Activity 8-1

Fill in the blanks for the F statistic reporting statement using the following format: F (dfbetween, dferror/residual) = computed F value, p = computed p value, MSE = computed Mean square error/residual value. F ([a],[b] ) = [c], p = [d], MSE = [e]

F (2,177 ) = 5.85, p = .003, MSE = 5118.62

Fill in the blanks for the APA style summary statement: An independent ANOVA was conducted comparing average daily minutes of SM use in the three sleep groups (IV levels: 6,7 and 8 hours), F ([a],[b]) = [c], p = [d], MSE = 5118.62] ,ω2= [e]. The means and standard deviations for each condition are in Table 1. Table 1. Descriptive statistics for each sleep group Sleep Group-Mean (SD) 6 hours-287.29 (72.58) 7 hours-257.36 (74.47) 8 hours-[f] The mean differences, Fisher LSD p values, d's and CIs are in Table 2. Fisher's LSD post hoc tests revealed that the group that slept for 6 hours used SM 43.68 minutes more than the group that slept for 8 hours. The group that slept for 6 hours also used SM 29.92 minutes more than the group that slept for 7 hours. Both of these differences were greater than expected by sampling error. The difference between the 7 and [g] hour sleep group ( 13.76 minutes) had a high p value and so we suspend judgment on this comparison. These results are consistent with the previous literature showing that SM use is associated with less sleep. This study was a true experiment so, it allows us to determine that sleep deprivation caused participants to use SM more. One possible methodological concern with this study is that we don't know if people really slept for the amount of time they were supposed to. However, these results are promising and certainly suggest that future research is needed in this area. Future studies should also use larger sample sizes to obtain more precise estimates of parameters because there is a great deal of variability in social media use and our estimates were imprecise. Table 2. Fisher LSD post hoc comparisons of sleep, ds , and CIs Comparison d p value Mean Difference 95% CI for d Lower Upper 6 vs. 7 hours .41 .02 29.92 .06.786 vs. 8 hours.62.00143.68.24.987 vs. 8 hours[h] [i] [j] [k] [l]

Fill in the blanks for the APA style summary statement: An independent ANOVA was conducted comparing average daily minutes of SM use in the three sleep groups (IV levels: 6,7 and 8 hours), F (2,177) = 5.85, p = .003, MSE = 5118.62] ,ω2= .05. The means and standard deviations for each condition are in Table 1. 8 hours 243.60 (67.40) The mean differences, Fisher LSD p values, d's and CIs are in Table 2. Fisher's LSD post hoc tests revealed that the group that slept for 6 hours used SM 43.68 minutes more than the group that slept for 8 hours. The group that slept for 6 hours also used SM 29.92 minutes more than the group that slept for 7 hours. Both of these differences were greater than expected by sampling error. The difference between the 7 and 8 hour sleep group ( 13.76 minutes) had a high p value and so we suspend judgment on this comparison. These results are consistent with the previous literature showing that SM use is associated with less sleep. This study was a true experiment so, it allows us to determine that sleep deprivation caused participants to use SM more. One possible methodological concern with this study is that we don't know if people really slept for the amount of time they were supposed to. However, these results are promising and certainly suggest that future research is needed in this area. Future studies should also use larger sample sizes to obtain more precise estimates of parameters because there is a great deal of variability in social media use and our estimates were imprecise. 7 vs. 8 hours .19 .29 13.76 -.17 .55

What does this p value tell you?

The probability of obtaining F > 5.85 if the null hypothesis is true and there are no methodological flaws in the study.

What is the p value for the omnibus F value?

.003

The omnibus F's p value suggests that the data are not consistent with the null hypothesis. An effect size will reveal the magnitude of the effect that different amounts of sleep had SM use. Depending on the software you are using, you may report eta squared, partial eta squared , or omega squared. In general, omega squared is a less biased estimate of effect size and so it is best to report that if it is available. Find one of these effect sizes in the output and report it.

.05

What F value would you expect if the null hypothesis were true and all three populations had equal means?

1

Your software provides a measure of the between group variability that is labeled Mean Square Sleep Group or Between or Treatment. What is this value for this study?

29928.37

The omnibus F is the ratio of between treatment to within treatment variability. What is the F value for this ANOVA?

5.85

Use statistical software to compute the ANOVA. Your software provides a measure of the within group variability labeled Mean Square (MS) Residual or Error or Within. What is this value for this study?

5118.62

Which of the three groups used social media the most?

6 hours

Use the statistical software to find the means and standard deviations for the three groups and record them in the table below. 6 hours: [a] 7 hours [b] 8 hours [c]

6 hours: M = 287.29, SD = 72.58 7 hours M = 257.36, SD = 74.47 8 hours M = 243.60, SD = 67.40

For each pairwise comparison, determine if the data are compatible with its null hypothesis. For each comparison, choose one of the following options. 6 vs 7 ; [a] 6 vs. 8 ; The data are not compatible with the null and lower sleep resulted in more social media use; 7 vs. 8 ; [b].

