Question

For each of the following situations, give the degrees of freedom for the group (DFG), for error (DFE), and for the total (DFT). State the null and alternative hypotheses, H₀ and H_a, and give the numerator and denominator degrees of freedom for the F statistic.

(a) A poultry farmer is interested in reducing the cholesterol level in his marketable eggs. He wants to compare two different cholesterol-lowering drugs added to the hen's standard diet as well as an all-vegetarian diet. He assigns 25 of his hens to each of the three treatments.

DFG	=
DFE	=
DFT	=

H₀:

All groups have the same mean cholesterol level.

All groups have different mean cholesterol levels.    

The all-vegetarian diet group has a higher mean cholesterol level.

At least one group has a different mean cholesterol level.

The all-vegetarian diet group has a lower mean cholesterol level.

H_a:

All groups have different mean cholesterol levels.

All groups have the same mean cholesterol level.

   The all-vegetarian diet group has a higher mean cholesterol level.

At least one group has a different mean cholesterol level.

The all-vegetarian diet group has a lower mean cholesterol level.

numerator df
denominator df

(b) A researcher is interested in students' opinions regarding an additional annual fee to support non-income-producing varsity sports. Students were asked to rate their acceptance of this fee on a seven-point scale. She received 99 responses, of which 30 were from students who attend varsity football or basketball games only, 18were from students who also attend other varsity competitions, and 51 who did not attend any varsity games.

DFG	=
DFE	=
DFT	=

H₀:

At least one group has a different mean rating.

The group of students who do not attend games has a lower mean rating.  

  The group of students who do not attend games has a higher mean rating.

All groups have different mean ratings.

All groups have the same mean rating.

H_a:

All groups have the same mean rating.

At least one group has a different mean rating.    

The group of students who do not attend games has a lower mean rating.

The group of students who do not attend games has a higher mean rating.

All groups have different mean ratings.

numerator df
denominator df

(c) A professor wants to evaluate the effectiveness of his teaching assistants. In one class period, the 45 students were randomly divided into three equal-sized groups, and each group was taught power calculations from one of the assistants. At the beginning of the next class, each student took a quiz on power calculations, and these scores were compared.

DFG	=
DFE	=
DFT	=

H₀:

At least one group has a different mean quiz score.

All groups have the same mean quiz score.   

The group taught by the oldest TA has a higher mean quiz score.

All groups have different mean quiz scores.

The group taught by the oldest TA has a lower mean quiz score.

H_a:

The group taught by the oldest TA has a lower mean quiz score.

At least one group has a different mean quiz score.  

  The group taught by the oldest TA has a higher mean quiz score.

All groups have different mean quiz scores.

All groups have the same mean quiz score.

numerator df
denominator df

Answer 1

(A)

there are 2 categories with a total of 25 hens

DFG = number of categories -1

= 2-1

= 1

DFE = total - number of categories

= 25 -2

= 23

DFT = DFG +DFE = 23+1 = 24

For ANOVA, we always assume that the null hypothesis is that all the means are same and alternate hypothesis is that there is at least one group with different mean.

H₀:

All groups have the same mean cholesterol level.

Ha

At least one group has a different mean cholesterol level.

numerator df = DFG = 1

denominator df = DFE = 23

(B)

there are 3 categories with a total of 99 students

DFG = number of categories -1

= 3-1

= 2

DFE = total - number of categories

= 99 -3

= 96

DFT = DFG +DFE = 2+96 = 98

For ANOVA, we always assume that the null hypothesis is that all the means are same and alternate hypothesis is that there is at least one group with different mean.

H₀:

All groups have the same mean rating

Ha

At least one group has a different mean rating.

numerator df = DFG = 2

denominator df = DFE = 96

(C)

there are 3 categories with a total of 45

DFG = number of categories -1

= 3-1

= 2

DFE = total - number of categories

= 45 -3

= 42

DFT = DFG +DFE = 2 + 42 = 44

For ANOVA, we always assume that the null hypothesis is that all the means are same and alternate hypothesis is that there is at least one group with different mean.

H₀:

All groups have the same mean quiz score

Ha

At least one group has a different mean quiz score.

numerator df = DFG = 2

denominator df = DFE = 42