Using ANOVA to Measure Stress in Humans
Analysis of variance (or simply ANOVA) is a statistical method used to test whether experiment or survey results are significant, or rather, whether to accept or reject the null hypothesis. An ANOVA test can be one-way or two-way. One-way ANOVA has only one independent variable while two-way ANOVA has two independent variables. Which one you choose to use to manipulate your data will depend on the type of analysis you are performing.
One-way ANOVA was performed to measure difference is stress measures among 4 groups of residents (Three Mile Island, Frederick, Dickerson, Oyster Creek).
The degrees of freedom for the ANOVA test were 3 and 117. The critical F-value for the 0.05 significance level is 2.682, and 3.954 for the 0.01 significance level.
A one-way ANOVA was run to test the hypothesis that the average total reported stress symptoms were equal for the four groups of residents. The result indicates that the stress symptom scores were not equal for all the four groups of residents F (3,117) = 3.827, which is greater than the critical F-value at a 5% significance level is 2.682. Thus, the stress levels are different for the four groups of residents.
The average stress levels of residents in the TMI group was 25.97, which is much higher than the stress levels of no nuclear group (M = 15.54), undamaged coal plant (M = 16.63), and undamaged nuclear plant (M = 16.16).
A one-way ANOVA was run to test the hypothesis that the average depression levels (measured on the Beck Depression Inventory) were equal for the four groups of residents. The result indicates that the depression levels were equal for all the four groups of residents F (3,117) = 2.104, which is lesser than the critical F-value at a 5% significance level is 2.682. Thus, the depression levels are not different for the four groups of residents.
The average depression levels of residents in the TMI group was 6.00, which is much higher than the depression levels of no nuclear group (M = 3.64), undamaged coal plant (M = 3.54), and undamaged nuclear plant (M = 3.50).
The effect size for stress levels is 0.32 and 0.541 for depression levels
If we had used 0.01 significance levels, our conclusion about stress levels would change and the conclusion about depression levels would remain the same.
We could do Tukey’s HSD test as a follow up to check on pairwise significance.
The coefficients are 1,1,0,0
The average depression levels of people living in the TMI area was 6.00, which would indicate minimal depression (Scores in the range of 0-13 are classified as minimal depression). We cannot conclude that the TMI accident made people severely depressed, as severe depression requires the Beck Depression Inventory to be in the range of 29-63.
The degrees of freedom of the F-test was 1 and 108. The critical F-value at the 0.05 significance level is 3.929. The computed value of the F-statistic is 184.29, which is greater than the critical F-value of 3.929. Thus, the difference is statistically significant.
The effect size for the difference in Q2 (a) is 0.63, which indicates a medium effect. It was computed using the following formula:
The researchers attempt to manipulate the perceived attractiveness was successful
The F-ratio has df = 1 in the numerator, as there are 2 groups to be compared. The df of the numerator is # of Groups -1, thus the df.
Group 1: M = 2.80
Group 2: M = 5.20
Group 3: M = 5.10
Group 1 vs 2: F(1,108) = 6.60, p< 0.025
The difference is statistically significant. The overall F would likely to be statistically significant.
The difference in the mean length of the sentence was in the predicted direction.
Effect size estimate:
Eta-squared = 0.0576 < 0.2
The effect size can be termed as small.
Tukey’s HSD test formula is:
We require MSwithin, n, and differences in the mean between the two groups.
(a) Change the decibel differences between low level and high level. Increase the sample sizes for the two groups. Increase the number of syllables.
(b) The researcher could add more groups.
Grand Mean MY = 550. Kim’s score = 610; Group average score = 565
Y – MY = 610- 550 = 60
Within group deviation: Y – Mi = 610 – 565 = 45
Between group difference: Mi– MY = 565 – 550 = 15
Effect component (alpha) = Mi– MY = 565 – 550 = 15
To further understand this topic, avail our hypothesis testing assignment help.
If all the numbers within a group were equal, the SSGroupwould become zero. When the means of the different groups are equal, but scores are not the same within the group, then SSBetween would become zero, and SSGroupwouldn’t be equal to zero.
The researcher would hope that MSBetweenwould be larger, as treatment would significantly change the measured metric of the group compared to the measured metric of the control group, increasing the MSBetween
By splitting the (Yij– MY) into the (Y – Mi) & (Mi– MY) terms, we are able to assign the variances in the score differences (Yij– MY) into the effect of group membership (Mi– MY) and difference of individual’s score from group score (Y – Mi). It can’t be taken as causality evidence but helps to seek out group differences.
The H0 for a one-way ANOVA is the mean of the interval variable is equal for all the groups in the ANOVA. If the H0 is rejected, it means that at least one group’s average metric is different than the average metric of the other groups.
We need a number of groups and statistical significance level, to choose the sample size to obtain adequate statistical power.
The α represents the effective component of the score. It shows how much the membership in the group leads to a difference from the grand mean.
We can’t conclude that
Eta squared can be computed as the ratio of between and SSTotal. It can be also calculated using the following formula:
Post hoc tests are more conservative, as it usually tests for more pairs for significant differences than planned contrasts, where the user chooses the pairs for testing, before testing itself.
The common post hoc procedures include Tukey’s HSD test, Bonferroni test. If you would like expert assistance with this topic, take our parametric tests assignment help.