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When you finish this chapter, you should be able to: Discuss the general idea of analysis of variance, discuss the general idea of analysis of variance, conduct a test of hypothesis to determine whether the variances of two populations are equal, organize data into a one-way and a two-way ANOVA table.
12 1 Copyright © 2004 by The McGrawHill Companies, Inc. All rights reserved. 12 2 When you have completed this chapter, you will be able to: Discuss the general idea of analysis of variance List the characteristics of the F distribution Conduct a test of hypothesis to determine whether the variances of two populations are equal. Organize data into a oneway and a twoway ANOVA table Copyright © 2004 by The McGrawHill Companies, Inc. All rights reserved. 12 3 Define the terms treatments and blocks Conduct a test of hypothesis to determine whether three or more treatment means are equal Develop multiple tests for difference between each pair of treatment means Copyright © 2004 by The McGrawHill Companies, Inc. All rights reserved. Characteristics of the Characteristics of the FDistribution FDistribution 12 4 There is a “family of FDistributions: There is a “family of FDistributions: Each member of the family is determined by two parameters: …the numerator degrees of freedom, and the … denominator degrees of freedom F cannot be negative, and it is a continuous distribution The F distribution is positively skewed Its values range from 0 to as F , the curve approaches the Xaxis Copyright © 2004 by The McGrawHill Companies, Inc. All rights reserved. Test for Equal Variances Test for Equal Variances 12 5 For the two tailed test, the test statistic is given by: F s1 s 22 for the two samples s 12 and are the sample variances s 22 The null hypothesis is rejected if the computed value of the test statistic is greater than the critical value Copyright © 2004 by The McGrawHill Companies, Inc. All rights reserved. 12 6 Colin, a stockbroker at Critical Securities, reported that the mean rate of return on a sample of 10 internet stocks was 12.6 percent with a standard deviation of 3.9 percent. The mean rate of return on a sample of 8 utility stocks was 10.9 percent with a standard deviation of 3.5 percent. At the .05 significance level, can Colin conclude that there is more variation in the internet stocks? Copyright © 2004 by The McGrawHill Companies, Inc. All rights reserved. Hypothesis Testing Hypothesis Testing 12 7 Step 1 Step 1 State the null and alternate hypotheses State the null and alternate hypotheses Step 2 Step 2 Select the level of significance Select the level of significance Step 3 Step 3 Identify the test statistic Identify the test statistic Step 4 Step 4 State the decision rule State the decision rule Step 5 Step 5 Compute the value of the test statistic Compute the value of the test statistic and make a decision and make a decision Do not reject H00 Do not reject H Copyright © 2004 by The McGrawHill Companies, Inc. All rights reserved. Reject H0 0 and accept and accept H Reject H H11 Hypothesis Test Hypothesis Test Step 1 Step 1 Step 2 Step 2 Step 3 Step 3 12 8 H : I2 U H : I2 U Select the level of significance Select the level of significance = 0.05 The test statistic is the Identify the test statistic Identify the test statistic F distribution State the null and alternate State the null and alternate hypotheses hypotheses Reject H0 if F > 3.68 The df are 9 in the numerator and 7 in the denominator. s 12 ( ) F = 1.2416 2 s2 (3 ) Step 4 Step 4 State the decision rule State the decision rule Step 5 Step 5 Compute the test Compute the test statistic and make statistic and make a decision a decision Do not reject the null hypothesis; there is insufficient evidence to show more variation in the internet stocks. Copyright © 2004 by The McGrawHill Companies, Inc. All rights reserved. ANOVA The F distribution is also used for testing The F distribution is also used for testing whether whether two or more two or more sample means sample means came from came from the same or equal the same or equal populations populations This this technique is called analysis of variance or ANOVA Copyright © 2004 by The McGrawHill Companies, Inc. All rights reserved. 