A one-way analysis of variance (ANOVA) showed no significant effect of deprivation condition on concentration, F(2,46) = 1.06, p = .36 (see Figure 1). What does this mean???
The one-way analysis of variance allows us to compare the values of parameters (often means) between groups. The assumptions to perform an ANOVA test are that the samples are taken from normal populations and that the populations have the same variance.
F is calculated by `F="MSB"/"MSW" ` . MSB is the mean-square between groups while MSW is the mean-square within groups. This ratio is then used as a test statistic. (If the null-hypothesis is true then F has the F-distribution with degrees of freedom (n-1,k-1) where n is the total number of pieces of data and k is the number of groups.)
I assume a study was done comparing groups -- some groups were deprived of something (sleep, caffeine, etc...) while the other groups were not deprived. The null hypothesis is that there is no difference between the groups, while the alternative hypothesis is that there is some difference.
After the critical value is determined using the F-distribution with the correct degrees of freedom, the test statistic F is located. Apparently the test statistic F=1.06 was not in the critical region so the null hypothesis was not rejected.
Alternatively, given a confidence level we compare the p-value to `alpha ` . Since p=.36 we cannot reject the null-hypothesis. (We would reject if our confidence level was 64% or less.)
Since the null hypothesis cannot be rejected, then there is no statistical difference between the groups.