Independent Samples ANOVA

Transcription

1 Independent Samples ANOVA In this example students were randomly assigned to one of three mnemonics (techniques for improving memory) rehearsal (the control group; simply repeat the words), visual imagery (form a mental image of what the word refers to) and peg method (memorize a rhyme, one is a gun, two is a shoe, three is a tree. and associate each word with the rhyming word). They then saw a list of 20 words, one word every three seconds. They then had to recall the words in the same order that they appeared in the list. The number of words correctly recognized in the proper order was recorded for each of the 15 participants in each of the three conditions. 1. Step 1: Write the null and alternative hypotheses and specify the probability of making a Type I error: H 0 : µ Rehearsal = μ Imagery = μ Peg H 1 : not H 0 α = Step 2: We will compare the reported p value to α. If p α, we will reject H 0 and conclude that the type of mnemonic likely had an effect on ordered recall.. Step : Calculate the test statistic: a. Open SPSS b. Either type the data (see the second to last page for the data) or open a data set. The class data set (for homework) is available from < The data set used in this example is available from < Save the data file somewhere and open it with SPSS. If you are typing the data, switch to the Variable View (click on that tab in the lower left) and create two variables. Name one of the variables mnemonic and name the other variable recall. Switch back to the Data View. For the mnemonic variable, enter 1 for rehearsal, 2 for imagery and for peg. c. Analyze General Linear Model Univariate

2 d. Move the dependent variable (recall) into the Dependent Variable box. e. Move the independent variable (mnemonic) into the Fixed Factor(s) box:

3 f. If the IV has more than two levels (ours has three levels), click the Post Hoc button

4 g. Move the IV(s) that have more than two levels from the Factor(s) box to the Post Hoc Tests for box. h. Select the desired type of multiple comparison (in our case, Tukey):

5 i. Click the Continue button j. Click the Options button k. Move everything from the Factor(s) and Factor Interactions Box into the Display Means for box l. In the Display section of the dialog box, check Descriptive Statistics, Estimates of Effect Size and Observed Power

6 m. Click the Continue button n. Click the OK button o. The SPSS output viewer will open p. Check the first part of the output to see if the levels of the independent variable are appropriately specified:

7 q. The next part of the output gives descriptive statistics for the dependent variable for each condition (level of the independent variable): r. This tells us that for the mean number of words recalled for the rehearsal mnemonic is 5.8 (mean column and rehearsal row), the sample standard deviation (s) is.2601, and the sample size (N) is 15. Likewise, this tells us that for the imagery condition there were 15 scores (in the N column of the Imagery row), that the sample mean ( ) is. (mean column and the Imagery row), and the sample standard deviation (s) is.92. s. The next part of the output is the ANOVA summary table: n. The only rows of interest are the ones with the IV (mnemonic), Error, and Corrected Total (not Total). The between-treatment information is on the row labeled with the IV. The within-treatment information is on the row labeled with Error. (These should make sense between-treatment variance measures the effect of the IV and error and the within-treatment variances measures error.). Step : Make a decision: Find the p value for the ANOVA (in the column labeled Sig. and the row labeled with the IV). For this output, that value is.000. If the p value is less than or equal to the α level, then you should reject H 0. Otherwise, you should fail to reject H 0.

8 Because p =.000 and α =.05, we reject H 0 and conclude that it is likely the case that at least one of the population means is different from at least one of the other population means. We would write: The mean number of words correctly recalled in the rehearsal (M = 5.8), imagery (M =.) and peg (M = 16.8) conditions are not likely all equal. The ANOVA revealed a main effect of type of mnemonic, F(2, 2) = 7.5, MS error = 1.581, η 2 =.69, p =.000, α = Because we rejected H 0 and the IV had more than two levels, we need to look at the multiple comparisons output. First, write the hypotheses: H 0 : μ Rehearsal = μ Imagery H 1 : μ Rehearsal μ Imagery H 0 : μ Rehearsal = μ Peg H 1 : μ Rehearsal μ Peg H 0 : μ Peg = μ Imagery H 1 : μ Peg μ Imagery 6. Find the Multiple Comparisons part of the output: 7. To see if the population means for rehearsal and imagery are likely different, find one of the rows (there are two of them) where one of the levels (e.g. rehearsal) is listed in the (I)

9 mnemonic column and the other level (e.g. imagery) is listed in the (J) mnemonic column. Look in the Mean Difference (I J) column. In this example, the intersection of that row and column has in it. If there is an asterisk (*) after the mean difference, then you should reject H 0 that those population means are equal. In this case, there is no asterisk after the mean difference, so we fail to reject H 0 there is insufficient evidence in this set of data to conclude that the population means of imagery and rehearsal are likely different. Repeat for the other two multiple comparisons.

10 Step 1 is identical to those used with SPSS. Step 2: We will return to it once we know the dfs. Step : Calculate the test statistic: Rehearsal Imagery Peg ΣX =05 ΣX =561 n N = = SS /15= /15= /15= =612. SS Total = ΣX 2 (ΣX) 2 / N = / 5 = 1996 SS Within-Treatment = SS Rehearsal + SS Imagery + SS Peg = = 612. SS Between-Treatment = SS Total SS Within-Treatment = = SS Between-Treatment = n SS of 5.8,. and 16.8 ΣX = = 27 ΣX 2 = = 5.2 SS Between-Treatment = 15 ( / ) = 18.6 df Total = N 1 = 5 1 = df Between-Treatment = k 1 = 1 = 2 (k is the number of conditions)

11 df Within-Treatment = Σ(n 1) = (15 1) + (15 1) + (15 1) = 2 MS Between-Treatment = SS Between-Treatment / df Between-Treatment = / 2 = MS Within-Treatment = SS Within-Treatment / df Within-Treatment = 612. / 2 = F = MS Between-Treatment / MS Within-Treatment =691.8 / = 7.5 η 2 = SS Between-Treatment / SS Total = 18.6 / 1996 =.69 Step 2: Determine the critical region From a table of critical F values, find F critical with 2 (df numerator) and 2 (df denominator) degrees of freedom with α =.05. F critical =.22 Step : Decide: If the observed or calculated value of F (= 7.5) is in the tail cut off by the critical F (from a table,.22; see step 2 above), then reject H 0, otherwise, fail to reject H 0. Reject H 0. It is likely the case that the type of mnemonic influences ordered recall. Multiple Comparisons: Step 1 (write the hypotheses) is the same as for SPSS Step 2: Find the critical q value We need the df Within-Treatment (2), the number of levels of the IV (k = ) and α (.05). The tabled value of q with those parameters is.7 Step : Calculate the statistic HSD = q (MS error / n) =.7 (1.581 / 15) =.89 Step : Decide If the difference of the sample means is at least as large as the HSD, reject H 0. Rehearsal vs Imagery: If Rehearsal - Imagery HSD, reject H <.89, fail to reject H 0. Rehearsal vs Peg: If Rehearsal - Peg HSD, reject H , reject H 0. Peg vs Imagery: If Peg - Imagery HSD, reject H , reject H 0.