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Comparing Means for Several Populations When we wish to test for differences in means for only 1 or 2 populations, we use one- or two-sample t inference. Testing for differences in more than 2 populations, or at several different levels (values) of a variable involves a different approach. This is called Analysis of Variance, or ANOVA. ANOVA partitions the total sum of squares into two parts: 1.within treatment variability 2.between treatment variability

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Comparing Means for Several Populations Example: Test 5 types of concrete for differences in moisture absorption. The 5 types of concrete are the five levels of the treatment. Within Variability – this seeks to quantify the variability in absorption for one particular type of concrete. Between Variability – this seeks to quantify the differences between the types of concrete. ANOVA seeks to answer the question “Are the differences between the 5 sample means what is expected purely from random variation alone?”

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Definitions An experimental unit is an object, or subject, that produces a sample measurement. The experimental conditions that define the different populations in a completely randomized design are called treatments. Testing for differences in the treatments is equivalent to testing for differences in the population means.

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Practice on Definitions See page 399 section 10.1 exercises.

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Graphical demonstration: Employing two types of variability

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Graphical demonstration: Employing two types of variability 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.

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Assumptions for ANOVA 1. The samples are independent –Selection of objects from any one population is unrelated to the selection of objects from any of the other populations. Selections are random. –Examples Different groups of people (no person in more than one group) Different types of music Different concentrations of chemicals Different models of automobiles

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Assumptions for ANOVA 2. Each population has the same standard deviation, But the values of the population standard deviations is not known before testing.

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Assumptions for ANOVA 3. Each sample has a mean that can be calculated. This mean is somehow representative of the population mean for its population.

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Assumptions for ANOVA 4. Each population is normally distributed –Quantitative data: sample size is at least 30 –However, we will assume normally distributed populations for all the problems we work.

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Assumptions for ANOVA The following assumptions are required for a 1-way ANOVA: The k populations are independent. Each population has common standard deviation, . Each population has a mean, i for i = 1, 2, …, k. Each population is normally distributed. So we now are testing whether all the treatment means are equal. H 0 : 1 = 2 = … = k H a : At least two of the population means are not equal

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Test Statistic If the null hypothesis is true, we expect the k sample means to have reasonably similar values. In other words, if the population means are equal, we would expect the variability among the sample means to be relatively small. Variability among the sample means is one of the things we will be testing for.

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Test Statistic If the null hypothesis is true, we do not expect the population means to be exactly the same, because there is a chance factor in our choice of sample experimental units. We need to take into account the variability due to chance among the sample means.

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Test Statistic This method is called “analysis of variance” of ANOVA because we are comparing two sources of variance: the variance among the sample means and the variation expected by chance among the sample means when the null hypothesis is true.

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Test Statistic Our test statistic is called F. F = Variability among the sample means Variability expected by chance

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Degrees of freedom For a sample, (or group) (k) df = n – 1 Total df = total number of units in the experiment – 1 Error df = Total df – Group df –Or Error df = N - k

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Minitab We will use Minitab to do our calculations. A typical Minitab display is on the next slide.

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ANOVA Table: Tensile Strength for 6 Machines Analysis of Variance for Tensile-Strength Source DF SS MS F P Machine 5 5.34 1.07 0.31 0.902 Error 18 62.64 3.48 Total 23 67.98 SSMachine = 5.34 (sample mean variability), k = 6 machines SSError = 62.64 (variability due to chance) Notice how much larger the “chance” variability is than the other. There is little to no evidence that the machines differ in mean tensile-strength. Look at that HUGE p-value!

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Another Minitab Example Example 102 page 369 Sociologist and GPA college students

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One-way ANOVA: GPA versus Group Source DF SS MS F P Group 3 1.519 0.506 2.99 0.044 Error 36 6.091 0.169 Total 39 7.610 S = 0.4113 R-Sq = 19.96% R-Sq(adj) = 13.29% Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev ---+---------+---------+---------+------ Lower Middle 10 2.5240 0.4362 (--------*--------) Poor 10 2.2640 0.3161 (-------*--------) Upper Middle 10 2.7170 0.4125 (--------*-------) Well-to-do 10 2.7560 0.4653 (--------*--------) ---+---------+---------+---------+------ 2.10 2.40 2.70 3.00 Pooled StDev = 0.4113

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Manual Calculation The formula for calculating F using the Mean Square Treatment is given on page 375.

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Manual Calculation To determine the p value when the f value is known, we need to use a table. Table 5 is on pages VII, VIII, IX in the table appendix. In general, Table 5 will provide only approximate p-values. To find precise values, technology is needed.

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ANOVA – What is expected from you? Be able to complete each of the following exercises: State the two hypotheses. What is the observed value of the test statistic? (F = ?) Is this valid? We will typically “assume” the method is ok. What is the p-value? State a conclusion. Using a table for comparisons, locate what mean(s) are significantly different if you accepted the alternative hypothesis. (Sect 10.3)

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Analysis of Variance results: Responses stored in Score. Factors stored in Hair Color. Factor means Hair ColornMeanStd. Error Dark Blond639.53.3936214 Dark Brunette632.6666681.2560962 Light Blond649.8333323.5158372 Light Brunette642.3333323.4123957 ANOVA table SourcedfSSMSF-StatP-value Treatments3908.8333302.944465.44374560.0067 Error20111355.65 Total232021.8334

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Example Page 423 # 1 One-way ANOVA: Score versus Hair Color Source DF SS MS F P Hair Color 3 908.8 302.9 5.44 0.007 Error 20 1113.0 55.7 Total 23 2021.8 H 0 : light_blond = dark_blond = … = dark_brunette H a : At least two population means are different. Accept Ha if p-value < 0.05 F = 5.44p-value = 0.007 At the 0.05 level of significance, there is sufficient evidence to conclude that there is a difference among mean pain thresholds for people possessing these four hair colors.

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10.3 Which means are different? Multiple Comparisons When an analysis of variance F-test indicates a significant difference among population means, (accept H a ), the next question is which means are different.

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Which means are different? We need to test each of the following pairs of hypotheses. Pair 1: H o : μ 1 -μ 2 =0 H a : μ 1 -μ 2 ≠0 Pair 2: H o : μ 1 -μ 3 =0 H a : μ 1 -μ 3 ≠0 Pair 3: H o : μ 2 -μ 3 =0 H a : μ 2 -μ 3 ≠0

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Which means are different? To test each pair of hypothesis, we are only testing two means for a difference between them. This is the two-sample t-statistic that we used in section 9-2. However, we will substitute MSE(Mean Square Error) for s 2 See page 416 for entire equation.

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Which mean is different? We can use StatCrunch to calculate the value of t and the p-value for each of the comparisons. We can then draw our conclusions based on the p-value for each pair (is it less than α? If so we accept the alternative hypothesis), and summarize our findings in a chart. This is how the revised section in the book does it. See example 10.4 p 418

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Let’s look further at the example on hair coloring.

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Multiple Comparisons Pairp-valuet-valueInterpretation LB v DB+2.110.0606NS LB v LBr+1.530.1569NS LB v DBr+4.590.0033LB > DBr DB v LBr-0.590.5691NS DB v DBr+1.890.1052NS LBr v DBr+2.660.0357LBr > DBr Let’s look further at the example on hair coloring

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Summary Ex 10.5 summarizes ideas from Chapter 10. See p 421

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When should we use the multiple comparison method? The sample data are obtained from the k populations using a completely randomized design An analysis of variance F-test indicates that there are some differences among the k population means. The objective is to determine which of the k population means differ. It is usually of interest to determine which mean might be the largest (or smallest).

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