Marcia Landell Applied Statistics Week 6: Analysis of Variance (ANOVA) Exercise 36 Analysis of Variance (ANOVA) I 1. A major significance is identifiable between the control group and the treatment group with the F value at 5% level of significance. The p value of 0.005 is less than 0.05 indicating that the control group and the treatment group are indeed different. Based on this fact‚ the null hypothesis is to be rejected. 2. Null hypothesis: The mean mobility scores for the control group and
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EXERCISE 36 Questions to be graded 1. The researchers found a significant difference between the two groups (control and treatment) for change in mobility of the women with osteoarthritis (OA) over 12 weeks with the results of F(1‚ 22) = 9.619‚ p = 0.005. Discuss each aspect of these results. * The F-value suggests that there is a significant difference between the results of the control and treatment groups. The P-value of 0.005 is < the alpha of 0.05. This suggest that the groups are significantly
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rationale for your answer. Would the null hypothesis be accepted or rejected? Answer: The 0.04 > 0.01 would indicate that there is no statistical significance and except the null and conclude that there is no difference between the groups. 6. Can ANOVA be used to test proposed relationships or predicted correlations between variables in a single group? Provide a rationale for your answer. Answer: OVA
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EXERCISE 36 6. Can ANOVA be used to test proposed relationships or predicted correlations between variables in a single group? Provide rationale for your answer. ANOVA cannot be used to test proposed relationships or predicted correlations between variables in a single group because it is designed to test for correlations and interactions amongst groups‚ i.e. in the test group of patients with OA you are testing the correlations between those who do not use GI and PMR and those that do. Although
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EXERCISE 36 Questions to be Graded 1. The researchers found a significant difference between the two groups (control and treatment) for change in mobility of the women with osteoarthritis (OA) over 12 weeks with the results of F (1‚ 22) = 9.619‚ p = 0.005. Discuss each aspect of these results. Answer: The F value suggests there is a significant difference between the results of the control and treatment groups. The P-value of 0.005 is < the alpha of 0.05.This suggest that the groups are significantly
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Exercise 36 Answers 1. Since the F value is significant‚ based on the p-value of 0.005 which is less than 0.05 which is sufficient to reject the null hypothesis. This suggests that there is a difference in the control and treatment groups. 2. Since the p- value is less than 0.05 and therefor the null hypothesis can be rejected. This presents that the mean‚ difficulty and mobility scores‚ must be different 3. The result was statistically significant with a probability score of p < 0.001.
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Exercise 36 1. The researchers found a significant difference between the two groups (control and treatment) for change in mobility of the women with osteoarthritis (OA) over 12 weeks with the results of F(1‚ 22) = 9.619‚ p = 0.005. Discuss each aspect of these results. The F-value is high enough at the 5% level of significance to suggest a significant difference between the control and treatment groups. The p-value 0.005 < 0.05 hence this suggests a rejection of the null hypothesis
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26 3.04 2.8 3.75 3.64 3.65 3.18 3.44 3.51 2.81 3.64 2.85 3.56 2.92 3.35 3.46 3.59 3.65 2.97 3.21 3.65 2.94 3.53 3.65 3.61 3.7 2.91 3.77 3.79 3.59 3.38 3.57 2.97 3.44 3.48 2.99 3.73 2.91 3.78 3.13 3.14 SUMMARY Groups Unemployed Part-time Full-time ANOVA Source of Variation Treatment Error Total Null hypothesis: Alternate hypothesis: Significance level: p-Value Decision: Count 25 45 130 0 0 Sum 82.110 152.050 450.130 0.000 0.000 Average 3.284 3.379 3.463 Variance 0.110 0.085 0.091 SS 0.771 18
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□ EXERCISE 20: Questions to be graded 1. Which patient scored the highest on the preoperative CVLT Acquisition? What was his or her T score? Pt. 3 had a pre-op CVLT score of 63. The T score was 50. 10x47.8/5.8+(50-10x47.8/5.8) 10x47.8=478 478/5.8=82.41 50-82.41= -32.41 -32.41+82.41=50 2. Which patient scored the lowest on postoperative CVLT Retrieval? What was this patient’s T score? Pt. 4 had a post-op CVLT retrieval score of 23. 10*47.8/5.8+(23-10*47.8/5.8 10x47.8=478
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Analysis of Variance (ANOVA) Indian Institute of Public Health Delhi MSc CR 2013-15 Outline of the session • Need for Analysis of Variance • Concept behind one way ANOVA • Example • Non-parametric alternative When dependent variable is continuous Type of Dependent variable Type of Independent variable Number of Groups Continuous Categorical More than two Non-parametric (Wilcoxon sign rank) Paired t – test Not normal Non-parametric (Wilcoxon sign
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