Statistical Hypothesis Testing in I/O and Cognitive Psychology - Prof. Kimberly Schneider, Study notes of Psychology

Download Statistical Hypothesis Testing in I/O and Cognitive Psychology - Prof. Kimberly Schneider and more Study notes Psychology in PDF only on Docsity! Review Activity – Exam 2  1-sample t test: (Why? The sample mean is being compared to a population/comparison mean, but the population standard deviation is unknown…see Ch. 7) An I/O psychologist is a consultant to a chain of ice cream stores. Reports show that the average number of complaints received by management for a 1-month period at this time of year is 6.5 (the variance is not known). Seven of this chain’s stores are randomly selected to keep records of complaints received over a 1-month period. The mean number of complaints in those 7 stores was 5.0 with an S2 of 1.15. Does this chain get a different amount of complaints than ice cream stores in general? Use an alpha = .05  Write your null and research hypotheses: Null:  = 6.5 (no difference between our sample stores and average complaintsno difference between our sample stores and average complaints) Research:  does not equal 6.5 (no difference between our sample stores and average complaintsthere is a significant difference).  Find the critical t value and draw a distribution with the cutoff point(s). Critical t value with 6 df, alpha = .05, 2-tailed test = + or – 2.447  Calculate your obtained t value (hint: start by finding SM…it becomes the denominator in your obtained t formula) SM = .405, t = (no difference between our sample stores and average complaints5-6.5) / .405 = -3.70  Draw a conclusion and interpret the findings. Reject the null hypothesis – there is a significant difference in complaints between our sample stores and others (no difference between our sample stores and average complaintsdifference between 5.0 and 6.5)  Dependent Samples t-test (Why? You are comparing two means, but they are both from the same group of people measured twice (before/after)…See Ch. 7) An I/O psychologist working at a company wants to know if workers’ health would improve if they were given extra days off. In this company, all workers take a physical exam twice a year and are given an overall health rating (higher is better health). Thus, the psychologist selected 5 workers at random, arranged to have them get an extra 2 days off per month over the period between exams, and then compared the health scores on the two exams. Their scores are: Worker Health Rating Before Health Rating After Extra Days Off Extra Days Off A 75 81 B 66 67 C 44 46 D 86 88 E 89 89  Write your null and research hypotheses Null: d = 0, there is no difference before/after, 1 = 2 Research: d < 0, there should be higher scores after days off than before (no difference between our sample stores and average complaints1 < 2)  Find the critical t value and draw a distribution with the cutoff point(s). T critical with 4 df, alpha = .05, 1-tailed = -2.132  Calculate your obtained t value (hint: start by finding difference scores for each person, then find the mean difference score and the S2 and SM for the difference scores to use in the t obtained formula) M difference =- 2.2, S2 = 5.2, SM = 1.02, t obtained = -2.16  Draw a conclusion and interpret the findings. Reject the null hypothesis. There is a significant improvement in health after days off (no difference between our sample stores and average complaintsdifference of 2.2 is significantly greater than 0).  Independent Samples t-test (Why? You need to compare two sample means and the two groups are unrelated to each other…See Ch. 9.) A cognitive psychologist conducted a study of the effects of sleep deprivation on short-term memory. Eight participants stayed in a lab for 2 days – four of them were randomly assigned to not sleep during that period, the other four were allowed to sleep when they were tired. At the end of two days, participants completed a short-term memory task. Group 1 (sleep-deprived) had a mean number of letters remembered of 7.75 (S2 = .91). Group 2 (normal sleep) had a mean number of letters remembered of 8.75 (S2 = 2.91). Does sleep deprivation reduce memory (use alpha = .05)?  Write your null and research hypotheses Null: 1 = 2 (no difference between our sample stores and average complaintsno mean differences between groups) Research: 1 does not equal 2  Find the critical t value and draw a distribution with the cutoff point(s). Critical t with 6 df, alpha = .05, 1 tailed = -1.943  Calculate your obtained t value (hint: start by finding S2Pooled, then S2M1 and S2M2, then S2Difference and Sdifference…you’ll finally get to t obtained) S2Pooled = [(no difference between our sample stores and average complaints3/6)(no difference between our sample stores and average complaints.91)] + [(no difference between our sample stores and average complaints3/6)(no difference between our sample stores and average complaints2.91)] = 1.91 S2M1 and S2M2 = 1.91/4 = .48 S2Difference = .48 + .48 = .96; Sdifference = .98 T Obtained = (no difference between our sample stores and average complaints7.75 – 8.75) / .98 = -1.02  Draw a conclusion and interpret the findings. Fail to reject the null. There is not a significant difference in the groups.  One-way ANOVA - Ch 9. Suggested practice problem is #5 from Set 1 at the end of the chapter. Complete parts a (overall ANOVA test) and e-i (follow-up planned comparisons with Bonferroni correction). #5a - Group M S S2 J 41 3.5 12.25 M 38 4.6 21.16 Q 14 3.8 14.44 W 37 4.9 24.01 n=8 (and total N=32) GM = 32.5 (take average of all group means) Step 1 – find S2within = 12.25 + 21.16 + 14.44 + 24.01 4 = 17.97 Step 2a – find S2m = (41-32.5)2 + (38-32.5)2 + (14-32.5)2 + (37-32.5)2 4.1 = 155 Step 2b – find S2between = 155(8) = 1240