# Does it make mathematically sense to aggregate data in order to reduce variance in statistical significance tests?

Let's consider an e-commerce site. We have an AB test for which we want to measure if the average revenue from treatment A is statistically significant different from B. i.e. the main goal is to determine if A is statistically significant better than B (or vice versa).

Both A, B variants generate a continuous stream of revenues for each purchased item. For most of the items the revenue is zero. Our measure is the average revenue per displayed item. i.e. The raw data is composed of many items with their associated revenue if purchased or zero otherwise.

The generated revenue has some patterns. For example, the sport category generates much more revenue than other categories. Another example is the day of week pattern in which we have much less revenue on weekends.

In order to reduce the variance I would like to account for the revenue patterns. One option is to aggregate the data. For example, calculate the avg revenue per category, calculate the difference in revenue between A & B for each category,  and then do a paired t-test on all the revenue differences. Or aggregate the results per day and then send to t-test the differences in revenue for each day.

The aggregation is done only to decrease variance, the research question is still, which treatment will generate statistically significant more revenue globally (i.e. considering all days and all categories)

My questions:

Suppose that I do aggregation per day

1. Does it make sense to aggregate and do paired t-test ? One of the downsides of this approach is that we lose the information about the standard deviation of each day

2. If the different aggregations have different sizes (e.g. each day has different revenue) should i weigh it somehow (i.e. give higher revenue days more weight) and if so how can i weigh in paired t-test?

3. Any other ideas of how to do it?

This question suggest that pairing is a good idea, but there are some crucial differences. for example, the data in this problem is already aggregated, and there is no solution for the different sizes of the aggregation.

• 1) What do you want to test? Better yet, what do you want to learn from your data? // 2) How would it help you to reduce the sample size?
– Dave
Oct 4, 2021 at 14:17
• Why do you want to reduce the variance? If you look at the aggregates, the groups would need to have exactly same size, otherwise taking each aggregate as a single sample would lead to invalid results.
– Tim
Oct 4, 2021 at 14:27
• @Dave 1) i want to conclude which treatment generate more revenue in total. 2) There is high variance between the different groups for example when i compare Tuesday to Friday there is huge difference but if i compare A on Tuesday to B on Tuesday they will be similar. Oct 4, 2021 at 14:33
• It sounds like you want to control for other variables, more-or-less what ANCOVA does (which considers variability from other sources, like the day of the week). That said, I am not so sure that you will be happy with your results unless you consider the time series nature of your data.
– Dave
Oct 4, 2021 at 14:38
• rather than doing a t-test, a linear regression might be appropriate where you control for the other variables eg day of week, item category and then do a test on the treatment coefficient. $revenue\sim treatment + category + \text{day_of_week}$. Oct 4, 2021 at 15:14

Aggregation then do a paired-t doesn't make sense for variance reduction and may result in large variance and biased std error. A linear regression or post-stratification might be appropriate.

Let's make some fake data.

library(tidyverse)
library(furrr)
options(future.globals.maxSize = 1073741824)

get_estimates <- function(strat_cnt = 1000) {

df <- tibble(
strat = c('s1', 's2'),
strat_y = c(0.05, 0.5), # two strat conversion rates are very different.
strat_cnt = rbinom(2, 1000, c(0.7, 0.3)) # strat sample proportion rate
) %>%
rowwise() %>%
mutate(y = list(rbinom(strat_cnt, 1, strat_y))) %>%
unnest(y) %>%
dplyr::select(strat, y) %>%
mutate(trt = if_else(runif(nrow(.)) > 0.5, 1, 0)) # random assign to A/B

df_agg <- df %>%
group_by(strat) %>%
summarise(
n = n(),
n0 = sum(trt),
n1 = sum(1 - trt),
s = n0 * n1 / n, # strat variance
y0 = sum(y * trt),
y1 = sum(y * (1 - trt)),
strat_diff = y1 / n1 - y0 / n0,
.groups = 'drop'
)

df_agg %>%
summarise(

# non_aggregate
non_aggregate = sum(y1) / sum(n1) - sum(y0) / sum(n0),

# agg_avg: aggregate by strat then average without weight
agg_avg = mean(strat_diff),

# post_strat: aggregate by strat then weight by strat proportion
weight_diff = weighted.mean(strat_diff, n),

# ols regression/ancova: if strat is categorical, then ols estimate equal to weight by strat variance
# lm(y ~ trt + strat, df)\$coefficients['trt']
ols_diff = weighted.mean(strat_diff, s)
)
}

plan(multisession, workers = 4)
SIM_CNT <- 100000
sim_estimate <- furrr::future_imap_dfr(
1:SIM_CNT,
~get_estimates(),
.options = furrr_options(seed = TRUE),
.progress = TRUE
)

sim_estimate %>%
pivot_longer(
cols = c('non_aggregate', 'agg_avg', 'weight_diff', 'ols_diff'),
names_to = 'key',
values_to = 'value'
) %>%
# filter(key == 'avg_day_diff') %>%
ggplot() +
geom_freqpoly(aes(value, color = key), bins = 100) +
labs(
x = 'diff', y = '', color = '',
title = 'aggregation then do a paired-t result in large variance')