PSY 216. Assignment 12 Answers. Explain why the F-ratio is expected to be near 1.00 when the null hypothesis is true.

Transcription

1 PSY 21 Assignment 12 Answers 1. Problem 1 from the text Explain why the F-ratio is expected to be near 1.00 when the null hypothesis is true. When H0 is true, the treatment had no systematic effect. In the formula for F: F = (systematic treatment effects + random, unsystematic difference) / random, unsystematic differences The systematic treatment effects term will be 0 if the null hypothesis is true. Thus, when H0 is true, F will equal F = (0 + random, unsystematic differences) / random, unsystematic differences Any value divided by itself equals Problem 3 from the text Several factors influence the size of the F-ratio. For each of the following, indicate whether it would influence the numerator or the denominator of the F-ratio, and indicate whether the size of the F-ratio would increase or decrease. a. Increase the differences between the sample means. This affects the numerator of the F-ratio. As the sample means become more different, the treatment has a larger and larger effect. The systematic effect of the treatment is in the numerator of the F-ratio. The size of the F-ratio would increase. b. Increase the size of the sample variance. This affects both the numerator and denominator of the F-ratio. Sample variance is the random, unsystematic change in the scores. Increasing sample variance will decrease the size of the F-ratio 3. Problem from the text The following data summarize the results from an independent-measures study comparing three treatment conditions: (Problem from the text)

2 I II III n = n = n = M = 1 M = M = N = 18 T = T = 30 T = 3 G = 2 SS = 30 SS =3 SS = 0 ΣX 2 = a. Use an ANOVA with α =.0 to determine whether there are any significant differences among the three treatment conditions. Step 1: State the hypotheses and α H0: μ1 = μ2 = μ3 H1: not H0 (or, at least one of the treatment means is different) α =.0 Step 2: Locate the critical region dfbetween Treatments = k (number of conditions) 1 = 3 1 = 2 dfwithin Treatments = N k = 18 3 = 1 Consult a table of critical F values with dfbetween Treatments = 2, dfwithin Treatments = 1, and α =.0. The critical value of F = Step 3: Perform the ANOVA SS Total = X 2 G2 N = = 189 SS Within = SS Inside each treatment = = 10 SS Between = T2 n G2 N = = = 8 18 dftotal = N 1 = 18 1 = 1 dfbetween = 2 (from step 2) dfwithin = 1 (from step 2) MSBetween = SSBetween / dfbetween = 8 / 2 = 2 MSWithin = SSWithin / dfwithin = 10 / 1 = F = MSBetween / MSWithin = 2 / =

3 Source SS df MS F Between Treatments Within Treatments 10 1 Total Step : Make a decision Because the calculated F is in the tail cut off by the critical F, we reject H0 and conclude that the treatment likely had an effect. b. Calculate η 2 to measure the effect size for this study. η 2 = SSBetween Treatment / SSTotal = 8 / 189 = 0. Because η 2 is larger than 0.2, this is a large effect. c. Perform multiple comparisons if appropriate. Because we rejected H0 and there were more than two conditions, it is appropriate to perform multiple comparisons. In this case, I am calculating the Tukey s Honestly Significant Differences. HSD = q MS Within n = 3. = 3.99 The value of q comes from a table of critical values of the Studentized Range with α =.0, three groups (number of conditions) and dfwithin = 1. Any pair of means that are at least 3.99 different are reliably different by Tukey s HSD test. Conditions 1 (M = 1) and 2 (M = ) are reliably different as are conditions 1 and 3 (M = ). However, conditions 2 and 3 are not reliably different from each other. d. Write a sentence demonstrating how a research report would present the results of the hypothesis test and the measure of effect size. The means and standard deviations of the three conditions can be found in Table 1. The analysis of variance revealed an effect of the treatment, F(2, 1) =.00, p <.0, α =.0, η 2 =.. According to Tukey s HSD, conditions 1 and 2, and 1 and 3 are reliably different, while conditions 2 and 3 are not reliably different from each other. Table 1

4 Means and standard deviations of the conditions Condition 1 Condition 2 Condition 3 M 1 s Problem 1 from the text The following summary table presents the results from an ANOVA comparing three treatment conditions with n = 8 participants in each condition. Complete all missing values. (Hint: Start with the df column.) Source SS df MS F Between Treatments Within Treatments 30 = 2 * 1 2 = = 1 / 3 3 = = (8 1) * 3 3 = 3 / 21 Total = 8 * 3 1 There are different ways that you could solve this. Here is one possible way: 1. dftotal = number of scores 1 = (8 participants per condition X 3 conditions) 1 = 2 1 = dfbetween Treatments = number of conditions 1 = 3 1 = 2 3. dfwithin Treatments = dftotal dfbetween Treatments = 23 2 = 21. dfwithin Treatments = Σ(number of people in each condition - 1) = (8 1) + (8 1) + (8 1) = 21 (Sum across the conditions). SSBetween Treatments = MSBetween Treatments X dfbetween Treatments (an algebraic rearrangement of MS = SS / df) SSBetween Treatments = 1 X 2 = 30. SSWithin Treatments = SSTotal SSBetween Treatments (an algebraic rearrangement of SSTotal = SSBetween Treatments + SSWithin Treatments) SSWithin Treatments = = 3. MSWithin Treatments = SSWithin Treatments / df Within Treatments = 3 / 21 = 3. F = MSBetween Treatments / MSWithin Treatments = 1 / 3 =. Problem 23 from the text New research suggests that watching television, especially medical shows such as Grey s Anatomy and House can result in more concern about personal health (Ye, 2010). Surveys administered to college

