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.
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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.