6 vs 7 ; The data are not compatible with the null and lower sleep resulted in more social media use 7 vs. 8 ; The data are compatible with the null and so we reserve judgment about this difference.

Find the standardized effect sizes (d) for each of the pairwise comparisons. These are easier for comparing your results to those of other studies. Identify them below. 6 vs 7 d = [a] 6 vs. 8 d = [b] 7 vs. 8 d = [c]

6 vs 7 d = .41 6 vs. 8 d = .61 7 vs. 8 d = .19

Which of the following is NOT a statistical assumption for ANOVAs:

Bimodality

Next, you need to consider the methodological rigor of the study as well as how well the study fits into the scientific literature. We did not give you a lot of previous research to work with, but this study is consistent with research showing that sleep deprivation is associated with social media use. Because this was an experiment, it allows you to determine that sleep deprivation caused social media use to increase. However, there are some possible methodological concerns with the study. Which of the following is the most concerning methodological issue with this study that would give you less confidence in the results?

Both of the above are methodological flaws.

Next, locate the confidence intervals around the mean difference or effect size (depending on your software) provided in the statistical output. Record these values . 6 vs 7 mean difference 95% CI = ([a]) 6 vs. mean difference 95% CI = ([b]) 7 vs. mean difference 95% CI = ([c])

6 vs 7 mean difference 95% CI = (mean difference: -17.12, 44.63; d: .05, .78) 6 vs. mean difference 95% CI = (mean differenece: -.95, 60.80; d: .25, .98) 7 vs. mean difference 95% CI = (mean difference: 12.80, 74.56; d: -.17, .55)

Partial eta squared and omega squared tell you the overall effect sleep had on SM use. However, you are more interested in the effect sizes for the pairwise comparisons. The mean differences for each pairwise comparison are simple measures of effect size. Identify them below. 6 vs 7 mean difference = [a] 6 vs. 8 mean difference = [b] 7 vs. 8 mean difference = [c]

6 vs 7 mean difference = 29.92 6 vs. 8 mean difference = 43.68 7 vs. 8 mean difference = 13.76

The omnibus F's p value suggests that the data are incompatible with the null hypothesis (i.e., the three groups use SM equal amounts). The pairwise comparisons among the three conditions are listed in the table below. Begin by entering the differences between the means. You can compute these values by hand, but your software provides them in the post hoc output. 6 vs 7 mean difference = [a] 6 vs. 8 mean difference = [b] 7 vs. 8 mean difference = [c]

6 vs 7 mean difference = 29.92 6 vs. 8 mean difference = 43.68 7 vs. 8 mean difference = 13.76

Next, you need to find the p value for each pairwise comparison and place it in the above table. There are many different types of post hoc tests. Here we are going to use Fisher's LSD. 6 vs 7 p = [a] 6 vs. 8 p = [b] 7 vs. 8 p = [c]

6 vs 7 p = .02 6 vs. 8 p = .001 7 vs. 8 p = .29

The null hypothesis for the ANOVA is that:

All three means are equal to each other

p values close to zero suggest that the data are:

Inconsistent with the null hypothesis

Generate a graph of the data from each group. You can create either boxplots or point and whisker plots. Does it look like all three means are approximately equal (i.e., the data are consistent with the null hypothesis)?

No

Are these data compatible with the null hypothesis that all three means are equal?

The data are not compatible with the null hypothesis

In the chapter, you read about a study where the researchers manipulated social media (SM) use and measured sleep and found that increased SM use led to reduced sleep. This finding may explain why people who use SM a lot tend to get less sleep. However, it is possible that the reverse is also true. Perhaps reduced sleep leads to increased SM use. To test this, you design a study in which you randomly assign college students to sleep for 6, 7, or 8 hours a night. You also ask students to install an app on their phone and computer that monitors the time they spend on SM during the week. The DV is the average time spent on SM. The data are on the textbook website. Why is an ANOVA an appropriate test for this study?

There are three groups and the dependent variable is measured on an interval/ratio scale.

Should you do post hoc tests? Explain your reasoning.

Yes

What could you do to increase the precision of the estimates (i.e., make the 95% CI narrower)? Select two.

increase the sample size decrease measurement error

If the amount of sleep influences SM use, the above value will be _________ (larger/smaller) than the within group variability.

larger

You should find that the confidence intervals are quite wide. This suggests that our estimates of the population parameters are:

rough estimates

The above value is a measure of expected

sampling error

Is this effect size small, medium or large?

small-medium