of pro football players x 1 245262255251244276240265257 252282256250264270275245275 253265270 Weight in lbs of pro Basketball players x 2 205200220210191215221216228 207225208195191207196181193 201

14. 1. Use your calculator to verify that 2. Let μ 1 be the population mean for  1 and let μ 2 be the population mean for  2. Find a 99% confidence interval for μ 1 – μ 2. 3. Examine this confidence interval and explain what it means in the context of the problem. At the 99% level of confidence, do professional football players tend to have a higher population mean weight than professional basketball players?

15. Interpretation of confidence Intervals If c% contains only negative values, then one is c% confident that μ 1 < μ 2 If c% contains only positive values, then one is c% confident that μ 1 > μ 2 If c% contains both positive and negative then no conclusion can be made, but reducing c may allow either of the above 2 conclusions to be made.

16. Confidence Interval for p 1 – p 2 Theorem 8-2 will likely look confusing. Essentially you must look for two samples with size n 1 and n 2, probability for success (and failure) for each trial, a point estimate for each ( ê 1 and ê 2 ) As long as four quantities are above five (sample size times point estimate for success and failure) then a confidence interval can be found.

17. Confidence Interval for p 1 – p 2 Theorem 8-2 will likely look confusing. Essentially you must look for two samples with size n 1 and n 2, probability for success (and failure) for each trial, a point estimate for each ( ê 1 and ê 2 ) As long as four quantities are above thirty (sample size times point estimate for success and failure) then a confidence interval can be found.

18. Confidence Interval for p 1 – p 2 For μ 1 – μ 2 The confidence interval will be ( ê 1 – ê 2 ) – E < p 1 – p 2 < ( ê 1 – ê 2 ) + E Where

19. Confidence Interval for p 1 – p 2 For μ 1 – μ 2 The confidence interval will be ( ê 1 – ê 2 ) – E < p 1 – p 2 < ( ê 1 – ê 2 ) + E Where

20. Most married couples have two or three personality preferences in common. Myers used a random sample of 375 couples and found that 132 had three preferences in common. Another random sample of 571 couples showed that 217 had two personality preferences in common. Let p 1 be the population proportion of all married couples who have three preferences in common and let p 2 be the population proportion of all married couples who have two personality preferences in common. Find a 90% confidence interval for p 1 – p 2.

21. Interpretation of confidence Intervals If c% contains only negative values, then one is c% confident that p 1 < p 2 If c% contains only positive values, then one is c% confident that p 1 > p 2 If c% contains both positive and negative then no conclusion can be made, but reducing c may allow either of the above 2 conclusions to be made.

22. Calculator Page 457 outlines the way to use the calculator (under STATS: TESTS) Choose 2 sample z interval, 2 sample t interval or 2 proportion z interval.