Question

8. Hypothesis testing with ANOVA

Opinions about whether caffeine enhances test performance differ. You design a study to test the impact of drinks with different caffeine contents on students’ test-taking abilities. You choose 21 students at random from your introductory psychology course to participate in your study. You randomly assign each student to one of three drinks, each with a different caffeine concentration, such that there are seven students assigned to each drink. You then give each of them a plain capsule containing the precise quantity of caffeine that would be consumed in their designated drink and have them take an arithmetic test 15 minutes later.

The students receive the following arithmetic test scores:

	Cola	Black Tea	Coffee
Caffeine Content (mg/oz)	2.9	5.9	13.4
	85	85	92	ΣX² = 147,641
	86	89	87	G = 1,755
	82	82	80	N = 21
	75	75	89	k = 3
	66	88	96
	78	76	83
	87	82	92
	T₁ = 559	T₂ = 577	T₃ = 619
	SS₁ = 338.86	SS₂ = 177.71	SS₃ = 185.71
	n₁ = 7	n₂ = 7	n₃ = 7
	M₁ = 79.8571	M₂ = 82.4286	M₃ = 88.4286

You plan to use an ANOVA to test the impact of drinks with different caffeine contents on students’ test-taking abilities. What is the null hypothesis?

The population mean test scores for all three treatments are different.

The population mean test scores for all three treatments are equal.

The population mean test scores for all three treatments are not all equal.

The population mean test score for the cola population is different from the population mean test score for the black tea population.

Calculate the degrees of freedom and the variances for the following ANOVA table:

Source	SS	df	MS
Between
Within	702.28
Total	973.14

The formula for the F-ratio is:

FF

=  =

MSbetweenMSbetween

/  /

MSwithinMSwithin

Using words (chosen from the dropdown menu), the formula for the F-ratio can be written as:

FF

=  =

/  /

Using the data from the ANOVA table given, the F-ratio can be written as:

FF

=  =

/  /

Thus:

FF

=  =

Use the Distributions tool to find the critical region for α = 0.05.

F Distribution

Numerator Degrees of Freedom = 26

Denominator Degrees of Freedom = 26

012345678F

At the α = 0.05 level of significance, what is your conclusion?

You can reject the null hypothesis; caffeine does appear to affect test performance.

You cannot reject the null hypothesis; you do not have enough evidence to say that caffeine affects test performance.

You can reject the null hypothesis; you do not have enough evidence to say that caffeine affects test performance.

You cannot reject the null hypothesis; caffeine does appear to affect test performance.

Answer 1

Null hypothesis: Ho:The population mean test scores for all three treatments are equal.

Source of Variation	SS	df	MS
Between Groups	270.86	2	135.43
Within Groups	702.29	18	39.02
Total	973.14	20

F =135.43/39.02 =3.47

Numerator Degrees of Freedom =(number of groups) k-1 =3-1 =2

Denominator Degrees of Freedom = N-k=21-3 =18

critical region for α = 0.05. is F >3.55

since test statistic <critical value

You cannot reject the null hypothesis; you do not have enough evidence to say that caffeine affects test performance.