PSYC204 Practice Exam Solutions

Practice Exam PSYC 204 (Winter 2023) Instructions 1. Answer all 7 questions. Either use the template to answer your questions or answer on a separate sheet of paper. If you answer on a separate sheet of paper, please order your answers as follows: • Question 1 • Question 2 • Question 3 and so on 2. Your exam must be submitted to myCourses as a single PDF (handwritten or typed or the combination of both), no later than April 21, 2023, 9:00 am (Montreal time). 3. Submit your exam on myCourses under the Assignments tab. Make sure that you submit your exam to Final Exam and not any other folder. 4. You will be allowed multiple submissions until the deadline. We will grade the latest submission before the deadline. 5. Make sure you receive an acknowledgement from myCourses that your submission has been received. If you encounter a problem with your submission, you must let your assigned TA know immediately. It will be too late to inform us once the deadline has passed. 6. It is recommended that you do not take the full 72 hours to complete your exam. Do not leave your submission until shortly before the deadline. 7. You must show your reasoning! An answer alone is not sufficient; for example, for a t statistic, you must calculate all separate components (numerator and denominator) and not simply state the t -value. When asked to conduct a hypothesis test, you answer should include all steps of the hypothesis test. 8. You are allowed to use your notes, book, slides, and other informational materials about the course content. This is an "open book/note" exam. It is allowed to use a simple calculator with a square root button while taking the quiz, either on your computer or handheld. 9. You are not allowed to work with others, keep and/or share exam content (via writing it down or taking pictures) with other people or online forums. It is not allowed to use advanced calculators or software that complete the problems for you (e.g., use SPSS to calculate a sample variance). All of these practices are considered cheating and have severe consequences. Where there is evidence of plagiarism or collaboration, the case will be forwarded to the Disciplinary Officer in the student's faculty. By starting to work on the exam, you confirm that you have read these instructions and adhere to standards of academic integrity. Question 1

Suppose that the weights of house cats in Montreal are approximately normally distributed. The distribution has a mean of 7.5 and a variance of 6.25. (a) For a random sample of ? = 49 Montreal house cats, what is the probability that the average weight is less than 8.5? 𝜎 𝑀 = √ 𝜎 2 𝑛 = 2.5 7 = .357 ? = 𝑀−𝜇 𝜎 𝑀 = 8.5−7.5 .357 = 1 .357 = 2.801 ?(𝑀 < 8.5) = .9974 (b) For a random sample of ? = 49 Montreal house cats, what is the probability that the average weight is between 7 and 8.5 pounds? 𝜎 𝑀 = √ 𝜎 2 𝑛 = 2.5 7 = .357 ?1 = 𝑀−𝜇 𝜎 𝑀 = 8.5−7.5 .357 = 1 .357 = 2.801 ?2 = 𝑀−𝜇 𝜎 𝑀 = 7−7.5 .357 = −.5 .357 = −1.400 ?(7 < 𝑀 < 8.5) = ?(𝑀 < 8.5) − ?(𝑀 < 7) = 0.9974 − 0.0808 = 0.9166 (c) What is the 80 th percentile in weight for a random sample of ? = 49 Montreal house cats? ?(? < 0.85) = .8023 𝑀 = 𝜇 + 𝜎 𝑀 ∗ ? = 7.5 + 0.357 ∗ 0.85 = 7.803

Question 3 A psychology researcher wants to evaluate the effect of a clinical treatment on ameliorating depression by lowering depression scores. Suppose the population mean for depression score is  = 40. She sampled 10 individuals to administer the treatment, and after the treatment the average depression score of the group is M = 35. Assume the sample variance is 18, compute a one-tailed hypothesis test at  = .05 to determine whether the treatment effect is significant and find the 95%- confidence interval for  . Step 1: Hypotheses H 0 : 𝜇 = 40 H 1 : 𝜇 < 40 Step 2: Critical value This is a one-tailed test, so we need to find the t-value corresponding to  = .05 in the left tail. First, we select the correct t-distribution based on df = n - 1 = 10 - 1 = 9 In the t-table, we find a critical value of: ? ??𝑖? = − 1.833 Step 3: Test statistic ? = ?̅ − 𝜇 0 ? ?̅ First we need to find the estimated standard error for M 𝑆 𝑀 = √ ? 2 ? = √ 18 10 = √1.8 = 1.342 Then we can compute the test statistic ? = 𝑀 − 𝜇 𝑆 𝑀 = 35 − 40 1.342 = −5 1.342 = −3.73 Step 4: Make a decision regarding H 0 The obtained test statistic (t = -3.73) is more extreme than the critical value ( ? ??𝑖? = -1.833), we can reject the null hypothesis and claim the treatment has an effect on lowering depression score. Additional step: Calculate the confidence intervals The lower confidence interval is 𝜇 = 𝑀 − ?(? 𝑀 ) = 35 − 2.262 × 1.34 = 31.97 The upper confidence interval is 𝜇 = 𝑀 + ?(? 𝑀 ) = 35 + 2.262 × 1.34 = 38.03