5 students measure television viewing habits and health concerns such as fear of developing the diseases and disorders seen on television. For the following data, students are classified into three categories based on their television viewing patterns and health concerns are measured on a 10-point scale with 0 indicating none. Television Viewing Little or none Moderate Substantial n1 = 10 T1 = 0 SS1 = n2 = 10 T2 = 0 SS2 = 3 n3 = 10 T3 = SS3 = 20. N = 30 G = 1 ΣX 2 = 933 T1 = = 0 SS1 = / 10 = = T2 = = 0 SS2 = / 10 = = 3 T3 = = SS3 = / 10 = = 20. N = n1 + n2 + n3 = = 30 G = T1 + T2 + T3 = = 1 ΣX 2 = = 933 a. Use an ANOVA with α =.0 to determine whether there are significant mean differences among the three groups. Step 1: State the hypotheses and α H0: μ1 = μ2 = μ3 H1: not H0 (or, at least one of the treatment means is different) α =.0 Step 2: Locate the critical region

6 dfbetween Treatments = k (number of conditions) 1 = 3 1 = 2 dfwithin Treatments = N k = 30 3 = 2 Consult a table of critical F values with dfbetween Treatments = 2, dfwithin Treatments = 2, and α =.0. The critical value of F = 3.3. Step 3: Perform the ANOVA SS Total = X 2 G2 N = = SS Within = SS Inside each treatment = = 100. SS Between = T2 n G2 N = = = 31. dftotal = N 1 = 30 1 = 29 dfbetween = 2 (from step 2) dfwithin = 2 (from step 2) MSBetween = SSBetween / dfbetween = 31. / 2 = MSWithin = SSWithin / dfwithin = 100. / 2 = 3.22 F = MSBetween / MSWithin = / 3.22 =.2 Step : Make a decision Because the calculated F is in the tail cut off by the critical F, we can reject H0 and conclude that watching TV has an effect on health concerns. b. Compute η 2 to measure the size of the effect. η 2 = SSBetween Treatment / SSTotal = 31. / = 0.20 Because η 2 is between 0.09 and 0.2, this is a medium effect. c. Use Tukey s HSD test with α =.0 to determine which groups are significantly different.

7 HSD = q MS Within = n 10 = 2.10 The value of q comes from a table of critical values of the Studentized Range with α =.0, three groups (number of conditions) and dfwithin = 2. Any pair of means that are at least 2.10 different are reliably different by Tukey s HSD test. Only conditions 1 (M = ) and 3 (M =.) are reliably different. 8. Enter the data from problem 21 into SPSS. Use SPSS to answer the following question: Do the data indicate any significant mean differences among the three groups of birds? Give H0, H1 and α. Is this a one-tailed or two-tailed test? Write a sentence or two in APA format that summarizes the results of the analysis. One possible explanation for why some birds migrate and others maintain year round residency in a single location is intelligence. Specifically, birds with small brains, relative to their body size, are simply not smart enough to find food during the winter and must migrate to warmer climates where food is easily available (Sol, Lefebvre, & Rodriguez-Teijeiro, 200). Birds with bigger brains, on the other hand, are more creative and can find food even when the weather turns harsh. Following are hypothetical data similar to the actual research results. The numbers represent relative brain size for the individual birds in each sample: Non- Migrating Step 1: Short- Distance Migrants Long- Distance Migrants 9 H0: μnon-migratingd = μshort-distance migrant = μlong-distance migrant H1: not H0 α =.0 Step 2: Skip use p value from SPSS output Two Tailed ANOVA is always two tailed

