Activity 13 Quiz 4 (20 Points) PDF

1/1/25, 4:30 PM Activity 13: Quiz 4 (20 Points) Quiz Details Quiz Instructions: This quiz will evaluate your understanding of experimental design, particularly focusing on within-subjects and between-subjects designs, as well as the a...

1/1/25, 4:30 PM Activity 13: Quiz 4 (20 Points) Quiz Details Quiz Instructions: This quiz will evaluate your understanding of experimental design, particularly focusing on within-subjects and between-subjects designs, as well as the analysis of variance (ANOVA) and interpretation of results. You will also be tested on hypothesis formulation, main and interaction effects, and effect size interpretation, using both research scenarios and statistical output. You are required to interpret findings from ANOVA tests, draw conclusions from statistical significance levels, and apply knowledge of interaction effects. The quiz will include a variety of question types, including research-based scenarios and multiple-choice questions focused on psychological research methods. Instructions The quiz consists of 14 multiple-choice questions. You have one (1) attempt and a 60-minute time limit for this quiz and You are encouraged to review the assigned textbook and lecture materials before starting. Some questions will require you to analyze statistical results from an ANOVA test, interpret post-hoc comparisons, or evaluate main and interaction effects within research scenarios. Use the feedback provided for each question to guide your understanding of core concepts. Make sure to focus on the context of each research scenario when interpreting statistical results because your ability to apply theory to real- world examples is key to succeeding in this quiz. Show Question Details Question 1.5 pts A psychologist wanted to examine the development of vocabulary skills for a group of 3-year-old children when they turn 4 and then again at age 5. What is this an example of? One-way within-subjects design In this case, there is only one group of subjects measured on one dependent variable, at two different time points. Because the scores for the subjects are the same, this is an example of a one-way within-subjects design. One-way between-subjects design Two-way between-subjects design None of the above Wrong answer comments Refer to the concepts related to between- and within-subjects designs in your chapter and lecture readings. Question 1.5 pts A researcher was interested in studying student motivation in elementary school and the way it impacts final school average. The researcher also believed that enthusiasm for school was also related to final school average. At the end of the school year, a random sample of 20 students from fourth through sixth grade was selected, and measures were taken for all variables. The average scores were compared for each group. What is this an example of? Two-way between-subjects design In this case, a group of three grade levels are compared for two factors: motivation and enthusiasm on final school average. This is a two- way between-subjects design. One-way between-subjects design One-way within-subjects design None of the above Wrong answer comments Refer to the concepts related to between- and within-subjects designs in your chapter and lecture readings. Question 1.5 pts A researcher was interested in studying student motivation in elementary school. A sample of 20 students at the fourth-grade level were administered an inventory measuring motivation. Measures were taken at the beginning of the school year, at the midpoint, and again at the end of the school year. What design did this researcher use? One-way within-subjects design https://calsouthern.instructure.com/courses/40535/quizzes/32464/read_only 1/9 1/1/25, 4:30 PM Activity 13: Quiz 4 (20 Points) In this case, one variable is measured at three different points over time by the same subject. This is an example of a one-way within- subjects design One-way between-subjects design Two-way between-subjects design None of the above Wrong answer comments Refer to the concepts related to between and within subjects designs in your chapter and lecture readings. Question 1.5 pts A researcher was interested in studying student motivation in elementary school and the way it impacts final school average across three different elementary levels. A sample of 20 