Question 12 pts ________ is a weighted average of the two estimates of ________. Standard deviation; pooled variance Pooled variance; standard deviation Pooled variance; variance Variance; pooled variance Flag this QuestionQuestion 22 pts By weighting the two estimates of variance in our calculation of pooled variance, we are able to allow the estimate based on a larger sample to contribute more to the calculation. We do this because: we increase power. it is the more trusted estimate of variance, as large samples tend to provide greater accuracy. we reduce the chance of outliers impacting our data. it helps to lower the critical cutoff for significance. Flag this QuestionQuestion 32 pts When calculating standard error for the independent-samples t test, we divide by N of each sample rather than the square root of Nbecause: the test statistic is based on N and things need to be consistent for meaningful comparisons. we have already used a correction and do not need to correct the estimate again. we are dealing with variance rather than standard deviation, so everything is squared. this helps to increase our estimate of variability and is more conservative. Flag this QuestionQuestion 42 pts A psychologist is interested in whether working memory is influenced by sleep loss. The psychologist administers a measure of working memory to two groups of subjects. The subjects in one group were kept awake for the entire night and the subjects in the other group maintained their normal sleep schedules. The independent variable in this study is: memory system being studied. sleep condition. performance on memory measure. group number. Flag this QuestionQuestion 52 pts The first step in calculating a confidence interval for an independent-samples t test is to: look up the t statistics for the lower and upper bounds. draw a normal curve with the sample difference between means in the center. write the percentages for the upper and lower bounds. convert the t statistics to raw score differences between means. Flag this QuestionQuestion 62 pts Dr. Jameson wanted to know if IQ scores differed between male and female participants in his study. He gave 28 participants an IQ test and then compared IQ scores for gender differences. He hypothesized that there would be a statistically significant gender difference in IQ scores. Contrary to Dr. Jameson’s hypothesis, there were no differences in IQ scores between men and women in his study. Which test was Dr. Jameson most likely to use to test his hypothesis? paired-samples t test z test single-samples t test independent-samples t test Flag this QuestionQuestion 72 pts A study found statistically significant results for a hypothesis tested with an independent-samples t test. The author of the study reported her effect size for the test as 1.24. What is true of the two sample means? The two sample means overlap 85%. The two samples likely come from the same distribution. The two sample means are 1.24 standard deviations apart. The two samples means do not indicate meaningful differences between groups. Flag this QuestionQuestion 82 pts In a(n) ________, each participant is assigned to only one group in order to compare mean differences. dependent-samples t test z test independent-samples t test single-sample t test Flag this QuestionQuestion 92 pts Random assignment helps to: equalize initial conditions. guarantee equal size groups. sensitize our measure of the dependent variable. remove all bias from a study. Flag this QuestionQuestion 102 pts When creating a distribution of differences between means, it is important that we: be consistent in the arithmetic used to calculate the difference between means. always get a positive difference value when comparing the two means. subtract the mean of the first group from the mean of the second group. subtract the mean of the second group from the mean of the first group. Flag this QuestionQuestion 112 pts An independent-samples t test was conducted to compare gas expenses of diesel truck owners to owners of regular gas trucks. Imagine the mean gas consumption in one week amounted to $68 for diesel trucks and $84 for regular gas trucks. Imagine also that values were calculated for sdifference of 6.9, 21.74 for pooled variance, and 82 for degrees of freedom total. What decision should she make based on these data? Fail to reject the null hypothesis and conclude we don’t know if there is a difference between these trucks in terms of gas consumption. Reject the null hypothesis and conclude that diesel trucks cost significantly less in gas per week than regular gas trucks. Reject the null hypothesis and conclude that diesel trucks cost significantly more in gas per week than regular gas trucks. Fail to reject the null hypothesis and conclude the test lacked adequate power. Flag this QuestionQuestion 122 pts For the following data, compute corrected variance, 1, 2, 3, 4, 5. 