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Lecture Slides. Elementary Statistics Twelfth Edition and the Triola Statistics Series by Mario F. Triola. In chapter 9, we introduced methods for comparing the means from two independent samples.

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**Lecture Slides**Elementary StatisticsTwelfth Edition and the Triola Statistics Series by Mario F. Triola**In chapter 9, we introduced methods for comparing the means**from two independent samples. Now we look to test the equality of three or more means by using the method of one-way analysis of variance (ANOVA). Review**1. The F distribution is not symmetric; it is skewed to**the right. 2. The values of F cannot be negative. 3. The exact shape of the F distribution depends on the two different degrees of freedom. ANOVA Methods Require the F Distribution**9-2 Key Concept**This section introduces the method ofone-way analysis of variance, which is used for tests of hypotheses that three or more population means are all equal. Because the calculations are very complicated, we emphasize the interpretation of results obtained by using technology.**Key Concept**Understand that a small P-value (such as 0.05 or less) leads to rejection of the null hypothesis of equal means. With a large P-value (such as greater than 0.05), fail to reject the null hypothesis of equal means. Develop an understanding of the underlying rationale by studying the examples in this section.**Definition**One-way analysis of variance (ANOVA) is a method of testing the equality of three or more population means by analyzing sample variances. One-way analysis of variance is used with data categorized with one factor (or treatment), which is a characteristic that allows us to distinguish the different populations from one another.**One-Way ANOVA**Requirements 1. The populations have approximately normal distributions. 2. The populations have the same variance σ2 (or standard deviation σ). 3. The samples are simple random samples of quantitative data. 4. The samples are independent of each other. 5. The different samples are from populations that are categorized in only one way.**Procedure**1. Use STATDISK, Minitab, Excel, StatCrunch, a TI-83/84 calculator, or any other technology to obtain results. 2. Identify the P-value from the display. Form a conclusion based on these criteria: If the P-value ≤ α, reject the null hypothesis of equal means. Conclude at least one mean is different from the others. If the P-value > α, fail to reject the null hypothesis of equal means.**Caution**When we conclude that there is sufficient evidence to reject the claim of equal population means, we cannot conclude from ANOVA that any particular mean is different from the others.**Example**Use the performance IQ scores listed in Table 12-1 and a significance level of α = 0.05 to test the claim that the three samples come from populations with means that are all equal.**Example - Continued**Here are summary statistics from the collected data:**Example - Continued**• Requirement Check: • The three samples appear to come from populations that are approximately normal (normal quantile plots OK). • The three samples have standard deviations that are not dramatically different. • We can treat the samples as simple random samples. • The samples are independent of each other and the IQ scores are not matched in any way. • The three samples are categorized according to a single factor: low lead, medium lead, and high lead.**Example - Continued**The hypotheses are: The significance level is α = 0.05. Technology results are presented on the next slides.**Example - Continued**The displays all show that the P-value is 0.020 when rounded. Because the P-value is less than the significance level of α = 0.05, we can reject the null hypothesis. There is sufficient evidence that the three samples come from populations with means that are different. We cannot conclude formally that any particular mean is different from the others, but it appears that greater blood lead levels are associated with lower performance IQ scores.**P-Value and Test Statistic**• Larger values of the test statistic result in smallerP-values, so the ANOVA test is right-tailed. • The figure on the next slide shows the relationship between the F test statistic and the P-value. • Assuming that the populations have the same variance σ2(as required for the test), the F test statistic is the ratio of these two estimates of σ2: • variation between samples (based on variation among sample means) • variation within samples (based on the sample variances).**Caution**When testing for equality of three or more populations, use analysis of variance. Do not use multiple hypothesis tests with two samples at a time.

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