students was randomly selected from fourth grade to sixth grade for motivation levels categorized as high, medium, and low. Measures were taken for motivation and final school average, and the average scores were compared across the groups. What is this an example of? Two-way between-subjects design Two factors were investigated for final school average: motivation and grade level. This is a two-way between-subjects design. One-way within-subjects design One-way between-subjects design None of the above Wrong answer comments Refer to the concepts related to between and within subjects designs in your chapter and lecture readings. Question 1.4 pts A human factors psychologist studied three different computer keyboard designs. A sample of 45 subjects was given material to type on a particular keyboard, and the number of errors was recorded. Descriptive statistics for the data are presented below, along with the analysis of variance (ANOVA) summary table and post hoc results. Keyboard A Keyboard B Keyboard C n 15 15 15 mean 5 25 30 ANOVA Source df SS MS F sig BG 2 70 35 14.7.023 WG 42 100 2.38 Total 44 ANOVA Effect Sizes Point Estimate 95% CI Eta-squared.60 (.33,.55) Post Hoc Tests Multiple Comparisons Keyboard Sig Tukey A B.06 C.01 https://calsouthern.instructure.com/courses/40535/quizzes/32464/read_only 2/9 1/1/25, 4:30 PM Activity 13: Quiz 4 (20 Points) B A.06 C.15 At the.05 level of significance, what is the decision (reject or fail to reject) and conclusion in the context of this scenario? d) Reject the null hypothesis. There is sufficient evidence to indicate a difference in the average errors across the three different keyboard types. The decision is to reject the null hypothesis. The conclusion is that there is sufficient evidence to indicate a difference in the average errors across the three different keyboard types (because p < alpha). Although p < alpha is correct, it is not the decision or conclusion within the context of this scenario. It is the statistical evidence as to why the null hypothesis is rejected. Reject the null hypothesis because p < alpha. Reject the null hypothesis. There is insufficient evidence to indicate a difference in the average errors across the three different keyboard types. Fail to reject the null hypothesis. There is insufficient evidence to indicate a difference in the average errors across the three different keyboard types. Wrong answer comments Refer to the concepts related to one-way ANOVA design in your chapter and lecture readings. Question 1.4 pts A human factors psychologist studied three different computer keyboard designs. A sample of 45 subjects was given material to type on a particular keyboard, and the number of errors was recorded. Descriptive statistics for the data are presented below, along with the analysis of variance (ANOVA) summary table and post hoc results. Keyboard A Keyboard B Keyboard C n 15 15 15 mean 5 25 30 ANOVA Source df SS MS F sig BG 2 70 35 14.7.023 WG 42 100 2.38 Total 44 ANOVA Effect Sizes Point Estimate 95% CI Eta-squared.60 (.33,.55) Post Hoc Tests Multiple Comparisons Keyboard Sig Tukey A B.06 C.01 B A.06 C.15 https://calsouthern.instructure.com/courses/40535/quizzes/32464/read_only 3/9 1/1/25, 4:30 PM Activity 13: Quiz 4 (20 Points) True or False: Approximately 60% of the variability in keyboard type can be explained by the errors made. True False Approximately 60% of the variability in errors can be explained by the type of keyboard. This is the correct interpretation of the effect size estimated by eta-squared in the results above. Wrong answer comments Refer to the concepts related to one-way ANOVA design in your chapter and lecture readings. Question 1.4 pts A human factors psychologist studied three different computer keyboard designs. A sample of 45 subjects was given material to type on a particular keyboard, and the number of errors was recorded. Descriptive statistics for the data are presented below, along with the analysis of variance (ANOVA) summary table and post hoc results. Keyboard A Keyboard B Keyboard C n 15 15 15 mean 5 25 30 ANOVA Source df SS MS F sig BG 2 70 35 14.7.023 WG 42 100 2.38 Total 44 ANOVA Effect Sizes Point Estimate 95% CI Eta-squared.60 (.33,.55) Post Hoc Tests Multiple Comparisons Keyboard Sig Tukey A B.06 C.01 B A.06 C.15 Based on the statistical results at the.05 level of