2.5 1.41 2.0 1.58 Flag this QuestionQuestion 132 pts Unnithan, Houser, and Fernhall (2006) were interested in whether playing the game Dance Dance Revolution (DDR) affected the heart rate of overweight and nonoverweight adolescents differently. A group of 22 adolescents, 10 classified as overweight and 12 as not overweight, played DDR for 12 minutes, during which time the researchers measured each participant’s heart rate. Which statistical test should the researchers use to analyze their data? z test independent-samples t test single-sample t test dependent-samples t test Flag this QuestionQuestion 142 pts In order to report the exact p value associated with the test statistic for an independent-samples t test, we need to: record the value from the t table under the 0.05 p level. compute Cohen’s d. record the value from the t table under the 0.01 p level. use software to obtain the value. Flag this QuestionQuestion 152 pts For an independent-samples t test, there were 8 participants in Group 1 and 11 participants in Group 2. The critical cutoffs for a two-tailed test with alpha of 0.05 are: -2.101 and 2.101 -2.120 and 2.120 -2.132 and 2.132 -2.110 and 2.110 Flag this QuestionQuestion 162 pts In addition to reporting the results of statistical hypothesis testing, it is also recommended that researchers report: effect size and confidence intervals. effect size and mean difference scores. effect size and median differences. confidence intervals and random selection results. Flag this QuestionQuestion 172 pts The benefits of the within-groups design are seen, in the calculation of the F statistic, in that: the denominator is smaller, resulting in a larger F value. the numerator is larger, resulting in a larger F value. the numerator is smaller, resulting in a smaller F value. the denominator is larger, resulting in a smaller F value. Flag this QuestionQuestion 182 pts A one-way within-groups ANOVA on 2 and 15 degrees of freedom results in an F statistic of 3.12. What decision should be made about the hypothesis based on this statistic? Reject the null hypothesis and conclude the independent variable has no effect. Fail to reject the null hypothesis and conclude there is no effect of the independent variable. Fail to reject the null hypothesis because the F statistic does not exceed the critical cutoff. Reject the null hypothesis and conclude there are differences among the levels of the independent variable. Flag this QuestionQuestion 192 pts ________ is the proportion of variance in the dependent variable that is accounted for by the independent variable. z t R2 Cohen’s d Flag this QuestionQuestion 202 pts A measure of the differences among group means is: between-groups variance. the proportionate reduction in error. within-groups variance. the z score. Flag this QuestionQuestion 212 pts According to Cohen’s conventions, an effect size of 0.019 for a one-way within-groups ANOVA is considered to be: medium-to-large. large. medium. small. Flag this QuestionQuestion 222 pts With two groups, the square root of F is equal to: 0.0. 1.0. 2t. t. Flag this QuestionQuestion 232 pts ________ is(are) an additional assumption that must be evaluated in a within-groups ANOVA. Mean differences Order effects Sample size Counterbalancing Flag this QuestionQuestion 242 pts An F statistic calculated on 2 and 26 degrees of freedom equals 2.23. Which decision would you make about a hypothesis tested at the p 0.05 level? Reject the null hypothesis and conclude there are differences among the groups. Reject the null hypothesis and conclude no significant differences are indicated among the groups. Fail to reject the null hypothesis and conclude there are differences among the groups. Fail to reject the null hypothesis and conclude no significant differences are indicated among the groups. Flag this QuestionQuestion 252 pts Pet preference was compared for four different types of animals (dogs, cats, birds, and fish), using a one-way within-groups ANOVA with seven participants. The standard error was calculated as 0.21. Given the following means for preference, where are the significant effects for preference among these groups? Dogs M = 4.43, Cats M = 4.18, Birds M = 2.73, Fish M = 4.07 Birds and fish are less preferred than dogs and cats, but there are no other differences present. Preference for birds is significantly lower than for the other three groups, and there are no other differences among the mean preferences. All post-hoc tests were significant at the 0.05 level. Dogs and cats are preferred over birds and fish, but there are no differences between the other groups. Flag this QuestionQuestion 262 pts Dr. Sanders was interested in investigating the effects of stress on memory. He exposed participants to public speaking stress and room temperature stress and gave them a memory measure immediately after exposure to the two stressors. Dr. Sanders also gave participants the memory measure at baseline, when they first entered the research laboratory. Dr. Sanders hypothesized that participants’ memories would be most affected by the two stressors compared to the baseline condition. What statistical test should Dr. Sanders use to test her hypothesis? between-groups ANOVA independent samples t test within-groups ANOVA single-sample t test Flag this QuestionQuestion 272 pts According to Cohen’s conventions, an effect size of 0.29 for a one-way within-groups ANOVA is considered to be: small. medium-to-large. medium. large. Flag this QuestionQuestion 282 pts Pet preference was compared for four different types of animals (dogs, cats, birds, and fish), using a one-way within-groups ANOVA with seven participants. The standard