Section 9 2 introduced the two-sample t-test method to test the hypothesis that the means of 2

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1 Section 9 3: ANOVA: Testing a claim that there is no difference in 3 or more population means Section 9 2 introduced the two-sample t-test method to test the hypothesis that the means of 2 normal populations are equal. H 0 : µ 1 = µ 2 using the two-sample test statistic: t = ( ) x 1 x 2 ( s 1 ) 2 + n 1 ( s 2 ) 2 n 2 ANOVA is a general technique to test the Null Hypothesis that the means of 3 or more populations are equal. The populations must be normal and all have the same standard deviations. =... H 1 : The population means ARE NOT ALL EQUAL µ 1 = µ 2 = µ 2 x µ 1 µ 2 µ 3 x Example An athletic trainer claims that the average resting heart rates of men in the age groups 15 to 24 years old, 25 to 34 years old and 35 to 44 years old are the same rate. Does each group have the same average heart rate? We will call Population 1 the 15 to 24 years olds with a mean of µ 1 We will call Population 2 the 25 to 34 years olds with a mean of µ 2 We will call Population 3 the 35 to 44 years olds with a mean of µ 3 The Null Hypotheses H 0 is a claim that all the age groups have the same average heart rate versus the alternate Hypotheses H 1 H 1 : At least one of theaverage heart rate is different from the others. Section 9 3 Page 1 of Eitel

2 Can we use Section 9-2 two-sample t-tests? You could try to test all possible pairs of means by the use of two-sample t-tests. Test 1 Test 2 Test 3 H 0 : µ 1 = µ 2 H 0 : µ 1 = µ 3 H 0 : µ 2 = µ 3 If a series of separate tests µ 1 = µ 2 and µ 2 = µ 3 and µ 1 = µ 3 all produced a DO NOT REJECT H 0 then you could conclude all the population means were equal µ 1 = µ 2 = µ 3. If we use a σ =.05 vale to conduct one two-sample t-test. H 0 : µ 1 = µ 2 then we would be able to say with a 95% certainty that the means µ 1 = µ 2 are equal. To conclude that µ 1 = µ 2 = µ 3 we would need to conduct 3 two-sample t-tests, µ 1 = µ 2 and µ 2 = µ 3 and µ 1 = µ 3, and have each test reach the correct conclusion. If we conduct the three two-sample t-tests. µ 1 = µ 2 and µ 2 = µ 3 and µ 1 = µ 3 then the certainty that all three are correct is.95 3 =.86 and the chances of a type I error becomes 1.86 or σ =.14 As the number of populations increases then the certainty of your conclusion decreases and the chances of a type 1 error increases. The number of separate tests required also increases as the number of populations increase. To address these problems ANOVA techniques were developed. These techniques allow us to perform only one test and allows the certainty of the conclusion to remain at 1 σ. For these reasons, ANOVA techniques are preferred in comparing three or more means for equality. This section will develop the techniques of One Way ANOVA where the populations are defined by only one factor that is used to differentiate the populations. Testing a Claim that the Means of Three or More Normal Populations are Equal using One Way ANOVA The computations required to perform ANOVA Tests are so tedious that the use of technology is the only practical way to conduct the tests. The use of software like EXCEL, MINI TAB or the use of a TI 84 Calculator are very common. This section will introduce the use of EXCEL as the tool to conduct ANOVA tests. You will NOT be required to use EXCELL. The reports table that is produced from the data table will be given to you. You will be expected to find the numbers in the results table and compare these numbers. That comparision will allow you to draw a conclusion about the claim the all the means are equal to each other µ 1 = µ 2 = µ 3 Section 9 3 Page 2 of Eitel

3 One Way ANOVA Requirements If the following requirements are met then the techniques of One Way ANOVA can be used to test if versus H 1 : At least one of the population mean is different form the others. given k samples of size n and a desired level of significance. 1. k populations are defined with only one factor is used to differentiate the populations. All other factors concerning the population must be the same ( or very similar) except for the one factor is used to differentiate the populations. 2. The k populations must be normally distributed. 3. The k populations must have the same variance or standard deviation σ 1 = σ 2 = σ A rule of thumb is that One Way ANOVA procedures can be used if the largest sample standard deviation is no more than two times as large as the smallest sample standard deviation. 4. A random sample is taken form each of the k populations. The sample size for each sample should be the same but this is not required. Each sample must be independent of the others. The subjects in one sample cannot be related to the subjects in the others sample in any way. The requirements for One Way ANOVA are robust. This means that you can be close to meeting the requirements and the procedure will still provide a good test of the hypothesis. The requirement of equal standard deviations can be relaxed if the sample size taken form each of the separate populations are equal. If the following requirements are met then the techniques of One Way ANOVA can be used to test versus H 1 : At least one of the population mean is different form the others. given k samples of size n and a desired level of significance. Conclusions If the F test statistic is less than the critical value Do NOT Reject H 0. If the F test statistic is greater than the critical value Reject H 0. Section 9 3 Page 3 of Eitel

4 Example 1 An athletic trainer claims that the resting heart rate of men in the age groups 15 to 24 years old, 25 to 34 years old and 35 to 44 years old are the same. It is known that the restring heart rates for the populations are normally distributed. An independent random sample of 4 people in each of the 3 age groups was taken and their heart rates were measured and recorded. Test the hypothesis that the populations have the same mean against the alternate hypothesis that at least one of the means is different from the other means with a α =.05 level of significance. Resting Heart Rates by Age Group in beats/min. 15 to 24 year olds to 34 year olds to 44 year olds Sample Mean Sample SD H 1 : At least one of the population mean is different from the others. Excel Results for α =.05 ANOVA Source of Variation SS df MS F Test Stat P value Critical f value Between Within TOTAL The F test statistic is F =.0604 and the F Critical Value is The F test statistic is less than the critical value and falls in the Do NoT Reject H 0 region. Conclusion. DO Not Reject. I am 95% confident that the means of the three populations are the same. The average resting heart rate for men in the age groups 15 to 24 years old, 25 to 34 years old and 35 to 44 years old Section 9 3 Page 4 of Eitel

5 Example 2 An student claims that the average test scores on test 1 are the same for the morning, afternoon and night sections of Stat 300. It is known that the test scores for the populations are normally distributed. An independent random sample of 4 test scores in each of the 3 section was taken and the scores are recorded in the table below. Test the hypothesis that the populations have the same mean against the alternate hypothesis that at least one of the means is different from the other means with a α =.05 level of significance. Sample Test Scores for Test 1 Morning Class Afternoon Class Night Class Sample Mean Sample SD Excel Results for α =.05 ANOVA Source of Variation SS df MS F Test Stat P value Critical f value Between Within TOTAL H 1 : At least one of the population mean is different from the others. The F test statistic is F = and the F Critical Value is The F test statistic is more than the critical value and falls in the Reject H 0 region. Conclusion based on H 0 : Reject Conclusion based on the problem: I am 95% confident that the means of the three populations ARE NOT the same. The average test scores for morning, afternoon and night sections of STAT 300 ate NOT the same. Section 9 3 Page 5 of Eitel

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