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:

	Water	Coffee	Espresso
Caffeine Content (mg/oz)	0	13.4	51.3
	78	92	83	ΣX² = 142,493
	88	87	83	G = 1,723
	72	80	76	N = 21
	72	89	79	k = 3
	64	96	86
	81	83	79
	82	92	81
	T₁ = 537	T₂ = 619	T₃ = 567
	SS₁ = 381.43	SS₂ = 185.71	SS₃ = 66
	n₁ = 7	n₂ = 7	n₃ = 7
	M₁ = 76.7143	M₂ = 88.4286	M₃ = 81.0000

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 equal.

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

The population mean test score for the water population is different from the population mean test score for the coffee population.

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

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

Source	SS	df	MS
Between
Within	633.14
Total	1,124.95

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 = 0.01.

F Distribution

Numerator Degrees of Freedom = 26

Denominator Degrees of Freedom = 26

012345678F

At the α = 0.01 level of significance, what is your conclusion?

You cannot 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.

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

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

Answer 1

Null hypothesis:

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

--

Number of treatment, k = 3

Total sample Size, N = 21

df(between) = k-1 = 2

df(within) = N-k = 18

df(total) = N-1 = 20

SS(between) = SS(total) - SS(within) = 491.81

MS(between) = SS(between)/df(between) = 245.905

MS(within) = SS(within)/df(within) = 35.1744

F = MS(between)/MS(within)

= 245.905/ 35.1744

= 6.991

ANOVA
Source of Variation	SS	df	MS	F
Between Groups	491.81	2	245.905	6.9910
Within Groups	633.14	18	35.1744
Total	1124.95	20

Critical value Fc = F.INV.RT(0.01, 2, 18) = 6.013

Conclusion:

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