Math 151 , Spring 2005, Day 31 Monday, April 18 Hit reloadAfter class

Day 31 (Friday, April 15): Reading: Continue Ch. 19, Confidence Intervals for Proportions. I was Wrong! Activstats Ch. 19 does a good job with confidence intervals for proportions.  Please read ahead, Ch. 20+21 thru p. 302 (Activstats is good here too.).
Please respond to my email about the textbook choice!
Do-overs for Exam 2 are due next class!
Hand in (All D&V)
Ch. 19p. 386  (including parts done ahead)
5, 6 Conclusions. Do these with the "Don't misstate..." section, pp. 361-2.
9 Cars
A.   Use the Normal table to find z* for a 99% CI (This is the z* for which 99% of the area is between -z* and +z*).  Find z* for a 90% CI. (Text p.358 shows that z* = 1.645 to 3 decimal places.  Normal table only gives 2 places.)  Check your results by comparing with the corresponding results in Table T p. A-53.
11 Ghosts  (Do rest.)
13 Teenage drivers  (Do if you didn't)
21 Rickets
3, 4 Conditions
16 Local news

ME, C, n pp. 356-7, 361-3.  Problems p. 368
7, 8  Relationships
23 Deer ticks
25 Graduation  The answers in the back use the 25% as the p to plug in.  Redo part a (only) using 50% as the p (what you would do if you had no idea what p would be.). How many subjects do you "save" by using the 25%?
26 Hiring
28 Hiring again
29  Pilot study
Hand in all (and only) the ABOVE problems.
= = = = = = = = = = 
Chapter 20:  Here are some of the problems I'll assign.  Read them to see what you'll need to know.  1,2,3,5, 6,7, 8, 9, 13, 18 (+more involving a two-sided alternative)
A. Use your greeen shoebox result to do a Two sided test against the null hypothesis p = .5.

Read,
  to 
discuss 
Optional 
I was Wrong! Activstats Ch. 19 does a good job with confidence intervals for proportions.  Also, Please read ahead, Ch. 20+21 thru p. 302.
Please respond to my email about the textbook choice!
Do-overs for Exam 2 are due next class!
You found the 68% and 95% CI's for your sample., Please add your results to our list:  (with your initials)
# of 1's, p-hat, SE(p-hat), p-hat + SE, ME for 95% = 1.96SE, p-hat + 1.96SE
Also draw your 68% CI on the graph circulating    |------o------|

Homework questions?
 Definition and computation of CI:  introduce table T, and Assumptions/conditions. Chapter 19 Day 29
Sample size for desired ME and C; Why this ME "works".  Day 30
Level C confidence interval estimate of population proportion p:
 "One -proportion  z-interval"


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



Tests:  (Chapter 20, for proportions) You have a hypothesis about the world. And some data.
Does the data lend support to the hypothesis, or is the data inconsistent with the hypothesis?
      (Retain / fail to reject the hypothesis)                       (Reject the hypothesis)
Easier to reject a hypotheses than to show that it's true.

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

Ho : 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.
Ho : p = .5 (proportion of 1's is 50%)
    Is your  sample (n = 30) far enough away from .5 to say that I'm lying? Suppose you believe I cheated on the 1's

HA : p < .5  (one-sided alternative hypothesis:  What you hope /fear /would like to prove)
How do we measure "far enough away?"
IF Ho  is true: how far out (weird) is your p-hat?
IF  p = .5, how far from  the real p is your p-hat?
                 Distribution of p-hats is approx.  N(p, SD(p-hat)), 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 Ho  is true, of observing the result we have (or one more extreme)--if we could do the experiment again...
     In our example: The probability of getting a p-hat of .4 or below, IF p = .5Sketch 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.

Alternative hypothesis:  Null hypothesis is often a particular parameter value.  Alternative is something "different."
       p < .5  You have reason to believe I skimped on the 1's.         One-sided
OR  p > .5   You have reason to believe I put in more 1's than 0's.  One-sided
OR  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!
Example:  If your alternative is HA : 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.


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