1. Testing Distributions Section 13.1.2

2. Starter 13.1.2 Elite distance runners are thinner than the rest of us. Skinfold thickness, which indirectly measures body fat, can show this. A random sample of 20 runners had a mean skinfold of 7.1 mm with a standard deviation of 1.0 mm. A random sample of 95 non-runners had a mean of 20.6 w/ sd of 9.0. Form a 95% confidence interval for the mean difference in body fat between runners and non- runners.

3. Today’s Objectives Students will graph the chi-square distributions with various degrees of freedom Students will calculate the chi-square statistic for a distribution sample Students will form hypotheses and determine whether a distribution is the same as or differs from a given distribution California Standard 19.0 Students are familiar with the chi- square distributions and chi- square test and understand their uses.

4. The Big Picture (so far) Chapter 11: Estimating means –One population: μ = some constant –Two populations: μ 1 = μ 2 Chapter 12: Estimating proportions –One population: p = some percent –Two populations: p 1 = p 2 Chapter 13: Comparing distributions –Distributions are categorical variables with more than 2 outcomes –Is a sample significantly different from population?

5. The chi-square distributions A family of distributions that take on only positive values Curves are skewed right A specific curve is determined by its degrees of freedom

6. Shapes of the Distribution Turn off all StatPlots, clear all Y= equations Set window to [0, 14] 1 x [-.05,.3].1 Find the X 2 pdf distribution on the TI Enter Y 1 = X 2 pdf(x, 3) and draw the graph Sketch the graph on a set of axes. –Label it X 2 3 Graph Y 2 = X 2 pdf(x, 4) and add to your sketch. Graph Y 3 = X 2 pdf(x, 8) and add to your sketch.

7. Properties of the chi-square distribution The area under the curve is 1 Each curve begins at x = 0, increases to a peak, then approaches the x axis asymptotically Each curve is skewed to the right. As degrees of freedom increases, the curve becomes more symmetric and less skewed –Note that it can never become approximately normal: the graph is not symmetric

8. The chi-square statistic We will test the hypotheses by forming the chi-square statistic The formula is: –Where “O” is the observed count and “E” is the expected count We will ask the probability that X 2 is as great as or greater than the value we found assuming the null hypothesis is true

9. Assumptions needed for X 2 test Data are from a valid SRS from the population. All individual expected counts are at least 1 No more than 20% of the expected counts are less than 5

10. Assumptions So Far… Procedure One population t for means Two population t for means One population z for proportions Two population z for proportions X 2 test for distributions Assumptions SRS, Normal Dist SRS, Normal, Independent SRS, large pop, 10 succ/fail SRS, large pop, 5 s/f each SRS, expect each count at least 1 –No more than 20% less than 5

11. Example Here is the age distribution of Americans in 1980. Put the percents in L 1 (as decimals) AgePopulation (thousands) Percent 0 to 2493,77741.39% 25 to 4462,71627.68% 45 to 6444,50319.64% 65 +25,55011.28% Total226,546

12. Example Continued Here is a sample of 500 randomly selected individuals. Put the counts in L 2 AgeCount 0 to 24177 25 to 44158 45 to 64101 65 +64 Total500

13. Example Continued Now let’s find the X 2 statistic: 1.For each age group, calculate the count you would expect if the 1980 distribution is true –Enter 500 x L 1 into L 3 (500 was the sample size) 2.For each age group, subtract the expected value from the actual count (observed) –Enter L 2 – L 3 into L 4 3.For each age group, square the (O-E) values and divide by E –Enter (L 4 )² / L 3 into L 5 4.Add all the terms to get X 2 –Sum(L 5 ) 5.You should have gotten X² = 8.214

14. The “Goodness of Fit” Test Here are the hypotheses for distributions: –H o : The distributions are the same –H a : The distributions are different To test the hypotheses, look at X 2 1.If the distributions are the same, then (O-E)=0, so X 2 will be zero (or very low) 2.If the distributions differ, then (O-E)²>0, so X 2 will be greater than the low number in case 1 3.The more the distributions differ, the greater will be X 2, the less will be the area under chi-square distribution to the right of X 2

15. Example Continued Form hypotheses for the age distributions –H o : The age distributions are the same –H a : The age distributions differ Find the X 2 statistic –We already found X 2 = 8.214 Find the area to the right of X 2 using n-1 d.f. –In this context, n is the number of categories –Enter Table E (p 842) at row 3 7.81 p>.025 –Use the calculator X 2 cdf(8.214, 999, 3) =.0418 Conclusion: There is good evidence (p<.05) to support the claim that the age distributions are different

16. Today’s Objectives Students will graph the chi-square distributions with various degrees of freedom Students will calculate the chi-square statistic for a distribution sample Students will form hypotheses and determine whether a distribution is the same as or differs from a given distribution California Standard 19.0 Students are familiar with the chi- square distributions and chi- square test and understand their uses.

17. Homework Read pages 702 – 709 Do problems 1 - 4