Prior to eating breakfast, 25 participants were randomly assigned to eat a large

Question

Prior to eating breakfast, 25 participants were randomly assigned to eat a large meal, 25 to eat a small meal, and 25 to eat nothing. Immediately following the meal, all participants took a memory test. The means and estimated population variance for the three groups on the memory test were: Large: M = 2.1, S^2 = 1.0; Small: M = 2.5, S^2 = 1.2; Nothing: M= 2.9, S^2 = 0.8. Using the .05 significance level, does amount of breakfast eaten affect memory? Use the five steps of hypothesis testing. Figure the effect size of this study and indicate whether it is nearer to small, medium, or large. Conduct a planned contrast for Large Meal versus Small Meal (using the .05 level). Explain your results.

Explanation / Answer

Null Hypothesis : All means are equal means meal doesn’t affect memory

Alternate Hypothesis : Meal affects memory

We use ANOVA for our hypothesis testing

f1 = Sum of squares between groups

df1 = degrees of freedom = number of groups - 1 = 2

f2 = Sum of Squares within groups

df2 = degrees of freedom = number of participants - number of groups = 72

F - ratio = (f1/df1)/(f2/df2)

Now

Calculate the grand mean = (2.1 + 2.5 + 2.9) / 3 = 2.5

f1 = ([2.5-2.1]^2   +   [2.5-2.5 ]^2   + [2.9-2.5]^2 ) * 25

f2 = Sum of (variances*sample size) = 1*25 + 1.2*25 + 0.8*25 = 3

Now F can be calculated very easily

With 0.05 significance level and 2 and 72 degrees of freedom (Use ANOVA table ) if calculated value of F is less than the cut off value then we cannot reject null hypothesis but if it lies in the rejection region then we can say that meal definitely affects memory.

If we can reject the null hypothesis from above calculation then we can say that meal have negative affect over memory test

For a planned contrast between small and large meal

Lets first do it by simple t test

Null Hypothesis : Difference of their means is zero

Alternate : Non zero mean difference

t - statistic = [(2.5-2.1) - 0]/ d

d = Standard error of mean difference = Sq.root[(S1^2/n1) + (S2^2/n2)]

S = Std dev and n is the sample size

We can also perform F test

F = S1^2 / S2^2 , S1 and S2 are the standard deviation

Now

Take the largest variance, and divide it by the smallest variance to get the f-value because placing the largest variance on top will force the F-test into a right tailed test, which is much easier to calculate than a left-tailed test.

Find your degrees of freedom. Degrees of freedom is your sample size minus 1. As you have two samples (variance 1 and variance 2), you’ll have two degrees of freedom: one for the numerator and one for the denominator.

Now use f table to check whether F value we calculated above lies in the rejection region or not with significance level of 0.05

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