T-test does not seem suitable for your purpose. For a t-test participants need to deliver information on a metric scale, but in your case you just counted how many people attentended, i.e. you counted how man participants were recruited, but you did not get any other information.

Before thinking about an adequate test, you should provide the hypothesis you want to test. Do you only want to know if there were more participants in 2019 as compared to 2020 or is it also relevant how many people attended each month and if there are differences?

Rainer Duesing thank you for your quick answer.

That is indeed what I was struggling with.

It is absolutely relevant how many people attended each month.

Use z-test since the samples are large.

Ette Etuk please explain how you would do a z-test with the data structure at hand?

Aren't you wanting to compare taking into account month (and therefore some seasonal adjustment), so that these are paired? The sample is small: n=5. Months is the unit of analysis, and the variable is those counts.

While they are counts, the distribution clearly is not Poisson (means about 4+ times the sd) so the usually things to do (logging, Poisson regression) are probably not the first choice.

> x1 t.test(x1-x2)

One Sample t-test

data: x1 - x2

t = -0.39356, df = 4, p-value = 0.714

alternative hypothesis: true mean is not equal to 0

95 percent confidence interval:

-32.21885 24.21885

sample estimates:

mean of x

-4

p.s. Not sure why Ette Etuk called this a large sample. Even if not matched there are only 10 data points.

Daniel Wright and D.J. Donker

since he is interested in the proportions, how many participants attended and compared between 2019 and 2020 for each month (as far as I understood), how about this idea:

As a kind of "omnibus" test, a chi-square test for crosstables (year*month) to test if observed proportions overall differ from expected proportions. Additionally you can calculate Cramer's V as some form of effect size.

If this test indicates deviation from independence, calculate binomial tests for the proportions between 2019 and 2020 for each month separately (alpha correction suggested).

The questionner says: " I want to analyze the number of people screened per month in 2019 and 2020. " So even in bold it is the number screened, and for lots of research questions that would be what is of interest (e.g., in COVID testing in the early days, it was how many you could get done, not a proportion, for the people organizing these tests ... the proportion is of more interest to the epidemiologists).

But if it is the proportion that they are interested in, they should delete this question, think about what they are interested in, and write a new question (and how they would estimate the denominator). The numbers would be those ten proportions, though depending on how the estimates of the denominator and the associate uncertainties would complicate things.

Let's wait what the D.J. Donker is interested in.

Daniel Wright Rainer Duesing

I'm interested in the numbers, not the proportions!

D.J. Donker, the syntax in your original question appears to be asking SPSS to compare the means of the month numbers--i.e., it has Months as the outcome variable. See if this gives you what you're after.

NEW FILE.

DATASET CLOSE ALL.

DATA LIST LIST / Year (F4.0) Month (F2.0) n (F5.0).

BEGIN DATA

19 3 58

19 4 48

19 5 58

19 6 44

19 7 37

20 3 30

20 4 56

20 5 55

20 6 52

20 7 72

END DATA.

* Independent samples t-test.

T-TEST GROUPS=Year(19 20)

/MISSING=ANALYSIS

/VARIABLES=n

/CRITERIA=CI(.95).

* I wonder if a paired t-test would be more appropriate.

* Restructure from LONG to WIDE so that paired t-test can be done.

SORT CASES BY Month Year.

CASESTOVARS

/ID=Month

/INDEX=Year

/GROUPBY=INDEX

/SEPARATOR="".

T-TEST PAIRS=n19 WITH n20 (PAIRED)

/CRITERIA=CI(.9500)

/MISSING=ANALYSIS.

Bruce Weaver , it was the paired, or one sample, t-test that I assumed and performed above, but the corr is negative. Whichever way, I think the power is likely too low. The result was t(4) = -0.40. The distribution of the variables (or their difference) seems, well, of secondary importances. The five numbers for the paired t-test are: 28 -8 3 -8 -35