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Degrees of freedom

J

Jammy

Member
This is in reference to Q 11 of May 2009 paper of IAI

We're given 3 treatments and 5 values each
We're asked to combine treatments 1 & 3, and then examine whether there is evidence to suggest that the average effect of treatments 1&3 is inferior to that of treatment 2.

Here, what would be the degrees of freedom using test statistic with distribution t_n1+n2-2 ?
Would it be 5+5-2 = 8 or 10+5-2 = 13 ?

What would be the 'n1' in the test statistic, 10 or 8?

Thanks a ton :)
 
So whenever we combine rows or columns, the degrees of freedom remains as per the original table?

We are essentially not putting any additional constraints on the possible values the blanks could take, so it should be that way.
 
In one of the questions, there 5 columns (Monday - Friday) and 4 rows (excellent, good, fair poor)

Normally, dof is 4*3= 12
We're asked to combine Monday-tuesday and wed-thurs-fri and also combine excellent-good and fair-poor

So our columns are now 1. Mon+Tues 2. Wed-Fri
Our rows are now 1. Excellent+Good 2. Fair+Poor

Now, what will be the dof?

This Q was in Nov 2008 paper, IAI
 
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In one of the questions, there 5 columns (Monday - Friday) and 4 rows (excellent, good, fair poor)

Normally, dof is 4*3= 12
We're asked to combine Monday-tuesday and wed-thurs-fri and also combine excellent-good and fair-poor

So our columns are now 1. Mon+Tues 2. Wed-Fri
Our rows are now 1. Excellent+Good 2. Fair+Poor

Now, what will be the dof?

This Q was in Nov 2008 paper, IAI
If you have just added columns 1. Mon+Tues 2. Wed-Fri then dof would be 1*3 as height is same.(dof remains same for Quality)
If you have just added Rows 1. Excellent+Good 2. Fair+Poor then dof would be 4*1 as breadth is same.(dof remains same for days)
But we need both so dof=1*1=1
 
Why the difference in the 2 cases discussed?

In the former case, dof did not change when we took average of 2 columns and made it 1 column

In the latter case, dof changed as per our operations i.e. dof is now in accordance with the new 2*2 table.

I'm not getting when dof changes and when it doesn't.
 
Last edited by a moderator:
Why the difference in the 2 cases discussed?

In the former case, dof did not change when we took average of 2 columns and made it 1 column

In the latter case, dof changed as per our operations i.e. dof is now in accordance with the new 2*2 table.

I'm not getting when dof changes and when it doesn't.
Where we took average of two columns please specify ?
 
This is in reference to Q 11 of May 2009 paper of IAI

We're given 3 treatments and 5 values each
We're asked to combine treatments 1 & 3, and then examine whether there is evidence to suggest that the average effect of treatments 1&3 is inferior to that of treatment 2.

Here, what would be the degrees of freedom using test statistic with distribution t_n1+n2-2 ?
Would it be 5+5-2 = 8 or 10+5-2 = 13 ?

What would be the 'n1' in the test statistic, 10 or 8?

Thanks a ton :)

Here
 
Okie!
first of all,we didn't took average of 2 columns and made it 1 column.
Rows!
we did concatenate(did not add) 2 Rows/group then checked the equality of mean with remains third row/group by t-dist. Hence so.
in later case we were checking for association between them by contingency table.Hence so.

Note:-ways of assuming dof for t-dist and Chi-squared-contingency-table are different.
 
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