HW question 9.20.e.iii (extra credit)

HW question 9.20.e.iii (extra credit)

by Thomas Newman -

Dear Students,

See question received via email below.  Can anyone answer it?  Seems like a good point!  Do we need to rewrite the question?

Tom

Subject: HW Ch 8 and 9 question

Hi Dr. Newman,

For our HW last week, I still don’t understand the answer to the extra credit question (9.20, e, iii). The answer key seems to do a back of the envelope calculation to determine the number of actual c-sections performed to prevent 1 neonatal/perinatal death but does not take into account that the risk in the vaginal delivery group also changed when looking at actual delivery method (not planned delivery method) so the original 100 number needed to plan is not a good starting point for the calculation.

If we imagine that we never calculated the number of c-sections needed to PLAN to prevent 1 death as in part ii, and instead start the calculation of number of c-sections actually COMPLETED to prevent 1 death (an as treated analysis) from scratch, I don’t understand why my approach is incorrect:

 

Based on the article table 4, 13 deaths occurs in the received vaginal delivery group and 3 in the received c-section group (ended up being the same numbers as the planned mode of delivery groups but there were crossovers among the neonatal deaths).

 

1041 assigned c-section                                   1042 assigned vaginal delivery

941 + 451 crossed over = 1392 had c/s            591 + 100 crossed over = 691 had vaginal delivery

 

Received c-section risk: 3 / 1392 = 0.002155

Received vaginal delivery risk: 13 / 691 = 0.0188

ARR = 0.0188 – 0.002155 = 0.016658

NNT = 1/ARR = 1/0.016658 = 60

Thanks!

Re: HW question 9.20.e.iii (extra credit)

by Jonathan Lee -
I agree that it doesn't seem to make sense to start with the number of C-sections planned when the benefit is assumed to be from C-sections completed. I did my calculation based on the idea that the difference of 10 deaths was due to the difference in the number of C-sections performed. In the planned c-section group there were 90%*1039 = ~935 c-sections In the planned vaginal delivery group there were 43%*1039 = ~447 c-sections So it took ~935-447 = 488 extra c-sections to prevent 10 events This equates to ~49 c-sections per event This is similar to the answer key answer but makes more sense to me intuitively.

Re: HW question 9.20.e.iii (extra credit)

by Michael Kohn -

Jonathan,

You calculation requires that the randomization groups be equal in size, which they are.  This is an implicit assumption when you use 13-3= 10 for the number of deaths prevented.  The calculation would not work if, for example, the planned C-section group were twice the size of the planned VBAC group.

If PPt and PPc represent the "per-protocol" proportions in the treatment and control groups respectively, then the general formula (PPt - (1-PPc))/ARR should give you the adjusted NNT assuming all differences in outcomes are due to the treatment.  Another reasonable assumption is that, if a patient is not treated "per protocol", she is a cross-over.

As you point out, either way you calculate it, the answer is 49 C-sections instead of vaginal deliveries to prevent one neonatal death.

49 may be the answer to this clinical epidemiology problem, but

The Answer to the Ultimate Question of Life, The Universe, and Everything is 42 -- Douglas Adams, The Hitchhiker's Guide to the Galaxy.

MAK

Re: HW question 9.20.e.iii (extra credit)

by Shama -

Hi, 

we also thought in similar way. we moved our 3 deaths from  c-section to vaginal delivery ( assuming no deaths in c-section)

so got 

Relative Risk of death vaginal delivery = 16/691= 0.023

NNT= 1/0.023= 43.

 

 

Re: HW question 9.20.e.iii (extra credit)

by Jennifer -

 

I agree with the student's answer from the original post from Dr. Newman. The question asks about "cesarean deliveries themselves" which I interpreted as to mean actual c-sections done. I did the analysis the same way but got a slightly different answer of 58 due to rounding.  

Actual

Mortality

No Mortality

Total

C-Section

3

1386

941+ 451 = 1392

Vaginal

13

685

100 + 591 = 691

 

  

 

 

 

3/1392 = 0.002

13/691 = 0.019

RRR = 0.9

ARR = 0.019-0.002 = 0.017

1/0.017 = 58

Number of Cesareans needed to prevent one neonatal death = 58

 

Re: HW question 9.20.e.iii (extra credit)

by Rae Wannier -

So I have to admit, that I see no problem with the original student's posted work, as they drew updated data directly from the article.  However, to offer an alternative viewpoint, I did not go back to the original article, but performed some calculations to come up with updated mortality rates for vaginal and cesarean deliveries using proportional attribution from the planned outcomes.  That is to say:

 

       MRvaginal = Mortality rate of performed vaginal deliveries

       MRcesarean = mortality rate of performed cesarean deliveries

 

MR planned Cesarean = MRcesarean*proportion receiving Cesarean +  MRvaginal*proportion receiving vaginal 

            => 0.003 = 0.90*MRcesarean + 0.10*MRVaginal

MR planned vaginal = MRcesarean*proportion receiving Cesarean +  MRvaginal*proportion receiving vaginal

            => 0.013 = 0.43*MRcesarean + 0.57*MRvaginal

 

In doing this I made one important assumption which is that the infant mortality rate from a vaginal delivery was the same whether or not the vaginal delivery was planned or an emergency delivery in the Cesarean group, and vice versa for the cesarean deliveries. (Clearly this was an imperfect assumption based upon the additional data drawn from the paper)

 

Then, by solving these two equations I came up with:

MRcesarean = 0.00087

MRvaginal = 0.0221

 

Then calculated:

ARR = Rvaginal - Rceasarian = 0.22 – 0.00087 = 0.0212

NNT = 1/ARR = 1/ 0.0212 = 47.2 women needed to treat with ceasarian instead of vaginal delivery to prevent one neonatal/perinatal death

 

I did get the same answer as the one provided in the answer key, though using a different methodology.  However, I believe I used the same underlying assumptions, just presented differently?

Re: HW question 9.20.e.iii (extra credit)

by Michael Kohn -

This reduces to MRv - MRc = ARRI/(0.90-0.43) .   Where "ARRI" is the ARR from the ITT analysis.  However you do the math, you end up inflating the ARRI by a factor of 1/0.47 , which reduces the NNT, so the adjusted NNT = NNTI * (0.47).  100 * 0.47 = 47.  The 49 that some of us are getting is because we used Excel which doesn't round 3/1039 to 0.003 (it's 0.00288739) or 10/1039 to 0.01 (it's 0.00962454).  This is not OCD; it's actually easier to do these kinds of calculation in Excel.

MAK