For each of the following situations, give the degrees of freedom for the group

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, Ho and Ha, 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 15 of his hens to each of the three treatments. DFE = DFT = Ho: O 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. O The all-vegetarian diet group has a lower mean cholesterol level. At least one group has a different mean cholesterol level Ha: O At least one group has a different mean cholesterol level All groups have the same mean cholesterol level. O The all-vegetarian diet group has a lower mean cholesterol level O All groups have different mean cholesterol levels. O The all-vegetarian diet group has a higher mean cholesterol level.

Explanation / Answer

we have the equation TSS=SSG+SSE

i.e, total sum of squares=sum of squares of groups +error sum of squares

their degrees of freedoms are also additive

if n is the total number of observations and k is the number of groups

then df of TSS is n-1

df of SSG is k-1

hence df of SSE is (n-1)-(k-1)=n-k

and for F statistic the numerator df is same as df of groups and denominator df is same as df of errors

a) here number of observations is n=15 and number of groups=k=3

so DFG=k-1=2

DFE=n-k=12

DFT=n-1=14

now null hypothesis

H0: all groups have the same mean cholesterol level

alternative hypothesis

H1: at least one group has a different mean cholesterol level

numerator df=DFG=2

denominator df=DFE=12

b) number of observations=number of responses=n=95

number of groups=k=3

hence DFG=k-1=2

DFE=n-k=92

DFT=n-1=94

null hypothesis H0:all groups have the same mean rating

alternative hypothesis H1: at least one group has different mean rating

numerator df=DFG=2

denominator df=DFE=92

c) here number of groups=k=3

number of observations=n=45

so DFG=k-1=2

DFE=n-k=42

DFT=n-1=44

H0: all groups have the same mean quiz score

H1: at least one group has different mean quiz score

numerator df=DFG=2

denominator df=DFE=42

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