12 9 ANOVA ANOVA requires the following requires the following 12 10 conditions… conditions… …the sampled populations follow the normal distribution …the populations have equal standard deviations …the samples are randomly selected and are independent Copyright © 2004 by The McGrawHill Companies, Inc. All rights reserved. Confidence Interval for the Confidence Interval for the Difference Between Two Means Difference Between Two Means X1 X2 t MSE 1 n1 n2 (1712.75) 2.228 975 MSE MSE 25 Copyright © 2004 by The McGrawHill Companies, Inc. All rights reserved. 48 12 35 1 ( 2.77, 5.73) Confidence Interval for the Confidence Interval for the Difference Between Two Means Difference Between Two Means 12 36 …continued Because zero is not in the interval, we conclude that this pair of means differs The mean number of meals sold in Aynor is different from Lander Copyright © 2004 by The McGrawHill Companies, Inc. All rights reserved. 12 37 ANOVA ANOVA For the twofactor ANOVA we test whether there is a significant difference between the treatment effect and whether there is a difference in the blocking effect! SSB B r2 k ( X) n …Let Br be the block totals (r for rows) …Let SSB represent the sum of squares for the blocks Copyright © 2004 by The McGrawHill Companies, Inc. All rights reserved. ANOVA ANOVA 12 38 The Bieber Manufacturing Co. operates 24 hours a day, five days a week. The workers rotate shifts each week. Todd Bieber, the owner, is interested in whether there is a difference in the number of units produced when the employees work on various shifts. A sample of five workers is selected and their output recorded on each shift. At the .05 significance level, can we conclude there is a difference in the mean production by shift and in the mean production by employee? Copyright © 2004 by The McGrawHill Companies, Inc. All rights reserved. 12 39 ANOVA ANOVA …continued Employee Day Evening Output Output Night Output McCartney 31 25 35 Neary 33 26 33 Schoen 28 24 30 Thompson 30 29 28 Wagner 28 26 27 Copyright © 2004 by The McGrawHill Companies, Inc. All rights reserved. 12 40 Hypothesis Test Hypothesis Test Difference between various shifts? Difference between various shifts? Step 1 Step 1 State the null and alternate State the null and alternate hypotheses hypotheses Step 2 Step 2 Select the level of significance Select the level of significance Step 3 Step 3 Identify the test statistic Identify the test statistic The test statistic is the F distribution Step 4 Step 4 State the decision rule State the decision rule Step 5 Step 5 Compute the test Compute the test statistic and make statistic and make a decision a decision Reject H0 if F > 4.46. The df are 2 and 8 F Copyright © 2004 by The McGrawHill Companies, Inc. All rights reserved. H 0: 1= 2= H : Not all means are equal = 0.05 SST k SSE ( k )( b 1) ANOVA ANOVA 12 41 …continued Compute the various sum of squares: SS(total) = 139.73 Using Using SST = 62.53 to get to get SSB = 33.73 these these SSE = 43.47 results results df(block) = 4, df(treatment) = 2 df(error)=8 Copyright © 2004 by The McGrawHill Companies, Inc. All rights reserved. 12 42 ANOVA ANOVA …continued Step 5 Step 5 F SST k SSE ( k )( b 62 53 43.47 3 1) = 5.754 Since 5.754 > 4.46, H0 is rejected. There is a difference in the mean number of units produced on the different shifts Copyright © 2004 by The McGrawHill Companies, Inc. All rights reserved. 12 43 Hypothesis Test Hypothesis Test Difference between various shifts? Difference between various shifts? Step 1 Step 1 State the null and alternate State the null and alternate hypotheses hypotheses Step 2 Step 2 Select the level of significance Select the level of significance Step 3 Step 3 Identify the test statistic Identify the test statistic The test statistic is the F distribution Step 4 Step 4 State the decision rule State the decision rule Reject H0 if F > 3.84 The df are 4 and 8 Step 5 Step 5 Compute the test Compute the test statistic and make statistic and make a decision a decision F Copyright © 2004 by The McGrawHill Companies, Inc. All rights reserved. H 0: 1= 2= H : Not all means are equal = 0.05 SST k SSE ( k )( b 1) ANOVA ANOVA …continued Step 5 Step 5 F SST k SSE ( k )( b 33.73 43.47 1) = 1.55 Since 1.55