For a study regarding mean cadence, two-way ANOVA was used. The two factors were walking device (none, standard walker, rolling walker) and dual task (being required to respond vocally to a signal or no dual task required). Results of two-way ANOVA showed that there was no evidence of interaction between the factors. However, according to the article, "the ANOVA conducted on the cadence data revealed a main effect of walking device." When the hypothesis regarding no difference in mean cadence according to which, if any, walking device was used, the sample F was 30.94, with d. f.N = 2 and d. f.D = 18. Further, the P-value for the result was reported to be less than 0.05. Required:

What is the conclusion regarding any difference in mean cadence according to the factor "walking device used"?

Step-by-step explanation:

With regards to the factor 'walking device used', the ANOVA conducted on the cadence data revealed a main effect of walking device, and also with the results of the experiment giving rise to a p - value less than 0.05, we can reject the null hypothesis which says, there is no effect of the walking device factor.

We can thus conclude that there is not enough statistics evidence to prove that there is no interaction between the two factors or that there is no effect of the walking device given the cadence data.

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below: