Levene's Test of Equality of Error Variances- Dependent Variable: Stress level after 3 months of therapy df1 df2 Sig. 8.18 42 4.65 Tests the null hypothesis that the error variance of the dependent variable is equal across groups Design: Intercept Baseline_Stress + Group Tests of Between Subjects Effects Dependent Variable: Stress level after months of therapy Type Sum Source of Squares Mean Square Corrected Model 314.796 104.932 Intercept 14.472 14.472 Baseline_Stress 135.062 135.062 Group 69.774 34.887 Error 64.138 5.831 Total 1523.000 Corrected Total 378.933 R Squared .831 (Adjusted R Squared .785) Partial Eta Squared .831 .184 .678 .521 17.996 2.482 23.164 5.983 .000 .414 .001 .017
Added by Heather H.
Close
Step 1
= 0.465 - Between Subjects Effects: Group Mean Square = 34.887, Error Mean Square = 5.831, Group F-value = 5.983, Group Sig. = 0.017 Now, we can interpret the results: Show more…
Show all steps
Your feedback will help us improve your experience
T. L. and 77 other Intro Stats / AP Statistics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Levene' Test of Equality of Error Variances Levene Statistic df1 df2 Sig: gpa Based on Mean Based on Median Based on Median and with adjusted df 516 28 674 .738 739 422 .422 28 25.521 Based on trimmed mean 521 671 Tests the null hypothesis that the error variance of the dependent variable is equal across groups: Dependent variable - gpa Design: Intercept, drink, race Interpretation: There are equal variances across groups, as assessed by Levene's Test of Homogeneity of Variance (p = .671 > .05). Conclusion:
Adi S.
Consider a two-way analysis of variance experiment with treatment factors A and B The results are summarized below Source of Variation df SS Factor A 4 86 Factor B 5 75 Interaction 20 75 Error 90 300 Total 119 536 Compute the mean square and F and test the null hypothesis that no interaction exists between factors A and B at α = .05
Lucas F.
Chapter 13: Experimental Design and Analysis of Variance Question (5 points) The following data are from a completely randomized design: (Data file Treatment can be found in the D2L) Compute the sum of squares between treatments. (.5 point) Compute the mean square between treatments. (.5 point) Compute the sum of squares due to error (1 point) Compute the mean square due to error (1 point) Set up the ANOVA table for this problem. Use the Excel Single Factor ANOVA test and provide a screenshot (1 point) Is there a significant difference between the treatment means? (1 point)
Sri K.
Recommended Textbooks
Elementary Statistics a Step by Step Approach
The Practice of Statistics for AP
Introductory Statistics
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD