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
Premium Degrees of freedom Scientific method Sample size
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.
Premium Statistical hypothesis testing Statistical power Statistical significance
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
Premium Statistical hypothesis testing Statistics Experimental design
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: 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. State the null
Premium Statistical hypothesis testing Statistical significance Statistical power
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
Premium Statistics Statistical hypothesis testing Analysis of variance
Exercise 11 Questions to be graded March 16‚ 2014 1. What demographic variables were measured at least at the interval level of measurements? Age‚ income‚ length of labor‚ return to work‚ and number of hours working per week 2. What statistics were used to describe the length of labor in the study? Were these appropriate? Mean and standard deviation were used to describe the length of labor. Yes‚ these were appropriate because a mean (M) can only be used to calculate on interval (and
Premium Standard deviation Statistics Arithmetic mean
EXERCISE IN STATISTICS Below are hypothetical data. (1) Organize them in bivariate tables to answer the problems below. Determine the statistics to use for each. 1. Are sex and occupation associated? 2. Are age and income correlated? 3. Are educational attainment and sex associated? 4. Are civil status and occupation associated? 5. Are occupation and income related N>E> you may use data transformation (from interval data to nominal data) Respondent No. Age Sex Civil Status Educational attainment
Premium Harshad number Chief executive officer Green Line
□ 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
Premium Temporal lobe Arithmetic mean Standard deviation
not = 440 Statistic 4.9 With 98% confidence Lower limit 3.05348411 Upper limit 24.7249703 With 98% confidence we cannot reject Ho since the statistic is inside the acceptance zone b) Check the same hypothesis with 95% confidence. With 95% confidence Lower limit 3.81574825 Upper limit 21.9200493 With 95% confidence we cannot reject Ho since the statistic is inside
Premium Normal distribution Variance Statistics
HLT362V Week 1 Homework EX#16 Answers for EXERCISE 16 page 122 (Questions 1- 4 are optional)• Mean and Standard Deviation Exercise 16: Mean and Standard Deviation 1. The researchers analyzed the data they collected as though it were at what level of measurement? a. Nominal b. Ordinal c. Interval/ratio d. Experimental Answer: c. The researchers analyzed the data as though it were at the interval/ratio level since they calculated means (the measure of central tendency that is appropriate
Premium Standard deviation Arithmetic mean Average