Find the size of each of two samples (assume that they are of equal size) needed to estimate the difference between the proportions of boys and girls under 10 years old who are afraid of spiders. Assume the "worst case" scenario for the value of both sample proportions. We want a 99% confidence level and for the error to be smaller than 0.02

Using the z-distribution, it is found that samples of size 4155 are needed.

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which z is the z-score that has a p-value of .

The margin of error is of:

99% confidence level, hence, z is the value of Z that has a p-value of , so .

The "worst case" scenario is an estimate of .

The sample size is n for which M = 0.02, hence:

Rounding up, samples of size 4155 are needed.

A similar problem is given at link

The sample size n = 4225

Step-by-step explanation:

We will use maximum error formula =

but we will find sample size "n"

Squaring on both sides , we get

Given 99% confidence interval (z value) = 2.56

given maximum error = 0.02

n≤    ( here S.D = p(1-p) ≤ 1/2

on simplification , we get n = 4225

Conclusion:

The sample size of two samples is  n = 4225

verification:-

We will use maximum error formula = =  = 0.0196

substitute all values and simplify we get maximum error is 0.02

its c or d   take a lucky guess my dude