Statistical Analysis: ANOVA on Display Panel and Emergency Conditions

This assignment presents an ANOVA analysis of display panel and emergency condition effects. The student begins by outlining the null and alternative hypotheses for (a) display panel effects, (b) emergency condition effects, and (c) the interaction between the two. The student then justifies the use of a two-way ANOVA due to the presence of two independent variables. The assignment proceeds to interpret p-values from the results, leading to the rejection of null hypotheses for both display panels and emergency conditions, while failing to reject the null hypothesis for the interaction effect. The conclusion highlights that the mean time taken is not equal for all panels and conditions but acknowledges the need for a post-hoc test, such as Tukey’s HSD, to determine which means differ from each other. The assignment demonstrates understanding of statistical concepts and their application in analyzing experimental data.

ANOVA results
Task 1: Set up null and alternative hypotheses to (a) test the
significance of display panel effect, (b) test the significance of
emergency condition effects, and (c) test the interaction a
between the two factors.
a)
H01: The mean time required is equal across display panels.
HA1: The mean time required is not equal across display panels.
b)
H02: The mean time required is equal across emergency conditions.
HA2: The mean time required is not equal across emergency conditions.
c)
H03: There is no interaction effect between display panels and
emergency conditions.
HA3: There is an interaction effect between display panels and
emergency conditions.
Task 2: Justify the type of ANOVA (one-way or two-way) you will
apply to test the hypotheses.
I will apply two-way ANOVA because the scenario has two independent
variables or two factors namely, display panels and emergency
conditions. We need to whether the time taken by a display is influenced
by emergency conditions. For such scenarios, two-ANOVA is suitable.
Task 3: Interpret the p values in the results table for an
accept/reject decision regarding the hypotheses.
Since p-value for ‘panel’ is less than 0.050 (as a rule of thumb, a null
hypothesis gets rejected if the p-values is less than 0.05), I will reject
the null hypothesis and accept the alternative hypothesis. In other

words, the mean time required is not equal across display panels. The p-
value for ‘condition’ is also less than 0.050, hence, I will reject the null
hypothesis and accept the alternative hypothesis, i.e. the mean time
required is not equal across emergency conditions. As for, the
interaction effect, the p-value (0.681) is bigger than 0.050. This means
that the null hypothesis cannot be rejected. In other words, there is no
interaction effect between display panels and emergency conditions.
Task 4: State your conclusions including what the ANOVA results
do not tell you and your needed actions as a researcher.
Conclusion: The mean time taken is not equal for all ‘panels’ and
‘conditions’. However, the ANOVA results that we had do not tell us
which mean differs from which mean. Therefore, as a researcher, I need
to use a post-hoc test such as Tukey’s HSD which provides a multiple
comparison table to help us decide which mean differs from which
mean.