Question

... sleep deprivation influences aggression. To test this assumption, volunteer participants were randomly assigned to sleep-deprivation periods of 0, 24, 48, and 72 hours and subsequently tested for aggressive behavior in a controlled social setting. Aggressive scores signify the total number of different aggressive behaviors, such as "put downs", arguments, or verbal interruptions, that were demonstrated by subjects during the test period. Compute the appropriate test to determine if sleep deprivation has an effect on level of aggression.

0	24	48	72
0	1	2	8
4	3	3	9
2	5	7	6
5	2	2	6
6	4	5	10
2	2	3	6
4	4	4	8
3	1	5	9

• What is your computed answer?
• What probability level did you use and why?

Answer 1

We need to check the difference in the agressive scores between different sleep-deprivation periods. We have four groups here of sleep deprivation periods, hence we will use the one-way ANOVA test here.
Let's perform one-way ANOVA in Excel.

Go to Data -> Data Analysis -> One-way ANOVA -> Select the data -> RUN

The output is:

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
0	8	26	3.25	3.642857
24	8	22	2.75	2.214286
48	8	31	3.875	2.982143
72	8	62	7.75	2.5
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	124.3438	3	41.44792	14.621	6.42E-06	2.946685
Within Groups	79.375	28	2.834821
Total	203.7188	31

The f-statistic = 14.621

The p-value <0.001, hence the p-value is significant. We will reject the Ho.

We can say that we have enough evidence to conclude that there is a significant difference in atleast two of the groups.

We used a 99% confidence level (alpha = 0.01) because we wanted to reduce the chances of a type-1 error.