The Easiest solution isn’t always the best solution, even in Math Should we always believe what we are taught in the classroom?

Purpose Statisticians use a well selected sample to estimate an unknown value of a population. The unknown value may be the mean income, or the proportion of defective products, or proportion of “yes” responses. Estimating an unknown population proportion is the topic of interest.

Background/symbols p= Population proportion (unknown) n= # of subjects/ objects randomly selected. X= # of subjects/ objects in the sample with Yes responses. p^ = sample proportion= x/n Traditionally p^ is used as an estimate of p. Is there a better alternative?

We often provide an interval estimate of p, p^ ± error of estimation: Confidence interval A well known interval is 95% confidence. To determine error, we need to understand how p^ value varies from sample to sample.

About p^ p = fixed value of a population, while p^= varies from sample to sample, and thus it has a distribution. What we know is Under certain conditions, p^ is normally distributed with a mean value of p, and standard deviation of √p(1-p)/n A normally distributed value can be changed to a standard normal score,called a z score. A well known result is that about 95% of the z scores fall between -2 and 2.

Lets standardize p^ = # of yes /n, ( p^- mean)/ std dev = z ( standard normal) ( p^- p)/ √p(1-p)/n = z ( p^- p)/ √p(1-p)/n = ± 2 ( p is unknown). Note: We need to solve the above equation for p Easy approach for non math majors : solve for p in numerator p= p^- 2 √p(1-p)/n, p= p^+ 2 √p(1-p)/n, (p^- 2 √p^(1-p^)/n, p^+ 2 √ p^(1-p^) / n) Makes an approximate 95% confidence interval of p. ( Mathematically incorrect)

Lets try again ( p^- p)/ √p(1-p)/n = ± 2 Solve it the right way by squaring both sides, and solving the quadratic equation for p. We get two solutions of p, ( mathematically tedious) Those solutions make the 95% interval of p.This interval is very tedious, and lacks logical explanation. we take the average of those solutions, we get p =( # of yes +2)/ (n+4)= our new estimate =p~

A new interval of p Recall old interval of p: (p^- 2 √p ^ ( 1-p^ )/n, p^+ 2 √ p ^ ( 1-p^ )/ n) An alternative interval of p (p~ -2 √ p~(1- p~)/n, p~+ 2 √ p~(1- p~)/n Recall p~ = ( # of yes +2)/ n+4 p^ = (# of yes)/n

How good is the new interval Simulation results coming soon.