significance, what is true? Participants using keyboard A did not have statistically less errors than keyboard B but did for keyboard C. Use the Tukey results to answer these questions. There is only a statistically significant difference in the average errors between keyboards A and C only at the.05 level of significance, since.01 <.05. Participants using keyboard B had statistically less errors than keyboard C. Participants using keyboard A had statistically less errors than keyboard B. https://calsouthern.instructure.com/courses/40535/quizzes/32464/read_only 4/9 1/1/25, 4:30 PM Activity 13: Quiz 4 (20 Points) Wrong answer comments Refer to the concepts related to one-way ANOVA design or post hoc tests in your chapter and lecture readings. Question 1.4 pts A research psychologist who lived in the south believed that on very hot summer days, it is difficult to think clearly. The6y have heard comments such as, “It’s not the heat, it’s the humidity”. To evaluate this claim scientifically, they designed a study that involved varying the temperature and humidity level. The heat varied from 70 degrees to 80 degrees to 90 degrees. The humidity varied from low to high. A measure of thinking or working proficiency, as measured by performance on a problem-solving task, were observed for n = 15 subjects in each of these conditions. The measure was scored from 0 to 100, where a higher score indicated better performance. The following matrix summarizes the statistics for each of the levels: Temperature (T) Humidity (H) 70º Room 80 º Room 90º Room Low Humidity n 15 15 15 M=80 (low) mean 85 80 75 High Humidity n 15 15 15 M=70 (high) mean 75 70 65 M=80 M=75 M=70 Which of the following null and alternative hypotheses describe whether there are any differences in thinking or working proficiency for the humidity main effect. Ho: µlow = µhigh H1: µlow ≠ µhigh The following null and alternative hypotheses describe whether there are any differences in thinking or working proficiency for the humidity main effect. Ho: µlow = µhigh H1: µlow ≠ µhigh b) Ho: µ70 = µ80 = µ90 H1: µ70 ≠ µ80 ≠ µ90 Ho: µTxH = 0 H1: µTxH≠ 0 None of the above Wrong answer comments Refer to the concepts related to the two-way ANOVA design in your chapter and lecture readings. Question 1.4 pts A research psychologist who lived in the south believed that on very hot summer days, it is difficult to think clearly. The6y have heard comments such as, “It’s not the heat, it’s the humidity”. To evaluate this claim scientifically, they designed a study that involved varying the temperature and humidity level. The heat varied from 70 degrees to 80 degrees to 90 degrees. The humidity varied from low to high. A measure of thinking or working proficiency, as measured by performance on a problem-solving task, were observed for n = 15 subjects in each of these conditions. The measure was scored from 0 to 100, where a higher score indicated better performance. The following matrix summarizes the statistics for each of the levels: Temperature (T) https://calsouthern.instructure.com/courses/40535/quizzes/32464/read_only 5/9 1/1/25, 4:30 PM Activity 13: Quiz 4 (20 Points) Humidity (H) 70º Room 80 º Room 90º Room Low Humidity n 15 15 15 M=80 (low) mean 85 80 75 High Humidity n 15 15 15 M=70 (high) mean 75 70 65 M=80 M=75 M=70 Which of the following null and alternative hypotheses describe whether there are any differences in thinking or working proficiency for the temperate main effect? Ho: µ70 = µ80 = µ90 H1: µ70 ≠ µ80 ≠ µ90 The following null and alternative hypotheses describe whether there any differences in thinking or working proficiency for the temperate main effect. Ho: µ70 = µ80 = µ90 H1: µ70 ≠ µ80 ≠ µ90 Ho: µlow = µhigh H1: µlow ≠ µhigh Ho: µTxH = 0 H1: µTxH≠ 0 None of the above Wrong answer comments Refer to the concepts related to the Two-Way ANOVA design in your chapter and/or lecture readings. Question 1.4 pts A research psychologist who lived in the south believed that on very hot summer days, it is difficult to think clearly. The6y have heard comments such as, “It’s not the heat, it’s the humidity”. To evaluate this claim scientifically, they designed a study that involved varying the temperature and humidity level. The heat varied from 70 degrees to 80 degrees to 90 degrees. The humidity varied from low to high. A measure of thinking or working