8 a. Step 3: In SPSS create two variables one for the IV and one for the DV. In variable view (Ctrl-T, click on the Variable View tab in the lower left corner, or click on View Variables), enter migratory in row one and brainsize in row two of the Name column. Click the cell at the intersection of the migratory row and Value column. A button with should appear in the cell. Click on it. Enter 1 in the value box and Non-Migrating in the label box. Click Add Enter 2 in the value box and Short-Distance Migrants in the label box. Click Add Enter 3 in the value box and Long-Distance Migrants in the label box. Click Add Click OK Switch to Data View (Ctrl-T, click on the Data View tab in the lower left corner, or click on View Data) Enter the data from the table into SPSS. For each non-migrating bird, enter a 1 in the migratory column and its brain size in the brainsize column. For each short-distance migrants bird, enter a 2 in the migratory column and its brain size in the brainsize column. For each long-distance migrants bird, enter a 3 in the migratory column and its brain size in the brainsize column. Click on Analyze General Linear Model Univariate Move brainsize into the Dependent Variables box Move migratory into the Fixed Factor(s) box (the IV box) Click the Post Hoc button Move migratory into the Post Hoc Tests for: box. Click in the check box to the left of Tukeys Click Continue Click on Options Move migratory into the Display Means for: box Click in the check box to the left of Descriptive statistics and Estimates of effect size

9 Click on Continue Click on OK Descriptive Statistics Dependent Variable:BrainSize MigrantType Mean Std. Deviation N Non-Migrating Short-Distance Migrants Long-Distance Migrants Total Dependent Variable:BrainSize Tests of Between-Subjects Effects Source Type III Sum of Squares df Mean Square F Sig. Partial Eta Squared Corrected Model a Intercept MigrantType Error Total Corrected Total a. R Squared =.20 (Adjusted R Squared =.83) Multiple Comparisons BrainSize Tukey HSD Mean 9% Confidence Interval (I) MigrantType (J) MigrantType Difference (I-J) Std. Error Sig. Lower Bound Upper Bound Non-Migrating Short-Distance Migrants.0000 * Long-Distance Migrants * Short-Distance Migrants Non-Migrating * Long-Distance Migrants Long-Distance Migrants Non-Migrating * Short-Distance Migrants Based on observed means. The error term is Mean Square(Error) =.33. *. The mean difference is significant at the.0 level. F(2, 1) = 19.29, p =.000, MSerror =.3, η 2 =.2

10 Step : Reject H0 because p =.000 which is less than α =.0 Multiple Comparisons: Step 1: H0: μnon = μshort H1: μnon μshort H0: μnon = μlong H1: μnon μlong H0: μshort = μlong H1: μshort μlong Step 2: Skip use p value from output Step 3: Skip Step : Non vs. short reject H0 (p =.003) Non vs. long reject H0 (p =.000) Short vs. long fail to reject H0 (p =.138) Summary: The mean and standard deviations of brain size for the three groups of birds (non-migrating, shortdistance migrants and long-distance migrants) are shown in Table 1. The analysis of variance revealed a significant effect of the type of bird on brain size, F(2, 1) = 19.28, p =.000, MSerror =.3, η 2 =.2. Tukey multiple comparisons revealed that non-migrating birds had a larger brain than both short-distance migrants (p =.003) and long-distance migrants (p =.000). There was insufficient evidence to suggest that the brain size of short- and long-distance migrants were reliably different (p =.138). Table 1 Brain Size of Different Types of Birds Group M SD Non-Migrating Short-Distance Migrant Long-Distance Migrant

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12 Hand calculations: ΣX = 90 ΣX 2 = 1398 M = 1 Non-Migrating Short-Distance Migrants Long-Distance Migrants SSNon = / = 8 ΣX = ΣX 2 = 20 M = 9 SSShort = 20 2 / = 3 ΣX = 3 ΣX 2 = 232 M = SSLong = / = 1 ΣX = = 180 ΣX 2 = =210 M = 10 SSTotal = / = 30 SSBetween-Treatments = n SSM SSM = ΣM 2 (ΣM) 2 / k ΣM = = 30 ΣM 2 = = 32 SSM = / 3 = 2 SSBetween-Treatments = 2 = 22 SSWithin-Treatments = ΣSSfor each group = = 98 SSTotal = 30 Check: SSTotal = SSBetween-Treatments + SSWithin-Treatments = dfbetween-treatments = k 1 = 3 1 = 2 dfwithin-treatments = Σ(n 1) = ( 1) + ( 1) + ( 1) = 1 dftotal = N - 1 = 18 1 = 1 Check: dftotal = dfbetween-treatments + dfwithin-treatments = = 1 MSBetween-Treatments = SSBetween-Treatments / dfbetween-treatments = 22 / 2 = 12 MSWithin-Treatments = SSWithin-Treatments / dfwithin-treatments = 98 / 1 =.3 F = MSBetween-Treatments / MSWithin-Treatments = 12 /.3 = η2 = SSBetween-Treatments / SSTotal = 22 / 30 =.2 HSD = q MS Within Treatment n = 3..3 = 3.83

13 MNon-Migrating MShort-Distance Migrants > HSD Reject H0: μnon-migrating = μshort-distance Migrants MNon-Migrating MLong-Distance Migrants > HSD Reject H0: μnon-migrating = μlong-distance Migrants MShort-Distance Migrants - MLong-Distance Migrants < HSD Fail to reject H0: μshort-distance Migrants = μlong-distance-migrants