Question

Suppose the Total Sum of Squares (SST) for a completely randomzied design with k=5 treatments and n=25 total measurements is equal to 450 In each of the following cases, conduct an FF-test of the null hypothesis that the mean responses for the 5 treatments are the same. Use α=0.05.

(a) The Treatment Sum of Squares (SSTR) is equal to 360 while the Total Sum of Squares (SST) is equal to 450.

The test statistic is F=

The critical value is F

The final conclusion is:
A. We can reject the null hypothesis that the mean responses for the treatments are the same and accept the alternative hypothesis that at least two treatment means differ.
B. There is not sufficient evidence to reject the null hypothesis that the mean responses for the treatments are the same.

(b) The Treatment Sum of Squares (SSTR) is equal to 315 while the Total Sum of Squares (SST) is equal to 450.

The test statistic is F=

The critical value is F=  

The final conclusion is:
A. We can reject the null hypothesis that the mean responses for the treatments are the same and accept the alternative hypothesis that at least two treatment means differ.
B. There is not sufficient evidence to reject the null hypothesis that the mean responses for the treatments are the same.

(c) The Treatment Sum of Squares (SSTR) is equal to 180 while the Total Sum of Squares (SST) is equal to 450

The test statistic is F=

The critical value is F=

The final conclusion is:
A. There is not sufficient evidence to reject the null hypothesis that the mean responses for the treatments are the same.
B. We can reject the null hypothesis that the mean responses for the treatments are the same and accept the alternative hypothesis that at least two treatment means differ.

Answer 1

a completely randomzied design with k=5 treatments and n=25

Degrees of freedom for Treatment (Denominator)= k-1 =4

Degrees of freedom forr total = n-1 =25-1 =24

Degrees of freedom for Error(Numerator) = 24-4 =20

Critical value of F for α=0.05 for Denominator DF : 4 and Numerator DF :20 = 2.8661

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(a) The Treatment Sum of Squares (SSTR) is equal to 360 while the Total Sum of Squares (SST) is equal to 450.

SSE : Error sum of Squares = Total Sum of Squares (SST) - Treatment Sum of Squares (SSTR) = 450 - 360 = 90

Mean Treatment Sum of Squares = Treatment Sum of Squares (SSTR) / Degrees of freedom for Treatment = 360/4=90

Mean Error Sum of Squares = Error Sum of Squares (SSTR) / Degrees of freedom for Error= 90/20=4.5

The test statistic is F= Mean Treatment Sum of Squares / Mean Error Sum of Squares = 90/4.5 =20

The test statistic is F= 20

Critical value of F for α=0.05 for Denominator DF : 4 and Numerator DF :20 = 2.8661

The critical value is F : 2.8661

As Test statistic - F: 20 Greater then Crtitical value of F: 2.8661 ; Reject null hypothesis:

The test statistic is F= 20

The critical value is F = 2.8661

The final conclusion is:
A. We can reject the null hypothesis that the mean responses for the treatments are the same and accept the alternative hypothesis that at least two treatment means differ.

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(b) The Treatment Sum of Squares (SSTR) is equal to 315 while the Total Sum of Squares (SST) is equal to 450.

SSE : Error sum of Squares = Total Sum of Squares (SST) - Treatment Sum of Squares (SSTR) = 450 - 315= 135

Mean Treatment Sum of Squares = Treatment Sum of Squares (SSTR) / Degrees of freedom for Treatment = 315/4=78.75

Mean Error Sum of Squares = Error Sum of Squares (SSTR) / Degrees of freedom for Error= 135/20=6.75

The test statistic is F= Mean Treatment Sum of Squares / Mean Error Sum of Squares = 78.75/6.75=11.6667

Critical value of F for α=0.05 for Denominator DF : 4 and Numerator DF :20 = 2.8661

The critical value is F : 2.8661

As Test statistic - F: 11.6667 Greater then Crtitical value of F: 2.8661 ; Reject null hypothesis:

The test statistic is F= 11.6667

The critical value is F : 2.8661

The final conclusion is:
A. We can reject the null hypothesis that the mean responses for the treatments are the same and accept the alternative hypothesis that at least two treatment means differ.

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(c) The Treatment Sum of Squares (SSTR) is equal to 180 while the Total Sum of Squares (SST) is equal to 450

SSE : Error sum of Squares = Total Sum of Squares (SST) - Treatment Sum of Squares (SSTR) = 450 - 180= 270

Mean Treatment Sum of Squares = Treatment Sum of Squares (SSTR) / Degrees of freedom for Treatment = 180/4=45

Mean Error Sum of Squares = Error Sum of Squares (SSTR) / Degrees of freedom for Error= 270/20=13.5

The test statistic is F= Mean Treatment Sum of Squares / Mean Error Sum of Squares = 45/13.5=3.3333

Critical value of F for α=0.05 for Denominator DF : 4 and Numerator DF :20 = 2.8661

The critical value is F : 2.8661

As Test statistic - F: 3.333 Greater then Crtitical value of F: 2.8661 ; Reject null hypothesis:

The test statistic is F= 3.3333

The critical value is F : 2.8661

The final conclusion is:
A. We can reject the null hypothesis that the mean responses for the treatments are the same and accept the alternative hypothesis that at least two treatment means differ.