error was calculated as 0.21. The groups’ means are provided below. What are the critical cutoffs for the two-tailed Tukey HSD tests for these data, assuming a p level of 0.05? Dogs M = 4.43, Cats M = 4.18, Birds M = 2.73, Fish M = 4.07 4.0 3.96 –4.0 and 4.0 –3.96 and 3.96 Flag this QuestionQuestion 292 pts Pet preference was compared for four different types of animals (dogs, cats, birds, and fish), using a one-way within-groups ANOVA with seven participants. The standard error was calculated as 0.21. Given the following means for preference, what is the Tukey HSDvalue when comparing data for dogs and cats? Dogs M = 4.43, Cats M = 4.18, Birds M = 2.73, Fish M = 4.07 0.544 0.198 1.714 1.190 Flag this QuestionQuestion 302 ptsMSbetween is obtained by: dividing SSbetween by dfbetween. dividing SSwithin by SSwithin. dividing SSbetween by SSwithin. dividing dfbetween by SSbetween. Flag this QuestionQuestion 312 pts A one-way within-groups ANOVA is also known as a ________ ANOVA. between-groups repeated-measures multiple groups single samples Flag this QuestionQuestion 322 pts The z, t, and F calculations have something in common: the numerator of the test statistic: contains a measure of sample variability. contains a measure of difference among groups. represents what would be expected to happen by chance. is a squared number. Flag this QuestionQuestion 332 pts Which of the following statements accurately captures why within-groups designs are preferred over between-groups designs? Loss of participants has a more significant negative impact on between-groups compared to within-groups designs. Within-groups designs take less time than between-groups designs. Control of extraneous variables is easier when using between-groups compared to within-groups designs. Variability due to participants’ differences is held constant across levels of the independent variable in the within-groups design, resulting in less within-groups variability. Flag this QuestionQuestion 342 pts The following figure reflects the results of a study by Roediger and Karpicke (2006). The authors investigated whether the test-enhanced learning effect (the demonstration that repeated testing improves memory for material) was due merely to repeated exposure to the material. They randomly assigned participants to one of two study conditions (study-study or study-test) and to one of three retention interval conditions (final test at a delay of 5 minutes, 2 days, or 1 week). The dependent variable was the proportion of idea units recalled from an encyclopedia passage. Figure: Testing and Memory Reference: Figure 1 (Figure: Testing and Memory) The figure reflects a main effect of study condition. Which statement best describes the main effect? On average, people in the study-test condition had better memory than people in the study-study condition. On average, people in the study-study condition had better memory than people in the study-test condition. On average, memory performance decreased as the retention interval increased. People in the study-test condition performed more poorly on the 5-minute recall test than did those in the study-study condition, but at longer retention intervals (2 days and 1 week); those in the study-test condition performed better than those in the study-study condition. Flag this QuestionQuestion 352 pts A researcher performs a 3 x 5 ANOVA examining how preferences for engagement with friends varied as a function of age (20, 30, or 40 year olds) and social media tools (Facebook, Twitter, LinkedIn, Google+, Instagram) with 26 participants in each cell of the study design. What is the degrees of freedom for the main effect of social media tool? 3 4 1 2 Flag this QuestionQuestion 362 pts Main effects refer to the: effect of one level of the independent variable on the dependent variable, disregarding other levels of the independent variable. combined effects of two dependent variables. effect of a single independent variable on the dependent variable, disregarding all other variables in the study. combined effects of two independent variables. Flag this QuestionQuestion 372 pts The following table reflects the results of a study by Roediger and Karpicke (2006). The authors investigated whether the test-enhanced learning effect (the demonstration that repeated testing improves memory for material) was due merely to repeated exposure to the material. They randomly assigned participants to one of two study conditions (study-study or study-test) and to one of three retention interval conditions (final test at a delay of 5 minutes, 2 days, or 1 week). The dependent variable was the proportion of idea units recalled from an encyclopedia passage. Table: Test-Enhanced Learning 5 minutes 2 days 1 week mean Study-study 0.80 0.55 0.42 0.59 Study-test 0.75 0.70 0.55 0.67 Mean 0.78 0.63 0.49 Reference: Table 1 (Table: Test-Enhanced Learning) The cells of this study reflect an interaction. Is this a quantitative or qualitative interaction? Why? This is a quantitative interaction because the effect of the study condition does not reverse depending on the retention interval. This is a qualitative interaction because the effect of the study condition reverses depending on the retention interval. With the 5-minute retention interval, study-study is better