KIN 4310 1nd Edition Lecture 19Outline of Last Lecture I. Two-way ANOVAII. FactorsIII. Main EffectsIV. Interaction EffectsV. ExampleVI. Interaction Effect of FactorsVII. F StatisticVIII. ANOVAIX. ANOVA – SPSS OutputX. Example: Two-way ANOVAOutline of Current Lecture I. WorkshopII. Testing the Significance of CorrelationsIII. Correlation StudiesIV. Correlation Studies StepsV. Hand StudyVI. ReviewThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.Current LectureI. Workshopa. Short-term memory function versus temporal perception: Are they related?b. H1 – STM and TP are correlated c. HO – There is no correlation between STM and TP d. Should we reject Ho? Why or why not?i. We should not reject HO because r = -0.01305 so it is not in the rejection zone because 1. n = 402. df = 383. critical values are 0.3044 and -0.3044II. Testing the Significance of Correlationsa.b. Note: If you have a population and there is no correlation, then the null hypothesis is truec. In this case, they were unlucky with the sample they got which gave them a strong negative correlation which is also a type 1 error.III. Correlation Studiesa. We know how to calculate the linear correlation coefficient, rb. Testing the significance of a correlation i. Determine critical values of r1. Table B4 in bookii. Calculate p-values of r1. Excel wont do this! Need better stats softwarec. Research Hypothesis: i. H1: r > 0ii. H1: r < 0iii. H1 does not equal 0d. Null Hypothesis:i. H0: p = 0 (p is the greek letter row)e. Critical Valuei. Table B4 on page 379f. Degrees of Freedomi. df = total number of data pairs – 2ii. = n - 2g. Note: If the r is significant, it is reasonable to assume it came from a population that was significant. The r distribution has the values from -1 to 1 and you are more likely to get an r that equals around 0 if the null is true and the population is uncorrelated. IV. Correlation Studies Stepsa. Step 1: Calculate test statistic, rb. Step 2: Look up critical value of rc. Step 3: Compare your r to rcritd. Step 4: Reject the null hypothesis?V. Hand Studya.b. r = 0.57c. df = n – 2d. = 48 – 2 = 46e. Using a two-tailed test with alpha = 0.05, find rcriti. Rcrit = 0.2875, -0.2875f. Note: There was a moderate correlation. We’re in the rejection zone. So these sample data say it is unlikely to happen by random chance. So there is a correlation in the population too.VI. Reviewa. Which statement is true about the F-distribution?i. It is negatively skewed? No, it is positively skewedii. 0 less than or equal to F less than or equal to 1. No, F has positive valuesiii. It can be approximated by the normal curve. No, this refers to tiv. It is used in analysis of variancev. The standard deviation is always 1. Falseb. The variation BETWEEN groups represents:i. How much the individual data vary with respect to their group means. No, this is the variation within groupsii. How much the group means vary with one another iii. How much the individual data vary with respect to the mean value of all dataiv. The denominator of the F-statistic. No, it’s the numerator.c. In an experimental study, 30 soldiers are randomly assigned to 4 groups. Each group receives a different style of training. The researchers want to know if the training style has an effect on the soldiers’ physical fitness test scores. What is the critical value for F (let alpha = 0.05)i. -2.69 No, because F is positively skewedii. 0 No, because F is positively skewed.iii. 2.98iv. There is not enough information. No, there is because we need df for the denominator, df for the numerator, and the type 1 error rate which is alpha – we have all those for table B3d. There is no correlation between aerobic fitness and grip strength. What is the probability of measuring r > 0.7293 within a group of 6 randomly selected people? – Here, the null is true which means there is no correlation.i. 0%ii. 2.5% Would be the answer if it was a two-tailed test.iii. 5%iv. 10%v. 95%vi. Note: n = 6, df =