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AGENDA I.Homework II.Exam 1 III.Tukey Test. HOMEWORK 9.26, 9.46, 9.58 (do not do the p-value part), and10.14 Due Friday, Feb. 26.

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Presentation on theme: "AGENDA I.Homework II.Exam 1 III.Tukey Test. HOMEWORK 9.26, 9.46, 9.58 (do not do the p-value part), and10.14 Due Friday, Feb. 26."— Presentation transcript:

1 AGENDA I.Homework II.Exam 1 III.Tukey Test

2 HOMEWORK 9.26, 9.46, 9.58 (do not do the p-value part), and10.14 Due Friday, Feb. 26

3 TUKEY TEST  Once the ANOVA model results in the rejection of the null hypothesis, the only conclusion we have is that not all means are equal. However, it does not tell us which mean(s) is (are) different.

4 TUKEY TEST (cont.)  To test this, we use a multiple comparison test called the Tukey Test. This test will allow us to compare each pair of means to find which pair(s) is (are) different.  Remember, we only do a Tukey test when we reject H 0 in ANOVA.

5 TUKEY TESY – How many pairs?  This test allows us to compare every pair of population means with a single level of significance.  So, the first step is to determine how many pairs of means there are. You can either write all of them down, or use the combination formula:  How many pairs are there if there are 3 means? 4 means? 5 means?

6 Tukey Test - Hypothesis  We have a pair of hypotheses for each pair of means in the following form: H 0 :  i =  j for each pair of means i  j H 1 :  i   j for at least one pair of means where i and j are the appropriate subscripts for the means that you are comparing.  This means we will have to go through as many hypothesis testing procedures, as there are pairs. Fortunately, the critical value is the same for all the pairs.

7 Tukey Test - Critical Value  The critical value for the Tukey Test is calculated as: where: MSE-mean square error from ANOVA N-number of all observations n i -the smallest sample size q , (r, N-r) -the studentized range distribution with r and (N-r) degrees of freedom

8 Tukey Test - Test Statistic  For each pair of hypothesis in the Tukey test; H 0 :  i =  j H 1 :  i   j The calculated test statistic is equal to the absolute difference between the respective sample means  We reject the null hypothesis, if for that particular pair, the test statistic is greater than the critical value, T:  Reject H 0 :  i =  j,if  Accept H 0 :  i =  j,if  The decision is made separately for each pair of hypotheses.

9 The Marketing department wants to compare the difficulty levels of four classes: MKT 300, MKT 310, MKT 317, and MKT students from each class are asked to rate that class using a 1(easy) - 10(difficult) rating scale. The average ratings for each class are as follows: MKT 300: 4.56, MKT 310: 3.92, MKT 317: 5.79, MKT 319: The mean square error (MSE) is 1.5. In the initial ANOVA analysis, the null hypothesis was rejected. Now the department wants to conduct a Tukey test as a follow up to the ANOVA analysis. TUKEY EXAMPLE

10 TUKEY EXAMPLE 2 Nike is studying how well students like various MSU logos. Four logos were tested (the block S, Gruff Sparty, the Spartan helmet, and a new Spartan Helmet) and 25 people were assigned to evaluate each logo (n i =25). A previously run ANOVA test has already revealed that there are significant differences among the four logos. Now the firm would like to know which logos differ. MSE = 121 and α = The average liking of each logo is as follows: 85.3 for the block S, 98.6 (for Gruff Sparty), 88.5 (for the Spartan helmet), and 72.6 (for the new Spartan Helmet).


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