Question

The following data were obtained from an independent-measures research study comparing three treatment conditions. I II...

The following data were obtained from an independent-measures research study comparing three treatment conditions.

I

II

III

n = 6

n = 4

n = 4                   

M = 2

M = 2.5

M = 5

N = 14

T = 12

T = 10

T = 20

G = 42

SS = 14

SS = 9

SS = 10

ΣX2tot = 182


Use an ANOVA with α = .05 to determine whether there are any significant mean differences among the treatments.

  1. The null hypothesis in words is
  1. There are no significant differences among the three treatment means.
  2. There is at least one significant mean difference among the three treatment means.
  3. All the pairs of means significantly differ from each other
  4. There are no significant differences between the means of groups 1 and 2
  1. The alternative hypothesis in symbols is
  1. H1: μ1 ≠ μ2≠ μ3
  2. H1: M1 ≠ M2≠ M3
  3. H1: μ1 = μ2 = μ3
  4. H1: M1 = M2 = M3
  5. In Anova, we do not state the alternative hypothesis in symbols
  1. The Critical F-value is:
  1. The F-statistic is:
  2. Your decision is
  1. Reject the null hypothesis and conclude that there are not significant differences among the conditions
  2. Reject the null hypothesis and conclude that there are significant differences among the conditions
  3. Fail to reject the null hypothesis and conclude that there are not significant differences among the conditions
  4. Fail to reject the null hypothesis and conclude that there are significant differences among the conditions

Homework Answers

Answer #1

The null hypothesis in words is

There are no significant differences among the three treatment means.

The alternative hypothesis in symbols is

In ANOVA, we do not state the alternative hypothesis in symbols

Number of treatment, k = 3

Total sample Size, N = 14

df(between) = k-1 = 2

df(within) = N-k = 11

df(total) = N-1 = 13

SS(between) = (T1)²/n1 + (T2)²/n2 + (T3)²/n3 - (Grand Sum)²/ N = 23

SS(within) = SS1 + SS2 + SS3 = 33

SS(total) = SS(between) + SS(within) = 56

MS(between) = SS(between)/df(between) = 11.5

MS(within) = SS(within)/df(within) = 3

F = MS(between)/MS(within) = 3.8333

Critical value Fc = F.INV.RT(0.05, 2, 11) = 3.982

The F-statistic = 3.833

Your decision is

Fail to reject the null hypothesis and conclude that there are not significant differences among the conditions.

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