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Comparing Proportions for Multiple Populations

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Presentation on theme: "Comparing Proportions for Multiple Populations"— Presentation transcript:

1 Comparing Proportions for Multiple Populations
Section Lesson

2 Starter 13.2.1 Group n mean Std dev Breast-fed 23 13.3 1.7 Bottle-fed
A study of iron deficiency in infants compared two groups. One had been breast-fed, the other had been fed formula from a bottle. The hemoglobin levels were measured at age 12 months. Here are the results: Assuming this was a properly randomized experiment, is there significant evidence that the mean hemoglobin level is different between the groups? Group n mean Std dev Breast-fed 23 13.3 1.7 Bottle-fed 19 12.4 1.8

3 Today’s Objectives Students will compare three or more proportions to determine if they are all the same or are significantly different. California Standard 19.0 Students are familiar with the chi- square distributions and chi- square test and understand their uses.

4 Why would you want to compare more than two proportions?
Consider an experiment with a control group and several treatment groups. For example, does aspirin prevent heart attacks, and if so what dosage is needed? Treatments might be 100 mg, 250 mg, 500 mg We would then ask if the success rates differ.

5 Can’t we just compare in pairs?
We could compare each pair of p-hat’s, but there are problems with this approach: If there are a lot of treatments, this is tedious. Perhaps the greatest and least p-hat’s differ, but neither differs from a central value. We want to know how likely it is that the values are spread as far as they are.

6 Example 13.3 An experiment compared two different anti-depressants and a placebo for effectiveness in helping cocaine addicts prevent a relapse.Here are the results: Is this good evidence that the drugs are effective (i.e. that proportions differ)? Group Treatment # subjects No relapse Proportion 1 Desipramine 24 14 .583 2 Lithium 6 .250 3 Placebo 4 .167

7 Arrange data in a two-way table with counts of success & failure
This is a 3x2 table (3 rows, 2 columns). This is called r x c; we will use it later for degrees of freedom The explanatory variable is the treatment. One of the three drugs on each row The response variable is success (no relapse) or failure. Success in first column, failure in second No Relapse Relapse Desipramine 14 10 Lithium 6 18 Placebo 4 20

8 Hypotheses of the test The null hypothesis is that there are no differences among the proportions. Ho: p1 = p2 = p3 The alternative is that they are not all the same: there is some difference. Ha: not all proportions are equal

9 Test with expected counts
If all proportions really are the same, how many successes would we expect for each treatment? P(success) = total # of successes in all treatments divided by total number of subjects in experiment. In other words, column total / table total So multiply number of subjects in a treatment by P(success) to get expected successes for treatment. In other words, P x row total So expected count is (column)(row)/(table).

10 Form a table of observed and expected counts
Calculate the # of expected counts for each treatment and summarize in a table. Observed Success Failure Expected Desipramine 14 10 8 ? Lithium 6 18 Placebo 4 20

11 Form a table of observed and expected counts
Calculate the # of expected counts for each treatment and summarize in a table. Now calculate the chi-square statistic. Observed Success Failure Expected Desipramine 14 10 8 16 Lithium 6 18 Placebo 4 20

12 Chi-square Statistic The formula is the same as we have seen:
“O” and “E” refer to observed and expected counts Use the (O – E) differences for both the successes and failures Degrees of freedom will be (r-1)(c-1)

13 Example Continued Calculate the chi-square statistic for the table:
Then find the p-value and draw a conclusion. Observed Success Failure Expected Desipramine 14 10 8 16 Lithium 6 18 Placebo 4 20

14 Example Concluded Put observed successes & failures in L1
Put expected successes & failures in L2 Let L3 be defined as (L1 – L2)²/L2 Sum L3 to get X² X² = (r-1)(c-1) degrees of freedom is (3-1)(2-1)=2 X²cdf(10.5, 999, 2) = .0052 Conclusion: There is strong evidence (p = .005) that the proportions of success among the three treatments are not the same.

15 Today’s Objectives Students will compare three or more proportions to determine if they are all the same or are significantly different. California Standard 19.0 Students are familiar with the chi- square distributions and chi- square test and understand their uses.

16 Homework Read pages 717 – 728 Do problems 13, 15
Re-do class example (13.3) on TI See p 726 for instructions


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