Question

In: Statistics and Probability

Using the partially completed analysis of variance and table of means provided below, answer the following questions. Use the .05 significance level. All three groups have the same sample size.

QUESTION 10

  1. Questions 10 through 13 refer to the following:

    Using the partially completed analysis of variance and table of means provided below, answer the following questions. Use the .05 significance level. All three groups have the same sample size.

    Source

    SS

    df

    MS

    F

    Treatment

    3.9302

    Error

    9.03

    Total

    26

    Group

    A

    B

    C

    sample mean

    278.3

    275.4

    274.5

    Find the value of LSD for comparing Groups A and B.

QUESTION 11

  1. Is the difference between groups A and B significant?

    yes

    no

    maybe?

    Antidisestablishmentarianism

QUESTION 12

  1. What's the value of SSTR?

QUESTION 13

  1. What's the value of SSE?

Solutions

Expert Solution

QUESTION 10

Source

SS

df

MS

F

Treatment

70.98

2

35.49

3.9302

Error

216.72

24

9.03

Total

287.70

26

DONE

Concept Base

Degrees of freedom

Treatment: Number if treatments (groups) – 1 = 3 – 1 = 2

Error: df for Total – df for Treatment = 26 (given) – 2 = 24

F = MSTr/MSE

So, MSTr = F x MSE = 3.9302 x 9.03 = 35.49

MSTr = SSTr/df, or

SSTr = MSTr x df = 35.49 x 2 = 70.98

Similarly, SSE = 9.03 (given) x 24 = 216.72

SST = SSTr + SSE = 70.98 + 216.72 = 287.70.

Value of LSD for comparing Groups A and B.

QUESTION 11

Value of LSD for comparing Groups A and B

Concept Base

Let n1 and n2 be the sample sizes for obtaining X1bar and X2bar.

LSDcrit = t(α/2)√[MSW{(1/n1) + (1/n2)}], where

t(α/2) = upper (α/2) percent point of t-distribution with

degrees of freedom = DF of MSW i.e., Error or Residual MS of ANOVA

Decision: The difference, d = |X1bar - X2bar| is significant if

d > LSDcrit, as given above.

Now, the calculations

d = 278.3 – 275.4 = 2.9

n1 = n2 = 9

t(α/2) = upper 2.5% percent point of t-distribution with 24 degrees of freedom

= 2.064 [from Standard t-distribution Tables]

MSW = 9.03 [from ANOVA Table]

{(1/n1) + (1/n2)} = 2/9 = 0.2222

LSDcrit = 2.064√(9.03 x 0.2222)

= 2.923

Here the difference between d and LSDcrit is marginal. Hence, we conclude that the difference between the means of Group A and Group B is not significant. Answer

QUESTION 12

SSTR = 70.98 [from ANOVA Table] Answer

QUESTION 13

SSE = 216.72 [from ANOVA Table] Answer

DONE


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