To perform the sample size and power calculations the desired type of effect to be observed needs to be clear. The way I understand it, the study will be looking at carriage level changes in contacts of children that get vaccine, looking at a continuous variable, so the t-test is the appropriate measure. Also, whether transmission occurs to previously uninfected individuals will be measured. This is a binary measure, where the chi-squared test is more appropriate. The calculations are done at the conventional 0.05 level, and considering for sample sizes that will provide statistical power of 0.8 and 0.9. Effect size (standardised difference between the control and treatment group) is considered as a variable ranging from 0.1-0.5. An effect size of 1 means that the control and treatment group differ by one standard deviation (for chi squared tests, this is Cohens w).

The effect in this study can only be seen by observing the effect on social contacts. So the ‘sample size’ is actually the number of observations of contacts of children rather than the children themselves. So, we want to estimate the number of people that form links in the study.

Household based study

Number of household estimates

If the study design is to sample individuals in households, with children as a ‘way in’, it seems sensible to estimate the number of children needed to reach the required sample size. The number of household contacts per child can be estimated using census data. The 2001 census provides tables containing the composition of households. Extracting the relevant types of households for the Bristol area (i.e. those that include children), gives the following distribution of household composition.

To get a sense for the number of ‘relevant’ families, we can consider only those that have one child and more than two adults in the household, two children and one or more adults in the household.

Comparing over the four study sites: Bristol, Southampton, Oxford and Wandsworth (London), and performing the same analysis reveals some differences between sites in terms of proportions of household composition.

There are proportionally more households with two or more children and two or more adults in Bristol, Southampton and Oxford when compared to Wandsworth in London. Something to bear in mind when sampling (bottom panel). Nevertheless, there are large numbers of households available to all sites with the exception of Oxford.

The numbers above are families with children with ages between 0 and 4. Restricting the age range to between 2 and 2 + 10 months, and assuming that the birth rate is constant over this time means we should scale the numbers by 10/48. So, the best estimate of the number of eligible families with children between 2 and 2 years and 10 months in the four sites are (rounded to the nearest integer):

One child, two+ adults Two+ children, two+ adults One adult, two+ children
Bristol 2884 4097 1720
Southampton 1558 2009 906
Oxford 941 1333 416
Wandsworth 1276 1660 699
Total 6659 9099 3741

If the response rate is pretty low (1-5%), that leaves us with between 194.9895833 - 974.9479167, approximately 200 - 1000 household responses in total. If only families with two or more children are targeted, we have 128.4041667 - 642.0208333, approximately 130-650 households. If half of those are already carriers, that reduces the available sample to between 64.2020833 - 321.0104167 65 - 320 households.

Having said all that, these numbers are coming from restrictive spatial regions. If the study sites are able to fulfill their limit of 100 families by reaching out further afield or pushing to get a higher response rate, we have about 500 households accross five sites. If half of children are already positive, we have 250 households.

Total expected contacts

Assuming that participants are chosen at random from these households, and that taking the very conservative estimate that two+ means two, it is possible to estimate the expected number of contacts. Children from families containing one child and two adults, and those from families with one adult and two+ children will have two contacts per child, and those from families with two+ children and two+ adults have three contacts per child. Therefore, a lower bound expectation for the total number of contacts is 480.9729167 ~480 with an upper bound of 2612.85625 ~2600.

Taking the other approach, that we will be able to sample 500 housholds, and an expected 250 of those will be included, we need to do a slightly different calculation. Of those 250 households that are useful, we would expect the following distribution.

One child, two+ adults Two+ children, two+ adults One adult, two+ children
85 117 48

So the conservative total expected number of contacts is (85+48)x2 + 117x3 = 617. Say 600.

We can calculate the expected number of contacts in a similar way, giving us: 2.4666595.

Power

All of the following numbers are for a power of 0.9 and sigma 0.05 for the chi-squared test (binary outcome). At the low end of the expectations for the number of contacts sampled, we can detect an effect size of 0.14, and at the higher end, an effect size lower than 0.1. An effect size of 0.14 is considered small. I think we should aim for a sample size above 1000, with which an effect size of less than 0.11 should be detectable.

What do we expect?

From reading the LAIV paper and comments from Adam, the increase in colonisation is between 4 and 23 fold (with 6 for Sp). For HIV, there is some debate, but a logarithmic relationship between viral load and transmission probability is often used as a baseline. If we expect a similar relationship to hold for bacterial pathogens, we could expect to see an increase of between 1.4-3 fold in transmission.

The effect size for this increase in transmission probability can be calculated. If a is the proportion by which transmission probability increases, the effect size should be a-1. So for a conservative 1.4-fold increase, we would expect an effect size of 0.4. For a two fold increase, we’d expect a huge effect size of 1 (vaccinated population one standard deviation higher than the control population).

Sanity check

I wanted to make sure the power calculation was right, so I did some checks, using another approach. Let’s assume that the rate of carriage is 0.5 in children and ~0.09 in adults of those children, and assume that each child has two adult contacts (this is not wildly wrong). This gives us one adult-positive child link. Given that 0.09 adults are positive, that suggests that the probability of transmission is 0.09.

Now, if we double the probability of transmission, to 0.18, we can treat both the control and treatment group as Bernouilli tests, which lead to a binomial distribution of outcomes. This way, we can calculate the power of the test (and therefore sample size) using differences of the two binomial distributions (conveniently found in the r stats library).

library(stats)
power.prop.test(p1=0.09,p2=0.18,power=0.9,alternative='two.sided')
## 
##      Two-sample comparison of proportions power calculation 
## 
##               n = 300.8815
##              p1 = 0.09
##              p2 = 0.18
##       sig.level = 0.05
##           power = 0.9
##     alternative = two.sided
## 
## NOTE: n is number in *each* group
power.prop.test(p1=0.09,p2=0.166,power=0.8,alternative='two.sided')
## 
##      Two-sample comparison of proportions power calculation 
## 
##               n = 302.1634
##              p1 = 0.09
##              p2 = 0.166
##       sig.level = 0.05
##           power = 0.8
##     alternative = two.sided
## 
## NOTE: n is number in *each* group

For a power of 0.9, we get an n of 601.763, close to what we had beforehand, but where the effect size argument is side-stepped, and the probability of transmission is doubled (!). For a lower power of 0.8, aiming for n around 600 gives us the ability to detect a 1.8 fold increase.

The reason for the difference between the two approaches is the use of the Binomial distribution rather than the chi-squared distribution (typically used in power calculations). I’d suggest that the binomial may be even more appropriate.

All this comes with the caveat that I’m not taking into account household limitations - a positive child only has so many possible contacts to infect, so the binomial distribution is cut off at 2. Correcting for that is possible, but I just don’t have the time to do those calculations. I leave it here as a note for later.

Conclusion from chi-squared power calculation

If the number of contacts sampled is 600, the study will be able to detect transmission increases of as little as 1.12 fold with a power of 0.8 and 1.4 fold with a power of 0.9.

New conclusion from binomial calculation

If the number of contacts sampled is 600, the study will be able to detect transmission increases of as little as 1.8 fold with a power of 0.8 and 2 fold with a power of 0.9.