Math 151 , Spring 2005, Day 32 Wednesday, April 20 Hit reloadAfter class, really

Homework questions? Day 31
 Sample size for desired ME and C Day 30
Level C confidence interval estimate of population proportion p:
 "One -proportion  z-interval"


Start here Wednesday:
  Why this ME "works". Day 30

Lots of machinery and vocabulary:
NULL Hypothesis H_o : (Straw man we collect evidence against. Status quo.)
Assume H_o is true.  Look at evidence (data).  Is it inconsistent with H_o ? Then Reject H_o .
  (How inconsistent with H_o is the data?  a little, somewhat, very?  how do we measure it?  Turn into numbers---)

H_o : a specific model for the population, with a specific parameter value.
example (suppose I hadn't told you...):  Green shoebox is full of 0's and1's.  I tell you Equal numbers.
H_o : p = .5 (proportion of 1's is 50%)  p_o for a general label.
    Is your  sample (n = 30) far enough away from .5 to say that I'm lying? Suppose you believe I undersupplied 1's:

H_A : p < .5  (one-sided alternative hypothesis:  What you hope /fear /would like to prove)
How do we measure "far enough away?"
IF H_o  is true: how far out (weird) is your p-hat?
IF  p = .5, how far from  the "real" p is your p-hat?  p_o = .5
                 Distribution of p-hats is approx.  N(), N( .5, sqrt(.5 ·.5/30)),  N( .5, .091)  (Usual assumptions.)
    Suppose you got 12 1's.  p-hat = .4. IF  p = .5, p-hat = .4 has a z-score of -.1/.091 = - 1.095 ~ -1.10 Sketch the Normal.
        If you know your z-scores, this is meaningful.  A more universal measure is the
P-value:  The probability, assuming H_o  is true, of observing the result we have (or one more extreme)--if we could do the experiment again...  Strength of evidence against H_o (thus for H_A)
   In our example: The probability of getting a p-hat of .4 or below, IF p = .5. Sketch on the curve.
                   The "tail" below z = -1.10.  From normal table, .1357 ~ 13.6%.  So
    P-value = .136.  Not so unusual; happens more than 1 in ten times (13-14 in a hundred).  Suggestive but not "significant" by most people's standards.
Start here Friday
Example:  U.S. Consumer Product Safety Comm. says 90% of American homes have smoke detector(s). Fire department 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)
--Hypotheses:  H_o : p = .9  (unchanged after campaign),    H_A : p > .9  (raised after campaign)
--Model:  Want to do One-proportion z-test.  Independence? Yes, random sample. n = 400. Population (city) > 4000 homes (10% rule).  .9x400 = 360, .1x400 = 40 so success/failure rule met.  Sample proportion p-hat modeled by Normal OK.
--Mechanics:    n = 400, successes x = 376.  p-hat = .940.   SD(p-hat) = sqrt(.9 ·.1/400)= .015 since we're assuming H_o true.  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 very low:  Strong evidence that the city proportion has risen.  (Reject H_o)
Detail:  IF H_o : p = .9  is true (the city proportion is really unchanged) THEN the chances of us seeing a proportion as high as .94 in another sample of 400 is only about 4 in a thousand.  (We prefer to believe that the alternative hypothesis is true than to believe that we got such a "lucky" sample.)

Suppose (as here) we do Reject H_o in favor of H_A.  We are pretty sure the true p is different from p_o but by how much?
    (The "size of the effect" = p_true - p_o.)  Still need an estimate of the true p. Use a CI to estimate p_true.
--Estimate:  How much has the proportion  changed?  Confidence Interval.  Now use SE(p-hat)= sqrt(.94 ·.06/400) = .0119 since were not assuming the null hypothesis is true.  95% CI 's ME = 1.96 · .0119 = 0.023324~ .023.
95% CI:  .94 + .023,  I'm 95% confident the real proportion is between 91.7% to 96.3%.

Statistical inference in a nutshell:
Am I surprised (If H_ois true)? (Do I reject null)
    How surprised? (give P-value)
  What would not surprise me?  (confidence interval)

More about the
Alternative hypothesis:  Null hypothesis is often a particular parameter value.  Alternative is something "different."
   Why? are you doing a test.   Back to shoebox:
       H_A : p < .5  You have reason to believe I skimped on the 1's.         One-sided
OR  H_A : p > .5   You have reason to believe I put in more 1's than 0's.  One-sided
OR  H_A : p not = .5  You believe the 0's and 1's are not equal, but don't know which way. Two-sided.

P-value concept needs refining:
    For One-sided alternatives, P-value is the single "tail" beyond our observed statistic,  in the direction of the alternative hypothesis.
   For a Two-sided alternative, P-value is "double the tail" beyond our observed statistic, because we could be "as or more extreme" in either direction!   (Measuring how weird our observation is, if  H_o is the case.)

Example:  Shoebox, 12/30 1's.  We got z = - 1.10.  If your alternative is H_A : p not = .5,
   there is probability .136 below z = - 1.10, and probability .136 above z = + 1.10,
   so the P-value = 2· .136 = .272.  1 in 4?   Not unusual at all.  Can't reject H_o .

 Example. Look at Therapeutic touch. Why? are you testing.
      Activstats 20-2, activity 2, H_A : p not = .5  The detection is different from chance (better, or worse)
      Step-by-step p. 390,  Ch. 21    H_A : p > .5  Can detect "energy field" better than chance.
    n = 150, observed successes 70.  p-hat = .467.   (notice, less able to detect than just chance)
        z-score is - .825 if you round SD to .04(text), -.809 if you round to .0408 (AS).  (so small differences in normal table.)
   H_A : p not = .5  There is .2090 to the left of -.81; approximately .21 .  Two-sided P value is .21 + .21 = .42.  No evidence for ability different from chance.
   H_A : p > .5   The statistic is actually on the wrong side  of the middle.  So the P-value is still the "up" side; the probability of seeing a p-hat greater than that observed.  So the P-value is 1- left tail value.  Using the above, P = 1-.21 = .79.  The chance of seeing a p-hat greater than what we observed (if they can't detect better than chance) is more than 3 out of 4.   Unsurprising result, no evidence that they can detect energy field.

One or two sided? p. 390   Statististicians differ philosophically.  Some much prefer 2 sided all the time. ("How can we really know which way things are changed/different?"  If it's clear you're expecting to see / looking to prove a particular direction, many (most? Me.) use one-sided. (D differs from V, I think.)  Say up front.  DON'T decide on one-sided/which side after you've seen the data.  That's cheating, statistically.
Skip for now "A Better Confidence Inteval..." p. 383.  We do a lot of approximating to get our CI's. This turns out to give trouble, especially for p's closeish to 0 or 1.  This is a nice "fix;" relatively new (1998).  Adds nothing to conceptual understanding.