The objective is to test ESP, for this, a psychic was presented with cards face down and asked to determine if each of the cards was one of four symbols: a star, cross, circle, square.

Be X: number of times the psychic identifies the symbols on the cards correctly is a size n sample.

p the probability that the psychic identified the symbol on the cards correctly

You have to calculate the sample size n to estimate the proportion with a confidence level of 95% and a margin of error of d=0.01

The CI for the population proportion is constructed "sample proportion" ± "margin of error" Symbolically:

p' ± $Z_{1-\alpha /2} * (\sqrt{\frac{p'(1-p')}{n} } )$

Where $d= Z_{1-\alpha /2} * (\sqrt{\frac{p'(1-p')}{n} } )$ is the margin of error. As you can see, the formula contains the sample proportion (it is normally symbolized p-hat, in this explanation I'll continue to symbolize it p'), you have to do the following consideration:

Every time the psychic has to identify a card he can make two choices:

"Success" he identifies the card correctly

"Failure" he does not identify the card correctly

If we assume that each symbol has the same probability of being chosen at random P(star)=P(cross)=P(circle)=P(square)= 1/4= 0.25

Let's say, for example, that the card has the star symbol.

The probability of identifying it correctly will be P(success)= P(star)= 1/4= 0.25

And the probability of not identifying it correctly will be P(failure)= P(cross) + P(circle) + P(square)= 1/4 + 1/4 + 1/4= 3/4= 0.75

So for this experiment, we'll assume the "worst case scenario" and use p'= 1/4 as the estimated probability of the psychic identifying the symbol on the card correctly.

The value of Z will be $Z_{1-\alpha /2}= Z_{0.975}= 1.96$