MATH 251, Probability and Statistics I, Fall 2007, Wed. Nov. 28, Day 39after class

Ch. 8, Inference for proportions.
Sample survey:  statistic is proportion who say "yes"out of n.  Let q = 1-p.

Our procedures will parallel those for means, using the fact that for "large" n, the sampling distribution of the sample proportion   is approximately Normal.

Level C confidence interval for population proportion p:  (Large n, approximate)

How large n?  For usual CI's (90%, 95%, 99%): Number of successes and number of failures both > 15.  (So n > 30, and p is not too close to the end of the range.)

What if your n is smaller?  (but n >10)
Modification:  "plus four estimate" p. 539:  Add 4 "fake" observations, so our new sample size is n + 4.  Call 2 of these observations "Successes" and 2 "Failures", and proceed as before!  Moore uses  (p-wiggle) for this modified estimate.
Example:   n = 16, with 5 "successes". =5/16 =.3125; but we'll use
= (5+2)/(16+4) = 7/20 = .35
SE_p-wiggle = sqrt[(.35· .65)/20] = .107.  So an approximate  95% CI for the population proportion p is .35 + 1.96·.107,   or  .35 + .209; roughly .14 to .56.  Wide because the sample size is small.

Test: (Large n) p. 540-3
How large n?  Expected number of successes, and failures >10:  np_o >10, nq_o>10.  Also:  Need that we don't "use up" too much of the population; population should be at least 10n.  (Pp.336-7, using Binomial for sample proportion, said 20n. This is looser.)
  H₀ : p = p_o  ; usual possibilities for H_a.
Base the test on the fact that IF H₀ is true, then  is approximately N().  Standardize  , and find the P-value from the z-normal table.  CI gives more information than test; also, it's rare to have a particular p to test against.  (Some appropriate situations:  Is a coin fair? Is Coke preferred to Pepsi? (Cf. sign test.) Is present proportion of Caesarian-section births (from a sample) different from proportion 20 years ago (presumably have "all")?)

Example:  U.S. Consumer Product Safety Comm. says 90% of American homes have smoke detector(s). Fire department in our city  runs a big publicity campaign to raise awareness and use.  Have they raised the level in this city?  Data:  Building inspectors visit 400 (random) homes, find 376 have detectors.   p is the proportion of detectors in the population (the whole city)
H_o : p = .9  (unchanged after campaign)
H_a : p > .9  (raised after campaign)
Assumptions: random sample. n = 400. Population (city) > 4000 homes (10n rule).  np_o= 400·.9 = 360, nq_o= 400·.1400 = 40 so success/failure rule met.  Sample proportion p-hat modeled by Normal OK.
Computations:  n = 400, successes x = 376.  = .940.   SD( ) = sqrt(.9 ·.1/400)= .015 since we're assuming H_otrue.  z=(.940-.9)/.015 = .04/.015 = 2.67.   One-sided alternative, evidence for it is to the right.  So P-value is proportion in the tail above z = 2.67; P=P(z> 2.67) = 1 - .9962 = .0038 ~ 0.4%
--Conclusions:  P-value is quite low, <.01 (but not <.001):  Strong evidence that the city proportion has risen.  (Reject H_o)
Problems?  We don't know that our city conformed to the American proportion before the campaign.  Would be better to have taken a sample before the campaign and another after, and compared (Sec. 8.2).
..
Plan ahead for desired CI Margin of Error (pp. 545-6)  Decide on desired Margin of error m,  and C (thus z*).  Guesstimate a p*. (p*=1/2 requires the largest sample size--safest.  Remember  you showed that p(1-p) had its maximum at p = .5).
Solve margin-of-error equation for n.   (Some results,  ex.8.6 p. 546.  If you wonder why so many polls feature a 3% margin of error and a sample size of 1060 or so, this tells you why. )

Next:  Sec. 8.2, Comparing 2 proportions (from independent samples)