1. 1 Always be mindful of the kindness and not the faults of others.

2. Categorical Data Sections 10.1 to 10.5 Estimation for proportions Tests for proportions Chi-square tests

3. 3 Example Researchers in the development of new treatments for cancer patients often evaluate the effectiveness of new therapies by reporting the proportion of patients who survive for a specified period of time after completion of the treatment. A new treatment of 870 patients with lung cancer resulted in 330 survived at least 5 years.

4. 4 Example Estimate , the proportion of all patients with lung cancer who would survive at least 5 years after being administered this treatment How much would you estimate the proportion as?

5. 5 Distribution of Sample Proportion Y: the number of successes in the n trials (independent and identical trials) What’s the distribution of Y? Sample proportion,

6. 6 Distribution of Sample Proportion When n  ≥ 5 and n(1-  ≥ 5, the distribution of Y can be approximated by a normal distribution. (approximate) (1-  ) Confidence Interval for  : Optional: (exact) C.I. for  for small sample

7. 7 Sample Size Where E is the largest tolerable error at (1-  confidence level.

8. 8 Test for a Large Sample When n    ≥ 5 and n(1-    ≥ 5, the test statistic is:

9. 9 Inference about 2 Proportions Notation: Population 1Population 2 Proportion  Sample sizen1n2 # of successesy1y2 Sample proportion

10. 10 Estimation for  Point estimate:

11. 11 Estimation for  (1-  ) Confidence Interval for two large samples:

12. 12 Example 10.6 A company markets a new product in the Grand Rapids and Wichita. In Grand Rapids, the company’s advertising is based entirely on TV commercials. In Wichita, based on a balanced mix of TV, radio, newspaper, and magazine. 2 months after the ad campaign begins, the company conducts surveys to determine consumer awareness of the product.

13. 13 Example 10.6: Data Set Grand RapidsWichita # of interviewed608527 # of aware392413 Q: Calculate a 95% C.I. for the regional difference in the proportion of all consumers who are aware of the product.

14. 14 Example 10.6 (conti.) Conduct a test at  =0.05 to verify if there are >10% more Wichita consumers than Grand Rapids consumers aware of the product.

15. 15 Test for  Large Samples) When n1  ≥ 5 and n1(1-  ≥ 5; n2  ≥ 5 and n2(1-  ≥ 5, the test statistic of Ho: p1-p2=d is Optional: Fisher Exact Test (p.511)

16. Minitab Z procedure for one proportion: Stat >> Basic Statistics >>1 proportion Z procedure for two proportions: Stat >> Basic Statistics >>2 proportion Sample size calculation: Stat >>Power & Sample size>>1 proportion or 2 proportion Stat >>Power & Sample size>>sample size for estimation 16

17. 17 Chi-Square Goodness of Fit Test More than two possible outcomes per trial  the multinomial experiment 1. The experiment consists of n identical trials. 2. Each trial results in one of k outcomes with probabilities     ...  k. Y=(Y 1,…,Y k ); Y i = the # of outcome i.

18. 18 Chi-square Goodness of Fit Test Goal:We are interested in testing a hypothesized distribution of Y (i.e. a set of  i ’s values). Hypotheses: Ho:  i =  io for all ivs. Ha: Ho is false

19. 19 Chi-square Goodness of Fit Test Test Statistic: ni = the observed Yi Ei = the expected Yi = n  io

20. 20 Chi-square Goodness of Fit Test Rejection Region: Reject Ho if where df=k-1. Note: This test can be trusted only when 80% of more cells of the Ei’s are at least 5.

21. 21 Example 10.10 CategoryHypothesized %Observed counts Marked decrease50120 Moderate decrease2560 Slight decrease10 Stationary of slight increase 1510

22. Minitab: Stat >> Tables >> Chi-Square Goodness-of-Fit Test(One Variable) 22 Example 10.11

23. 23 Contingency Table(Example 10.12) n ij Age Category Severity of skin disease 1234 Total n i* 1153218570 2829231878 3120252268 Total n *j 24816645216 = n

24. 24 Contingency Table 2 categorical variables: row and column indexed by i and j, respectively If they are independent, then

25. 25 Test for Independence of 2 Var’s Hypotheses: Ho: the row and column variables are independent Ha: they are dependent Test Statistic:

26. 26 Test for Independence of 2 Var’s Rejection Region: Reject Ho if where df=(r-1)(c-1). Note: This test can be trusted only when 80% of more cells of the are at least 5.

27. Minitab: Stat >> Tables >> Cross Tabulation and Chi-square Tabulated Statistics: C1, Worksheet columns Rows: C1 Columns: Worksheet columns 1 2 3 4 5 All A 32 87 91 46 44 300 3.56 9.67 10.11 5.11 4.89 33.33 42 107 78 34 39 B 53 141 76 20 10 300 5.89 15.67 8.44 2.22 1.11 33.33 42 107 78 34 39 C 41 93 67 36 63 300 4.56 10.33 7.44 4.00 7.00 33.33 42 107 78 34 39 All 126 321 234 102 117 900 14.00 35.67 26.00 11.33 13.00 100.00 Cell Contents: Count % of Total Expected count Pearson Chi-Square = 72.521, DF = 8, P-Value = 0.000 Likelihood Ratio Chi-Square = 79.263, DF = 8, P-Value = 0.000 27 Example 10.12