Question

question 4
using four methods to teach anova, do these four samples differ enough from each other to reject the null hypothesis that type of instruction has no effect on mean test performance?

method to teach anova	mean	sd	n
method 2 (co - teachers)	4.61	0.715	31
method 3 (computer)	4.61	0.688	36
method 4 (lab)	4.38	0.793	32

since we are comparing more than 2 groups, we will use anova to test whether the data provide evidence that test score is related to teaching method.
one of the conditions that allows us to use anova safely is that of equal (population) standard deviations. can we assume that this condition is met in this case?
a. o yes, since 0.793 - 0.360 < 2.
b. o no, since 0.793/0.360 > 2
c. o no, since the four sample standard deviations are not all equal.
d. o no, since the population standard deviations are not given, so we cannot check this condition.

Explanation:

Step1: Recall ANOVA assumption

One of the assumptions for ANOVA is that population standard - deviations are equal. A rule - of - thumb is that if the ratio of the largest to the smallest sample standard deviation is less than 2, we can assume equal population standard deviations.

Step2: Calculate the ratio

The largest sample standard deviation is $0.793$ and the smallest is $0.360$. Calculate the ratio $\frac{0.793}{0.360}\approx2.203> 2$.

Answer:

B. No, since 0.793/0.360 > 2