than study-test, but with longer retention intervals, study-test is better than study-study. This is a qualitative interaction because the effect of the study condition does not reverse depending on the retention interval. This is a quantitative interaction because the effect of the study condition reverses depending on the retention interval. With the 5-minute retention interval, study-study is better than study-test, but with longer retention intervals, study-test is better than study-study. Flag this QuestionQuestion 382 pts The following table reflects the results of a study by Roediger and Karpicke (2006). The authors investigated whether the test-enhanced learning effect (the demonstration that repeated testing improves memory for material) was due merely to repeated exposure to the material. They randomly assigned participants to one of two study conditions (study-study or study-test) and to one of three retention interval conditions (final test at a delay of 5 minutes, 2 days, or 1 week). The dependent variable was the proportion of idea units recalled from an encyclopedia passage. Table: Test-Enhanced Learning 5 minutes 2 days 1 week mean Study-study 0.80 0.55 0.42 0.59 Study-test 0.75 0.70 0.55 0.67 Mean 0.78 0.63 0.49 Reference: Table 1 (Table: Test-Enhanced Learning) Based on the cells of this study, which effects appear to be present? an interaction between study condition and retention interval a main effect of study condition and an interaction between study condition and retention interval a main effect of study condition and a main effect of retention interval a main effect of study condition, a main effect of retention interval, and an interaction between study condition and retention interval Flag this QuestionQuestion 392 pts In a factorial ANOVA, each cell in the design represents a(n): quantitative or qualitative interaction. unique combination of level of the independent variables. main effect of one of the independent variables. specific participant. Flag this QuestionQuestion 402 pts The following source table depicts partial results of a fictional study investigating whether students’ stress levels vary as a function of the type of residence they inhabit (house, apartment, dorm room) and the noise volume to which they are subjected (soft, medium, loud). Six participants were recruited for each cell of the study. Table: Residence and Noise Source SS df MS F Residence 7.64 2 Noise 2 1.11 Residence × Noise Within 14.36 0.319 Total 27.84 Reference: Table 4 (Table: Residence and Noise) Using the source table and information provided, calculate the sum of squares for the community variable. 2.22 5.31 4.81 3.62 Flag this QuestionQuestion 412 pts The table that follows reflects the results of a study by Forys and Dahlquist (2007) investigating the effects of coping style and cognitive strategy on dealing with pain. Participants were first classified as having a monitoring or avoiding coping style. They were then randomly assigned to one of two cognitive strategy conditions, distraction or sensation monitoring. Participants were then instructed to use the cognitive strategy while submerging their hand in ice water. The researchers measured pain tolerance as the number of seconds participants were able to keep their hand in the ice water. Table: Coping with Pain Distraction Sensation Monitoring Mean Monitoring 84.5 93.3 88.9 Avoiding 136.8 85.6 111.2 Mean 110.7 89.5 Reference: Table 2 (Table: Coping with Pain) The cells of this study reflect a main effect of cognitive strategy. Which statement best describes the main effect? The effect of cognitive strategy depended on the coping style of the participant. Those with an avoiding coping style kept their hand in the ice water longer when using a distraction strategy, but those with a monitoring coping style kept their hand in the ice water longer when using a sensation-monitoring strategy. People using both distraction- and sensation monitoring cognitive strategies were able to keep their hand in the ice water for longer than 60 seconds, on average. People using a sensation-monitoring strategy were able to keep their hand in the ice water for longer. People using a distraction strategy were able to keep their hand in the ice water for longer. Flag this QuestionQuestion 422 pts The table that follows reflects the results of a study by Forys and Dahlquist (2007) investigating the effects of coping style and cognitive strategy on dealing with pain. Participants were first classified as having a monitoring or avoiding coping style. They were then randomly assigned to one of two cognitive strategy conditions, distraction or sensation monitoring. Participants were then instructed to use the cognitive strategy while submerging their hand in ice water. The researchers measured pain tolerance as the number of seconds participants were able to keep their hand in the ice water. Table: Coping with Pain Distraction Sensation Monitoring Mean Monitoring 84.5 93.3 88.9 Avoiding 136.8 85.6 111.2 Mean 110.7 89.5 Reference: Table 2 (Table: Coping with Pain) The cells of this study reflect an interaction. Is it a quantitative or qualitative interaction? Why? This is a qualitative interaction because the effect of coping style does not reverse depending on the cognitive