# Question about z-tests comparing two proportions

i saw a question online and i am confused with its answer. I will be grateful if anyone can explain the theory clearly to me.

Question: In a monthly survey, last month got 55% approval from 1000 ppl and this month gets 58% from 1500 ppl. The question is, Is there any significant change in this 2 month?

This question should use z-test between 2 proportion to answer, right? my problem is that, the answer provided in the website use "SE of average" instead of "SE of sum" to calculate the SE of difference.

I thought I was clear the difference between SE of average and SE of sum after I check the definition, but I get confuse again now. Could you help me?

• Probably you could edit your question and indicate test statistic and the formula you are thinking about. Commented Oct 11, 2021 at 14:16
• Hint: are proportions, like 55% and 58%, sums or averages?
– whuber
Commented Oct 11, 2021 at 14:23

There are several methods of comparing two binomial proportions using normal approximation. In R, the procedure proptest will perform one of them. For your data, its P-value exceeds 5%, so there is no statistically significant difference at the 5% level.

.55*1000; .58*1500
[1] 550
[1] 870

prop.test(c(550,870), c(1000,1500), cor=F)

2-sample test for equality of proportions
without continuity correction

data:  c(550, 870) out of c(1000, 1500)
X-squared = 2.2007, df = 1, p-value = 0.1379
alternative hypothesis: two.sided
95 percent confidence interval:
-0.069681406  0.009681406
sample estimates:
prop 1 prop 2
0.55   0.58


Note: I declined use of a continuity correction (with parameter cor=F) on account of your sample sizes in the thousands.