Question

Question 1

For an experiment comparing more than two treatment conditions you should use analysis of variance rather than separate t tests because ________.

Group of answer choices

a test based on variances is more sensitive than a test based on means

separate t tests would require substantially more computations

conducting several t tests would inflate the risk of a Type I error

There is no difference between the two tests, you can use either one.

Question 2

A restaurant manager is considering changing the secret sauce on their hamburgers. To decide which recipe to use, she recruits 15 customers and assigns them to 1 of 3 sauce groups (A, B, or C). She then serves them a plain hamburger with the sauce for their group and asks them to rate their satisfaction with the taste of the sauce on a scale from 1 (disgusting) to 5 (delicious) to see if there is a significant (α = .05) difference in taste. What is the alternative hypothesis for this Analysis of Variance?

Group of answer choices

μA ≠ μB ≠ μC

there is a significant difference in taste between at least one pair of groups

μA = μB = μC

there is no significant difference in taste between at least one pair of groups

Question 3

A restaurant manager is considering changing the secret sauce on their hamburgers. To decide which recipe to use, she recruits 15 customers and assigns 5 different customers to each of the 3 sauce groups (A, B, or C). She then serves them a plain hamburger with the sauce for their group and asks them to rate their satisfaction with the taste of the sauce on a scale from 1 (disgusting) to 5 (delicious) to see if there is a significant (α = .05) difference in taste. What is the critical value for this Analysis of Variance?

Group of answer choices

F crit = 99.42

F crit = 19.41

F crit = 6.93

F crit = 3.88

Question 4

A restaurant manager is considering changing the secret sauce on their hamburgers. To decide which recipe to use, she recruits 15 customers and assigns 5 different customers to each of the 3 sauce groups (A, B, or C). She then serves them a plain hamburger with the sauce for their group and asks them to rate their satisfaction with the taste of the sauce on a scale from 1 (disgusting) to 5 (delicious) to see if there is a significant (α = .05) difference in taste. Based on the data and results below, what is the decision and conclusion for this Analysis of Variance? HINT: SS for Group A = 2.8, SS for Group B = 5.2, SS for Group C = 4.
A B C
1 3 3
2 4 5
1 2 4
3 1 3
2 2 5

Practice by placing your calculations in a table like this:

Source	SS	df	MS	F
Between
Within
Total

Group of answer choices

6.467 Reject the null hypothesis; results suggest there is a significant difference in taste among the three sauces

6.467 Reject the null hypothesis; results suggest there is not a significant difference in taste among the three sauces.

6.467 Fail to reject the null hypothesis; results suggest there is not a significant difference in taste among the three sauces.

1.078 Reject the null hypothesis; results suggest there is a significant difference in taste among the three sauces.

1.078 Reject the null hypothesis; results suggest there is not a significant difference in taste among the three sauces.

1.078 Fail to reject the null hypothesis; results suggest there is not a significant difference in taste among the three sauces.

Answer 1

Question 1

For an experiment comparing more than two treatment conditions you should use analysis of variance rather than separate t tests because ________.

Answer :

separate t tests would require substantially more computations

----------------

Question 2

Alternative hypothesis for this Analysis of Variance:

Answer:

there is a significant difference in taste between at least one pair of groups

----------------

Question 3

Critical value for this Analysis of Variance:

df1 = 3-1=2

df2 = 15-3 = 12

F crit = F.INV.RT(0.05, 2, 12) = 3.88

Answer:

F crit = 3.88

-----------------------

Question 4

	Group 1	Group 2	Group 3	Total
Sum	9	12	20	41
Count	5	5	5	15
Mean, Sum/n	1.8	2.4	4
Sum of square, SS	2.8	5.2	4

Number of treatment, k =    3

Total sample Size, N =    15

df(between) = k-1 =    2

df(within) = N-k =    12

df(total) = N-1 =    14

SS(between) = (Sum1)²/n1 + (Sum2)²/n2 + (Sum3)²/n3 - (Grand Sum)²/ N =    12.9333

SS(within) = SS1 + SS2 + SS3 =    12

SS(total) = SS(between) + SS(within) =    24.9333

MS(between) = SS(between)/df(between) =    6.4667

MS(within) = SS(within)/df(within) =    1

F = MS(between)/MS(within) =    6.4667

ANOVA
Source of Variation	SS	df	MS	F
Between Groups	12.9333	2	6.4667	6.4667
Within Groups	12.0000	12	1.0000
Total	24.9333	14

Answer:

6.467 Reject the null hypothesis; results suggest there is a significant difference in taste among the three sauces