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Copyright © 2012 by Nelson Education Limited. Chapter 9 Hypothesis Testing III: The Analysis of Variance 9-1.

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Presentation on theme: "Copyright © 2012 by Nelson Education Limited. Chapter 9 Hypothesis Testing III: The Analysis of Variance 9-1."— Presentation transcript:

1 Copyright © 2012 by Nelson Education Limited. Chapter 9 Hypothesis Testing III: The Analysis of Variance 9-1

2 Copyright © 2012 by Nelson Education Limited. The basic logic of hypothesis testing as applied to analysis of variance (ANOVA) Perform the ANOVA test using the five-step model Limitations of ANOVA In this presentation you will learn about: 9-2

3 Copyright © 2012 by Nelson Education Limited. ANOVA (analysis of variance) can be used in situations where the researcher is interested in the differences in sample means across three or more categories. Examples: ◦ How do urban, suburban, and rural families vary in terms of number of children? ◦ How do people with less than high school, high school, and post-secondary education vary in terms of income? ◦ How do younger, middle-aged, and older people vary in terms of frequency of religious service attendance? Basic Logic 9-3

4 Copyright © 2012 by Nelson Education Limited. Can think of ANOVA as extension of t test for more than two groups. ANOVA asks “are the differences between the samples large enough to reject the null hypothesis and justify the conclusion that the populations represented by the samples are different?” –The H 0 is that the population means are the same: H 0: μ 1 = μ 2 = μ 3 = … = μ k –If the H 0 is true, the sample means should be about the same value. Basic Logic (continued) 9-4

5 Copyright © 2012 by Nelson Education Limited. If H 0 is false, there should be substantial differences between the sample means of the categories, combined with relatively little difference within (sample standard deviations should be low in value) categories. When we reject the H 0, we are saying there are differences between the populations represented by the samples. Basic Logic (continued) 9-5

6 Copyright © 2012 by Nelson Education Limited. Students taking introductory biology at a large university were randomly assigned to one of three sections: 1.the first section was taught by traditional “lecture-lab” method 2.the second section by “all-lab” method 3.the third section by “videotaped lectures and labs” method.  At the end of the semester, random samples of final exam scores were collected from each section. Basic Logic: An Example 9-6

7 Copyright © 2012 by Nelson Education Limited. In Scenario 1 (Table 9.1) Means and standard deviations of the groups are very similar. These results would be quite consistent with the null hypothesis of no difference. Basic Logic: An Example (continued) 9-7

8 Copyright © 2012 by Nelson Education Limited. In Scenario 2 (Table 9.2) There are large differences in scores between groups (means) but small differences in scores within each group (standard deviations). These results would contradict the null hypothesis, and support the notion that final exam scores do vary by teaching method. Basic Logic: An Example (continued) 9-8

9 Copyright © 2012 by Nelson Education Limited. 1. Calculate total sum of squares (SST): OR Highlighted formula provides a quicker way to calculate the statistic. Six Steps in Computation of ANOVA 9-9

10 Copyright © 2012 by Nelson Education Limited. 2. Calculate sum of squares between (SSB): Six Steps in Computation of ANOVA (continued) 9-10

11 Copyright © 2012 by Nelson Education Limited. 3. Calculate sum of squares within (SSW): OR Highlighted formula provides a quicker way to calculate the statistic. Six Steps in Computation of ANOVA (continued) 9-11

12 Copyright © 2012 by Nelson Education Limited. 4. Calculate degrees of freedom (Formulas 9.5 and 9.6): Six Steps in Computation of ANOVA (continued) 9-12

13 Copyright © 2012 by Nelson Education Limited. 5. Calculate the mean squares (Formulas 9.7 and 9.8): Six Steps in Computation of ANOVA (continued) 9-13

14 Copyright © 2012 by Nelson Education Limited. 6. Calculate F ratio (Formula 9.9): Six Steps in Computation of ANOVA (continued) 9-14

15 Copyright © 2012 by Nelson Education Limited. The computational routine for ANOVA can be summarized as: Six Steps in Computation of ANOVA: Summary 9-15

16 Copyright © 2012 by Nelson Education Limited. The grade point average of students in three (co-ed, all-male, and all-female) residences has been monitored by the administration of a university. The GPA from random samples of 14 students from each residence was collected. Computation of ANOVA: An Example 9-16

17 Copyright © 2012 by Nelson Education Limited. Does GPA vary significantly by type of residence? Computation of ANOVA: An Example (continued) Co-Ed All-Male All-Female 3.5 2.02.0 4.0 1.64.0 3.0 3.03.5 3.0 2.03.5 2.7 3.23.2 3.1 3.43.1 3.2 2.13.0 3.3 2.61.8 3.5 2.82.5 3.2 1.82.8 4.0 2.43.1 3.8 3.12.8 3.6 1.02.6 2.8 0.82.5 9-17

