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.