strategy employed. This is a quantitative interaction because the effect of coping style does not reverse depending on the cognitive strategy employed. This is a qualitative interaction because effect of cognitive strategy reverses depending on the coping style. Those with a monitoring coping style withstood pain longer when using the sensation-monitoring strategy, but those with an avoiding coping style withstood pain longer when using the distraction strategy. This is a quantitative interaction because the effect of cognitive strategy reverses depending on the coping style. Those with a monitoring coping style withstood pain longer when using the sensation-monitoring strategy, but those with an avoiding coping style withstood pain longer when using the distraction strategy. Flag this QuestionQuestion 432 pts A researcher performs a 3 x 5 ANOVA examining how preferences for engagement with friends varied as a function of age (20, 30, or 40 year olds) and social media tools (Facebook, Twitter, LinkedIn, Google+, Instagram) with 26 participants in each cell of the study design. What is the degrees of freedom for the interaction? 6 5 8 7 Flag this QuestionQuestion 442 pts The following figure reflects the results of a study by Forys and Dahlquist (2007) investigating the effects of coping style and cognitive strategy on dealing with pain. Participants were first classified as having a monitoring or avoiding coping style. They were then randomly assigned to one of two cognitive strategy conditions, distraction or sensation monitoring. Participants were next instructed to use the cognitive strategy while submerging their hand in ice water. The researchers measured pain tolerance as the number of seconds participants were able to keep their hand in the ice water. Figure: Strategies for Dealing with Pain Reference: Figure 2 (Figure: Strategies for Dealing with Pain) The figure reflects a main effect of cognitive strategy. Which statement best describes the main effect? People using a sensation-monitoring strategy were able to keep their hand in the ice water for longer. People using a distraction strategy were able to keep their hand in the ice water for longer. People using both distraction and sensation-monitoring cognitive strategies were able to keep their hand in the ice water for longer than 60 seconds, on average. The effect of cognitive strategy depended on the coping style of the participant. People with an avoiding coping style kept their hand in the ice water longer when using a distraction strategy, but those with a monitoring coping style kept their hand in the ice water longer when using a sensation-monitoring strategy. Flag this QuestionQuestion 452 pts Forys and Dahlquist (2007) investigated the effects of coping style and cognitive strategy on dealing with pain. Participants were first classified as having a monitoring or avoiding coping style. Participants were then randomly assigned to one of two cognitive strategy conditions, distraction or sensation monitoring. Participants were next instructed to use the cognitive strategy while submerging their hand in ice water. The researchers measured pain tolerance as the number of seconds participants were able to keep their hand in the ice water. How would you label the ANOVA used to analyze this data? 2 × 2 within-groups ANOVA 2 × 2 between-groups ANOVA 4 × 2 within-groups ANOVA 4 × 2 between-groups ANOVA Flag this QuestionQuestion 462 pts The degrees of freedom for each main effect are calculated as: number of groups minus one. number of IVs minus one times number of cells. number of participants minus one. number of cells minus one. Flag this QuestionQuestion 472 pts The following source table depicts partial results of a fictional study investigating whether students’ stress levels vary as a function of the type of residence they inhabit (house, apartment, dorm room) and the noise volume to which they are subjected (soft, medium, loud). Six participants were recruited for each cell of the study. Table: Residence and Noise Source SS df MS F Residence 7.64 2 Noise 2 1.11 Residence × Noise Within 14.36 0.319 Total 27.84 Reference: Table 4 (Table: Residence and Noise) Using the source table and information provided, calculate the mean square interaction. 2.44 0.08 0.91 3.82 Flag this QuestionQuestion 482 pts The following source table depicts partial results of a fictional study investigating whether students’ stress levels vary as a function of the type of residence they inhabit (house, apartment, dorm room) and the noise volume to which they are subjected (soft, medium, loud). Six participants were recruited for each cell of the study. Table: Residence and Noise Source SS df MS F Residence 7.64 2 Noise 2 1.11 Residence × Noise Within 14.36 0.319 Total 27.84 Reference: Table 4 (Table: Residence and Noise) Using the source table and information provided, calculate the effect size for the interaction. 0.13 0.04 0.20 0.35 Flag this QuestionQuestion 492 pts An effect size of 0.058 for a two-way ANOVA is considered to be: small. medium. large. an error. Flag this QuestionQuestion 502 pts If an analysis of variance includes both within-groups factors and between-groups factors, we call it a: mixed-design ANOVA. complex-design ANOVA. two-way ANOVA. between-within ANOVA.