You could look at $2 \times 2$ table with columns Y and O for age
groups and rows P and N for whether a purchase was made.
Y O Tot
P 35 43 88
N 98 82 180
Tot 143 125 268
To see
if the proportions of sales for Y and O groups are significantly different, you have your choice of several kinds of tests. The following ones are shown here:
If you have lots of data you could do a chi-squared test.
With fewer observations, you might want to use Fisher's exact test.
(If you get an error message from a chi-squared test, saying that you
don't have enough data, then use Fisher's exact test.)
A test of two proportions of purchases.
Here is how to run the various tests in R. Most of the tests can use a matrix of the data as follows:
MAT = matrix(c(35, 53, 98, 82), byrow=T, nrow=2)
MAT
[,1] [,2]
[1,] 35 53
[2,] 98 82
Chi-squared tests. Here are two slightly different versions of essentially the same test.
chisq.test(MAT)
Pearson's Chi-squared test
with Yates' continuity correction
data: MAT
X-squared = 4.5194, df = 1, p-value = 0.03351
I don't believe it is good idea to do the Yates correction
for samples as large as this. The correction can make the
chi-squared statistic smaller, and thus the P-value larger.
Without correction, the test is as follows:
chisq.test(MAT, cor=F)
Pearson's Chi-squared test
data: MAT
X-squared = 5.0894, df = 1, p-value = 0.02407
Fisher's exact test. This test uses the row and column totals and (and one of the four individual counts) along with a
hypergeometric distribution. Before modern computation,
Fisher's test was used mainly for small counts because
of difficulties computing cumulative hypergeometric probabilities.
Typically, as here, the P-value for Fisher's exact test
is between the P-values for the chi-squared test, with and
without the Yates correction.
fisher.test(MAT)
Fisher's Exact Test for Count Data
data: MAT
p-value = 0.02729
alternative hypothesis: true odds ratio is not equal to 1
95 percent confidence interval:
0.3177046 0.9568553
sample estimates
odds ratio
0.5537864
Test of two proportions. As implemented in R, prop.test is a test to see if two proportions
$\hat p_y = 35/143 = 0.245$ and $\hat p_o = 53/135 = 0.424$ are consistent with each other or
significantly different.
This test, comparing proportions of people in each age group who
made purchases, may be easier to explain in a report for
non-statisticians.
Notice that the two numbers of purchases and the two group sizes
are used for the data in this test. [As of 5/10/20, instructions in the R documentation for prop.test give information about the use of a matrix that is incompatible with R version 3.4.4.]
prop.test(c(35,53), c(143,135))
2-sample test for equality of proportions
with continuity correction
data: c(35, 53) out of c(143, 135)
X-squared = 6.3486, df = 1, p-value = 0.01175
alternative hypothesis: two.sided
95 percent confidence interval:
-0.26344067 -0.03223402
sample estimates:
prop 1 prop 2
0.2447552 0.3925926
Without the continuity correction, the P-value is a little smaller:
prop.test(c(35,53), c(143,135), cor=F)$p.val
[1] 0.008081579
For comparison, here is a test of two proportions from Minitab statistical software, which uses a normal approximation, seemingly with no continuity correction. [Computations agree with the formula given in the
NIST link near the beginning of this Answer.]
Test and CI for Two Proportions
Sample X N Sample p
1 35 143 0.244755
2 53 135 0.392593
Difference = p (1) - p (2)
Estimate for difference: -0.147837
95% CI for difference: (-0.256240, -0.0394342)
Test for difference = 0 (vs ≠ 0):
Z = -2.67 P-Value = 0.008
Note: I am just showing you tests in R that are appropriate for your data.
I am not advocating that you should do them all.
For your own data, you should decide which one test to do.
The P-values of these tests will generally be pretty
much the same (here, significant at levels roughly from 3% to 1%). Even so, it would be cheating to do them all and pick the
one with the smallest P-value.
