(1 pt) The sample size needed to estimate the difference between two population

Question

(1 pt) The sample size needed to estimate the difference between two population proportions to within a margin of error E with a significance level of can be found as follows. In the expression E=zp1(1p1)n1+p2(1p2)n2 we replace both n1 and n2 by n (assuming that both samples have the same size) and replace each of p1, and p2, by 0.5 (because their values are not known). Then we solve for n, and get n=(z)22E2. Finally, increase the value of n to the next largest integer. Use the above formula and a standard normal table to find the size of each sample needed to estimate the difference between the proportions of boys and girls under 10 years old who are afraid of spiders. Assume that we want a 96% confidence level and that the error is smaller than 0.06. n=

Explanation / Answer

proportions p1 = p2= p =0.5

Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)

Z a/2 at 0.04 is = 2.054

Sample Proportion = 0.5

ME = 0.06

n = ( 2.054 / 0.06 )^2 * 0.5*0.5

= 292.98 ~ 293

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