18 Copyright © 2012 by Nelson Education Limited. Co-Ed All-Male All-Female 3.52.02.0 4.01.64.0 3.03.03.5 3.02.03.5 2.73.23.2 3.13.43.1 3.22.13.0 3.32.61.8 3.52.82.5 3.21.82.8 4.02.43.1 3.83.12.8 3.61.02.6 2.80.8 2.5 ΣX = 46.7 31.8 40.4 =3.34 2.27 2.89 ΣX 2 = 158.01 80.62 121.14 = 2.83 Computation of ANOVA: An Example (continued) 9-18

19 Copyright © 2012 by Nelson Education Limited. The difference in the means suggests that GPA does vary by type of residence. GPA seems to be highest in co-ed residence and lowest in all-male residence. Are these differences statistically significant? Computation of ANOVA: An Example (continued) 9-19

20 Copyright © 2012 by Nelson Education Limited. Six Steps in Computation of ANOVA: 1. SST (Formula 9.10) = (158.01+80.62+121.14)-(42)(2.83) 2 = 359.77 -(42)(8.01) = 359.77 -336.42 = 22.35 Computation of ANOVA: An Example (continued) 9-20

21 Copyright © 2012 by Nelson Education Limited. 2. SSB (Formula 9.4) = 14(3.34-2.83) 2 + 14(2.27-2.83) 2 + 14(2.89- 2.83) 2 = 14(0.26) + 14(0.31) + 14(0.0036) = 3.64 + 4.34 + 0.05 = 8.03 Computation of ANOVA: An Example (continued) 9-21

22 Copyright © 2012 by Nelson Education Limited. 3. SSW (Formula 9.11) = 22.35– 8.03 = 14.32 4. Degrees of freedom (Formulas 9.5 and 9.6) dfw= n - k = 42 - 3 = 39 dfb= k - 1 = 3 - 1 = 2 Computation of ANOVA: An Example (continued) 9-22

23 Copyright © 2012 by Nelson Education Limited. 5. Mean Squares (Formulas 9.7 and 9.8) MSW = SSW/dfw =14.32/39 = 0.37 MSB = SSB/dfb = 8.03 /2 = 4.02 Computation of ANOVA: An Example (continued) 9-23

24 Copyright © 2012 by Nelson Education Limited. Computation of ANOVA: An Example (continued) 6. F ratio (Formula 9.9) = 4.02 / 0.37 = 10.86 9-24

25 Copyright © 2012 by Nelson Education Limited. Independent Random Samples Level of Measurement is Interval-Ratio –The dependent variable (e.g., GPA) should be I-R to justify computation of the mean. Populations are normally distributed. Population variances are equal. *ANOVA will tolerate some deviation from its assumptions as long as sample sizes are roughly equal. Performing the ANOVA Test Using the Five-Step Model Step 1: Make Assumptions and Meet Test Requirements* 9-25

26 Copyright © 2012 by Nelson Education Limited. H 0 : μ 1 = μ 2 = μ 3 –The H 0 states that the population means are the same. H 1 : At least one population mean is different. Step 2: State the Null Hypothesis 9-26

27 Copyright © 2012 by Nelson Education Limited. Sampling Distribution = F distribution Alpha = 0.05 dfw = (n – k) = 39 dfb = k – 1 = 2 F(critical) = 3.32 (Note, the exact dfw (39) is not in the table but dfw = 30 and dfw = 40 are. Choose the larger F ratio as F critical). Step 3: Select Sampling Distribution and Establish the Critical Region 9-27

28 Copyright © 2012 by Nelson Education Limited. F (obtained) = 10.86 Step 4: Calculate the Test Statistic 9-28

29 Copyright © 2012 by Nelson Education Limited. F (obtained) = 10.86 F (critical) = 3.32 –The test statistic, F (obtained), falls in the critical region. W e reject the null hypothesis, H 0, of no difference. At least one of these residences is significantly different than the other residences. Step 5: Make Decision and Interpret Results 9-29

30 Copyright © 2012 by Nelson Education Limited. 1.Requires interval-ratio level measurement of the dependent variable 2.Statistically significant differences are not necessarily important. Limitations of ANOVA 9-30

31 Copyright © 2012 by Nelson Education Limited. 3.The alternative (research) hypothesis, H 1, is not specific. It only asserts that at least one of the population means differs from the others. –Thus, we must use other (e.g., post hoc) statistical techniques for more specific differences. Limitations of ANOVA (continued) 9-31


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