Confused about the sample size in large sample confidential interval and population proportion intertal

My professor said when sample size is large, it will has the following formula, but how big is the sample can be considered to be a large sample and use the following formula?

I did some questions in my book,and it seems like most of the question don't use the formula above. For example in a survey of 2003 American adults, 25% said they believed in astrology. a. Calculate and interpret a confidence interval at the 99% confidence level for the proportion of all adult Americans who believe in astrology.

The solution shows that this question actually uses

I am really confused, I think 2003 is pretty big, so it can use the first formula....tomorrow is the final, and I still don't know which one to use... Really hope someone can answer my question, I will be really thankful

So... I drafted this R code for you to play around:

p <- seq(0.01, 0.99, by=0.01)

q <- 1-p

z <- 1.96

n <- 2000

lower.short <- p - z * sqrt(p*q/n)
upper.short <- p + z * sqrt(p*q/n)

lower.long <- (p + z^2/(2*n))/(1 + z^2/n) - z*(sqrt(p*q/n + z^2/(4*(n^2))))/(1+z^2/n)
upper.long <- (p + z^2/(2*n))/(1 + z^2/n) + z*(sqrt(p*q/n + z^2/(4*(n^2))))/(1+z^2/n)

plot(lower.short, lower.long, type="l")
abline(0,1,col="red",lwd=3)


This is the agreement of the lower bound using the approximation (lower.short) and the original formula (lower.long) with a sample size of only 200:

The red line indicates perfect agreement. You'll see most deviation happens at the extreme proportions, but otherwise tolerable. I would suggest if the sample size is more than 200, it's quite safe to use the approximation.

The textbook may want to insist on using the original formula, however, doing so doesn't mean it's wrong to use the approximated one.