Comment: There is one sense in which seeing the ad
seems to have prompted a greater response. That is
the response rate (rather than the dollar amount spent).
There are various tests of $H_0: p_1 = p_2$ vs.
$H_a: p_1 > p_2,$ where the $p_i$ are the respective response rates.
Output from Minitab for two such tests is shown below.
Test and CI for Two Proportions
Sample X N Sample p
1 50 1000 0.050000
2 29 1000 0.029000
Difference = p (1) - p (2)
Estimate for difference: 0.021
95% lower bound for difference: 0.00669270
Test for difference = 0 (vs > 0):
Z = 2.41 P-Value = 0.008
Fisher’s exact test: P-Value = 0.011
The first test (P-value 0.021) uses a normal approximation
of the binomial proportions, which should give
reasonably accurate results for such large samples.
Fisher's exact test (P-value 0.011) uses a hypergeometric distribution. Both tests are significant.
If looking at response rates is of interest to you, you can
find particulars in an elementary applied statistics text
or online.
As @NickCox suggests, we would have to know the $(50+29 = 79)$ individual dollar amounts in order confidently to explore two-sample tests for
amount spent. However, it seems each purchase in each group averages around $15, so looking at response rates might tell
you what you really want to know about the effect of exposure
to the ad.
Note: Just as an experiment, I simulated a dataset assuming
that the 50 nonzero sales in Group 1 are distributed
$\mathsf{Norm}(\mu=16, \sigma=3)$ and that the 29 nonzero sales in Group 2 are distributed $\mathsf{Norm}(\mu=14, \sigma=3).$ Dollar amounts were rounded to integers:
table(x1)
x1
0 10 11 12 13 14 15 16 17 18 19 20 21 22
950 2 1 7 6 8 5 4 5 2 2 5 1 2
table(x2)
x2
0 8 9 10 11 12 13 14 15 16 18 21 22
971 1 2 1 3 4 2 3 4 3 4 1 1
A Welch two-sample t test in R gave P-value 0.0033, as follows:
Welch Two Sample t-test
data: x by g
t = 2.7117, df = 1803.6, p-value = 0.003379
alternative hypothesis:
true difference in means is greater than 0
95 percent confidence interval:
0.1415163 Inf
sample estimates:
mean in group 1 mean in group 2
0.768 0.408
In spite of all the zeros and additional ties that result from
rounding dollars to integers, a one-sided, two-sample Wilcoxon (rank sum) test in R gave P-value 0.007. with no error messages.
wilcox.test(x ~ g, alt="g")$p.val
[1] 0.00725217
A one-sided permutation test with the pooled 2-sample t statistic
as metric (but not assuming normality) gave P-value about 0.003.
Unless your non-zero dollar values are much different
from my simulated ones, I do not expect a problem finding
a valid two-sample test to compare dollar amounts.
