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

2. 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.

3. 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?

4. 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.

5. 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.

6. 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)

7. 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~

8. 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

9. How good is the new interval Simulation results coming soon.