Question

question 5
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?
select one answer. 10 points

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 sat score is related to study strategy.
the following hypotheses were tested:
$h_0:mu_1=mu_2=mu_3=mu_4$
$h_a:mu_1,mu_2,mu_3,mu_4$ are not all equal
the analysis was run on the data and the following output was obtained:

source	ss	df	ms	f	p
error	8	16	0.50
total	138	19

which of the following is a valid conclusion based on this output?
a. the data provide strong evidence that the four mean scores (representing the four teaching strategies) are not all equal.
b. the data do not provide sufficient evidence that scores are related to teaching strategy.

Explanation:

Step1: Recall ANOVA decision - rule

In ANOVA, if the p - value is less than a chosen significance level (commonly 0.05), we reject the null hypothesis.

Step2: Check p - value

The p - value from the ANOVA output is less than 0.001. Since 0.001 < 0.05, we reject the null hypothesis $H_0:\mu_1=\mu_2=\mu_3=\mu_4$.

Answer:

A. The data provide strong evidence that the four mean scores (representing the four teaching strategies) are not all equal.