proficiency, as measured by performance on a problem-solving task, were observed for n = 15 subjects in each of these conditions. The measure was scored from 0 to 100, where a higher score indicated better performance. The following matrix summarizes the statistics for each of the levels: Temperature (T) Humidity (H) 70º Room 80 º Room 90º Room Low Humidity n 15 15 15 M=80 (low) mean 85 80 75 High Humidity n 15 15 15 M=70 https://calsouthern.instructure.com/courses/40535/quizzes/32464/read_only 6/9 1/1/25, 4:30 PM Activity 13: Quiz 4 (20 Points) (high) mean 75 70 65 M=80 M=75 M=70 Which of the following null and alternative hypotheses describe whether there are any differences in thinking or working proficiency for the interaction effect? Ho: µTxH = 0 H1: µTxH≠ 0 The following null and alternative hypotheses describe whether there are any differences in thinking or working proficiency for the interaction effect. Ho: µTxH = 0 H1: µTxH≠ 0 Ho: µ70 = µ80 = µ90 H1: µ70 ≠ µ80 ≠ µ90 Ho: µlow = µhigh H1: µlow ≠ µhigh None of the above Wrong answer comments Refer to the concepts related to the two-way ANOVA design in your chapter and lecture readings. Question 1.4 pts True or False: Based on the graph below, we can conclude that the effect of room temperature on thinking or working proficiency most likely depends on the levels of humidity. True Refer to the concepts related to two-way ANOVA or interaction in your chapter and lecture readings. False Because the two lines are about parallel, the effect of room temperature on thinking or working proficiency appears to stay the same across the levels of humidity. Therefore, there is no interaction. Question 1.4 pts True or False: Based on the graph below, we can conclude that the effect of room temperature on thinking or working proficiency most likely depends on the levels of humidity. https://calsouthern.instructure.com/courses/40535/quizzes/32464/read_only 7/9 1/1/25, 4:30 PM Activity 13: Quiz 4 (20 Points) True Because the two lines intersect, the effect of room temperature on thinking or working proficiency most likely depends on the levels of humidity. Therefore, there is an interaction effect. False Refer to the concepts related to two-way ANOVA or interaction in your chapter and lecture readings. Question 1.4 pts Suppose the following results were obtained for the scenario above: Tests of Between-Subjects Effects Source df SS MS F sig Partial eta-squared Temp (T) 2 120 240 42.8.001.32 Humidity (H) 1 60 60 10.7.002.24 TxW 2 60 120 21.4.001.27 Error 84 15 5.6 Total 89 Post Hoc Tests Multiple Comparisons Temp Sig Tukey 70 80.20 90.01 80 70.20 90.10 90 70.01 80.10 Tukey Humidity Low high.04 https://calsouthern.instructure.com/courses/40535/quizzes/32464/read_only 8/9 1/1/25, 4:30 PM Activity 13: Quiz 4 (20 Points) True or False: Based on the results above at the.05 level of significance, we can conclude that the effect of room temperature on thinking/working proficiency depends on the humidity level. True Because F(2,84) = 21.4, p=.001, we reject the null hypothesis at the.05 level of significance because p < alpha. The effect of room temperature on thinking or working proficiency depends on the levels of humidity. Therefore, there is an interaction effect. False Refer to the concepts related to two-way ANOVA or interaction in your chapter and lecture readings. Question 1.4 pts Suppose the following results were obtained for the scenario above: Tests of Between-Subjects Effects Source df SS MS F sig Partial eta-squared Temp (T) 2 120 240 42.8.001.32 Humidity (H) 1 60 60 10.7.002.24 TxW 2 60 120 21.4.001.27 Error 84 15 5.6 Total 89 Post Hoc Tests Multiple Comparisons Temp Sig Tukey 70 80.20 90.01 80 70.20 90.10 90 70.01 80.10 Tukey Humidity Low high.04 True or False: Based on the results above, approximately 27% of the variability in the average thinking or working proficiency score can be explained by room temperature. True Refer to the concepts related to Two-Way ANOVA or interaction in your chapter and lecture readings. False Based on the results above, approximately 27% of the variability in the average thinking or working proficiency score can be explained by both room temperature and humidity levels. This is the reported partial-eta-squared reported for the interaction. https://calsouthern.instructure.com/courses/40535/quizzes/32464/read_only 9/9