5. Hypothesis testing with ANOVA opinions about whether caffeine enhances test p
ID: 3220767 • Letter: 5
Question
5. 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: Black Tea Coffee Cola Caffeine Content (mg/oz) 5.9 13.4 EX2 147,641 92 85 G 1,755 87 N 21 3 75 89 76 83 78 82 92 T1 559 T2 577 T3 619 SS1 338.86 SS 177.71 SS3 185.71 n3 n1 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? Q The population mean test score for the cola population is different from the population mean test score for the black tea population Q The population mean test scores for all three treatments are equal. Q The population mean test scores for all three treatments are different. Q The population mean test scores for all three treatments are not all equal. calculate the degrees of freedom and the variances for the following ANovA table: MS 702.28 Total 973.14Explanation / Answer
Answer to part a)
The null hypothesis is : that all the three groups have same mean
The correct answer choice is "Second"
.
Answer to part b)
SSbetween = SStotal - SSwithin
SSbetween = 973.14 - 702.28 = 270.86
.
df between = number of groups - 1
number of groups = 3
so df between = 3 - 1 = 2
df total = N -1
N = 7+7+7 = 21
df total = 21 - 1 = 20
.
df within = df total - df between
df within = 20 - 2 = 18
.
MS between = SS between / df between
MS between = 270.86 / 2 = 135.43
MS within = SS within / df within
MS within = 702.28 / 18 = 39.02
.
Answer to part c)
F = MS between / MS within
F = 135.43 / 39.02
F = 3.4708
.
Answer to part d)
F = variance due to both chance and caffeine / variance due to chance
.
Answer to part e)
F = 135.43 / 39.02
.
Answer to part f)
F = 3.47
.
Answer to part g)
The P value for F = 3.47 , df between = 2 and df within = 18 , is 0.053
Since the P value 0.053 > alpha 0.05 , we fail to reject the null hypothesis, we do not have sufficient evidence to say that caffeine affects the test performance
Thus the correct answer choice is "third statement"
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.