1. Lecture 10 Inference about the difference between population proportions (Chapter 13.6) One-way analysis of variance (Chapter 15.2)

2. Testing p 1 – p 2 There are two cases to consider: Case 1: H 0 : p 1 -p 2 =0 Calculate the pooled proportion Then Case 2: H 0 : p 1 -p 2 =D (D is not equal to 0) Do not pool the data

3. Example 13.9 (Revisit Example 13.8) –Management needs to decide which of two new packaging designs to adopt, to help improve sales of a certain soap. –A study is performed in two supermarkets: –For the brightly-colored design to be financially viable it has to outsell the simple design by at least 3%. Testing p 1 – p 2

4. Solution –The hypotheses to test are H 0 : p 1 - p 2 =.03 H 1 : p 1 - p 2 >.03 –We identify this application as case 2 (the hypothesized difference is not equal to zero). Testing p 1 – p 2 (Case 2)

5. Compute: Manually The rejection region is z > z  = z.05 = 1.645. Conclusion: Since 1.15 < 1.645 do not reject the null hypothesis. There is insufficient evidence to infer that the brightly-colored design will outsell the simple design by 3% or more. Testing p 1 – p 2 (Case 2)

6. Confidence Interval for confidence interval :

7. Estimating p 1 – p 2 Estimating the cost of life saved –Two drugs are used to treat heart attack victims: Streptokinase (available since 1959, costs $460) t-PA (genetically engineered, costs $2900). –The maker of t-PA claims that its drug outperforms Streptokinase. –An experiment was conducted in 15 countries. 20,500 patients were given t-PA 20,500 patients were given Streptokinase The number of deaths by heart attacks was recorded.

8. Experiment results –A total of 1497 patients treated with Streptokinase died. –A total of 1292 patients treated with t-PA died. Estimate the cost per life saved by using t-PA instead of Streptokinase. Estimating p 1 – p 2

10. Interpretation –We estimate that between.51% and 1.49% more heart attack victims will survive because of the use of t-PA. –The difference in cost per life saved is 2900-460= $2440. –The cost per life saved by switching to t-PA is estimated to be between 2440/.0149 = $163,758 and 2440/.0051 = $478,431 Estimating p 1 – p 2

11. 15.2 One-way ANOVA Analysis of variance compares two or more populations of interval data. Specifically, we are interested in determining whether differences exist between the population means. We obtain independent samples from each population. Generalization of two sample problem to two or more populations

12. Examples Compare the effect of three different teaching methods on test scores. Compare the effect of four different therapies on how long a cancer patient lives. Compare the effect of using different amounts of fertilizer on the yield of a crop. Compare the amount of time that ten different tire brands last.

13. Example 15.1 –An apple juice manufacturer is planning to develop a new product -a liquid concentrate. –The marketing manager has to decide how to market the new product. –Three strategies are considered Emphasize convenience of using the product. Emphasize the quality of the product. Emphasize the product’s low price. One Way Analysis of Variance

14. Example 15.1 - continued –An experiment was conducted as follows: In three cities an advertisement campaign was launched. In each city only one of the three characteristics (convenience, quality, and price) was emphasized. The weekly sales were recorded for twenty weeks following the beginning of the campaigns. One Way Analysis of Variance

15. See file Xm15 -01 Weekly sales

16. Solution –The data are interval. –The problem objective is to compare sales in three cities. –We hypothesize that the three population means are equal. One Way Analysis of Variance

17. H 0 :  1 =  2 =  3 H 1 : At least two means differ To build the statistic needed to test the hypotheses use the following notation: Solution Defining the Hypotheses

18. Independent samples are drawn from k populations (treatments). 12k X 11 x 21. X n1,1 X 12 x 22. X n2,2 X 1k x 2k. X nk,k Sample size Sample mean First observation, first sample Second observation, second sample X is the “response variable”. The variables’ value are called “responses”. Notation

19. Terminology In the context of this problem… Response variable – weekly sales Responses – actual sale values Experimental unit – weeks in the three cities when we record sales figures. Factor – the criterion by which we classify the populations (the treatments). In this problems the factor is the marketing strategy. Factor levels – the population (treatment) names. In this problem factor levels are the marketing strategies.

20. Rationale Behind Test Statistic Two types of variability are employed when testing for the equality of population means –Variability of the sample means –Variability within samples Test statistic is essentially (Variability of the sample means)/(Variability within samples)

21. The rationale behind the test statistic – I If the null hypothesis is true, we would expect all the sample means to be close to one another (and as a result, close to the grand mean). If the alternative hypothesis is true, at least some of the sample means would differ. Thus, we measure variability between sample means.

22. The variability between the sample means is measured as the sum of squared distances between each mean and the grand mean. This sum is called the Sum of Squares for Treatments SST In our example treatments are represented by the different advertising strategies. Variability between sample means

23. There are k treatments The size of sample j The mean of sample j Sum of squares for treatments (SST) Note: When the sample means are close to one another, their distance from the grand mean is small, leading to a small SST. Thus, large SST indicates large variation between sample means, which supports H 1.

24. Solution – continued Calculate SST = 20(577.55 - 613.07 )2 + + 20(653.00 - 613.07) 2 + + 20(608.65 - 613.07) 2 = = 57,512.23 The grand mean is calculated by Sum of squares for treatments (SST)

25. Is SST = 57,512.23 large enough to reject H 0 in favor of H 1 ? Large compared to what? Sum of squares for treatments (SST)

26. 20 25 30 1 7 Treatment 1Treatment 2 Treatment 3 10 12 19 9 Treatment 1Treatment 2Treatment 3 20 16 15 14 11 10 9 The sample means are the same as before, but the larger within-sample variability makes it harder to draw a conclusion about the population means. A small variability within the samples makes it easier to draw a conclusion about the population means.

27. Large variability within the samples weakens the “ability” of the sample means to represent their corresponding population means. Therefore, even though sample means may markedly differ from one another, SST must be judged relative to the “within samples variability”. The rationale behind test statistic – II

28. The variability within samples is measured by adding all the squared distances between observations and their sample means. This sum is called the Sum of Squares for Error SSE In our example this is the sum of all squared differences between sales in city j and the sample mean of city j (over all the three cities). Within samples variability

29. Solution – continued Calculate SSE Sum of squares for errors (SSE)  (n 1 - 1)s 1 2 + (n 2 -1)s 2 2 + (n 3 -1)s 3 2 = (20 -1)10,774.44 + (20 -1)7,238.61+ (20-1)8,670.24 = 506,983.50

30. Is SST = 57,512.23 large enough relative to SSE = 506,983.50 to reject the null hypothesis that specifies that all the means are equal? Sum of squares for errors (SSE)

31. mean squares To perform the test we need to calculate the mean squares as follows: The mean sum of squares Calculation of MST - M ean S quare for T reatments Calculation of MSE M ean S quare for E rror

32. Calculation of the test statistic with the following degrees of freedom: v 1 =k -1 and v 2 =n-k Required Conditions: 1. The populations tested are normally distributed. 2. The variances of all the populations tested are equal.

33. And finally the hypothesis test: H 0 :  1 =  2 = …=  k H 1 : At least two means differ Test statistic: R.R: F>F ,k-1,n-k The F test rejection region

34. The F test H o :  1 =  2 =  3 H 1 : At least two means differ Test statistic F= MST  MSE= 3.23 Since 3.23 > 3.15, there is sufficient evidence to reject H o in favor of H 1, and argue that at least one of the mean sales is different than the others.