Question 4 Sam and Sacha want to settle an argument about which animal is cuter: The North American Opossum or the Ferret. They recruited 23 participants and showed 10 of them a picture of a ferret, while the other 13 were shown a picture of an opossum. Participants gave a rating, on a scale of 0- 100, of how cute they thought the picture was. Sam’s hypothesis was that participants would find the opossum pictures cuter. The data collected are displayed below together with some descriptive statistics. Ferret 62 59 59 79 94 66 47 83 70 86 M = 70.5 SD = 14.65 Opossum 89 99 68 66 59 63 92 51 71 75 96 77 69 M = 75 SD = 14.90 (a) Does Sam have enough evidence to support his hypothesis (at an alpha level of .05)? Follow the steps involved in hypothesis testing and make a decision regarding the null hypothesis based on the comparison between the critical value and the test statistic. Step 1: Hypotheses H 0 : 𝜇 ??????? − 𝜇 ?????? = 0 H 1 : 𝜇 ??????? − 𝜇 ?????? > 0 Step 2: Critical value This is a one-tailed test, so we need to find the t-value corresponding to 𝛼 =.05 in the right tail. First, we select the correct t-distribution based on ?? = ? 1 + ? 2 − 2 = 23 − 2 = 21 . In the t-table, we find a critical value of: ? ??𝑖? = 1.721 Step 3: Test statistic ? = ?̅ 1 − ?̅ 2 √ ? 𝑝 2 ? 1 + ? 𝑝 2 ? 2 First, we need to calculate the pooled variance: 𝑆𝑆 1 = 𝑆? 1 2 ∗ (? 1 − 1) ≈ 1,930.5 𝑆𝑆 2 = 𝑆? 2 2 ∗ (? 2 − 1) ≈ 2,664 ? 𝑝 2 = 𝑆𝑆 1 +𝑆𝑆 2 ?? 1 +?? 2 ≈ 218.79 Now, we can calculate the estimated standard error: 𝑆? = √ ? ? 2 𝑛 1 + ? ? 2 𝑛 2 = √ 218.79 10 + 218.79 13 = 6.22 Now, we can plug these values in the formula for t:

Question 5 An organizational psychologist wants to test whether there is an effect of background noise on productivity. They recruit 20 employees. They then ask the participants to complete different tasks under two different conditions: (1) while there is background noise and (2) while sitting in silence. The number of tasks completed in each condition is recorded. On average, participants completed 12.3 tasks (SD = 3.2) in the background noise condition and 15.2 words (SD = 2.8) in the silence condition. The average difference in the tasks completed between the two conditions was 2.9 tasks with a standard deviation of 3.1 words. (a) Conduct the appropriate statistical test (using an alpha level of .05) to answer the psychologist ’s research question. Step 1: Hypotheses H 0 : 𝜇 𝐷 = 0 H 1 : 𝜇 𝐷 ≠ 0 Step 2: Critical Value This is a two-sided test, so we need to find the t-value corresponding to .05 First, we select the correct t-distribution based on df = n - 1 = 20 - 1 = 19. We can use the t-table to find the critical t-value for df = 19. This gives us the critical values: tcrit = ±2.093. Step 3: Test statistic ? = ? ̅ 𝐷 − 𝜇 𝐷 ? ? ̅ 𝐷 We need to calculate the estimated standard error: ? ? ̅ 𝐷 = ? 𝐷 √𝑛 = 3.1 √20 ≈ 0.69 …and now the test statistic ? = ? ̅ 𝐷 −𝜇 𝐷 ? 𝑋 ̅ 𝐷 = 2.90 .69 ≈ 4.19 Step 4: Make a decision regarding H 0 The obtained test statistic ( ? = 4.19 ) is more extreme than the critical value ( ? ??𝑖? =2.093 ), we therefore reject the null hypothesis and conclude a significant increase in productivity in the silence condition. (b) In a related research article, you read the following text: “We found a significant difference in productivity betwe en the background noise ( M = 9.11, SD = 3.90) and silence conditions ( M = 7.66, SD = 2.60): t (8) = 2.58, p = .04, d = 0.859.” Calculate r 2 as an additional effect size measure.

? 2 = ? 2 ? 2 + ?? = 2.58 2 2.58 2 + 8 ≈ 0.45

Thus, the regression equation is given by ? ̂ = −10.166 + 1.0986 ? .

Question 7 You buy a big bag of skittles; it seems like your favorite flavor is underrepresented (!) — there seem to be fewer lemon skittles in the bag than the other flavors. You decide to dump out all the skittles and count them, are the flavors all equally represented? Red Orange Yellow Green Purple 25 20 18 12 25 (a) Based on the data in the table above, do you have enough evidence to reject the null hypothesis that all flavors are equally represented at an alpha level of .01? (5 Points) Step 1: Hypotheses H 0 : Skittle flavors are equally represented. H 1 : Skittle flavors are not equally represented. Step 2: Critical value We have to find the critical chi-square value corresponding to 𝛼 =.01 and ?? = ? − 1 = 5 − 1 = 4 : 𝜒 ??𝑖? 2 = 13.28 Step 3: Test statistic ? ? = ? ∗ ? = 1 5 ∗ 100 = 20 Red Orange Yellow Green Purple f o 25 20 18 12 25 f e 20 20 20 20 20 𝜒 2 = ∑ (? ? −? 𝑒 ) 2 ? 𝑒 = (25−20) 2 20 + (20−20) 2 20 + (18−20) 2 20 + (12−20) 2 20 + (25−20) 2 20 = 1.25 + 0 + 0.2 + 3.2 + 1.25 = 5.9 Step 4: Make a decision regarding H 0 We fail to reject the null hypothesis because the empirical chi-square value ( 𝜒 2 = 5.9 ) is not larger than the critical chi-square value (13.28). This means that we do not have enough evidence to reject the null hypothesis that all flavors are equally represented at an alpha level of .01 (b) Compute an appropriate effect size measure for the example above. This is an example of a chi- square goodness of fit test. We can, therefore, calculate Cohen’s w as the effect size for the example above. ? = √ 𝜒 2 𝑛 = √ 5.9 100 ≈ 0.24