Math 151 , Day 35, Wednesday, November 19, 2008 ...Hit reload .

Your shoebox results so far:  Write your samples, and their xbars (one on each pad--yellow or white), say whether you think the mean of that box is > 20 (Y/N).

Exam 4 this Friday, next class .  Covers  Ch. 10 (continuous only, Ch. 11 up to p. 286 only, Ch. 14 all, Ch. 16 thru p. 391. (i.e. thru Day 34 HW).    Sample Exam  (Handed out Fri. Outside my door..)  Solutions .
   Sign up Today for early (>9:30) start:  Confirm with me any other time to take it.

Homework questions?  sample size:      Day 34

Exam Review (big ideas only!)
Probability:  Prob. of event = proportion of times it would happen in a long sequence of repetitions of the experiment.
   Discrete and Continuous Random Variables
   Probability that a single individual chosen at random has some characteristic = proportion in population with that characteristic.
Take a Sample    from a        Population (hope sample is like population)
          Statistic    estimates      Parameter   (summary numbers calculated from above)
            xbar       estimates             µ
Sample means Xbar:  all possible xbars from all possible SRS's (of size n) from the population. 
                     Population has mean µ, s.d. sigma.  X, sample of size 1 , has distribution of population.
   Behavior:  Let n get larger.  Xbar(s) will get closer and closer to µ.  -->LLN (Law of large numbers)
     Fixed n.  Xbars distribution: "Sampling distribution of the mean"
                      Mean = µ  , S.d. is sigma ÷ (square root of n)
                      If X is Normal, so is Xbar.
                      In any case, if n is "large", Xbar is approx. Normal  -->Central Limit Theorem
Estimating unknown  µ using an xbar:  (estimating a parameter using a statistic)
     CI (confidence interval): level C,  margin of error m.
         No particular CI is guaranteed to successfully capture µ; But the Method will produce a "true" result C% of the time.
      For  µ:   (SRS,  pop. Normal, sigma known)
              xbar + m,   m = z* (sigma ÷ (square root of n)),    +z*  cut off central C% in standard normal dist.
        Plan ahead:  for desired C and m, known sigma, find n needed to achieve those.

In practice:  Need SRS or reasonable facsimile!!!  Large n allow substituting s from data for unknown parameter sigma.  Large n allows using CI formula even if pop. not very normal (CLTh).

Links: 
Normal and Xbar, compared
Sampling dist; Central limit th.   Rice U. Applet 
TableA(Normal),    TableC(t distribution)
ConfidenceInterval.xls        

Questions?    Sample exam Solutions . 

"Statistics means never having to say you're certain."
Confidence interval Estimation made our best guess at an unknown population mean.
Testing will investigate a claim made that the unknown mean is actually a particular value.
~~~~~~~~~~~~~~~~
 
Ch. 15: "Significance tests use an elaborate vocabulary, but the basic idea is simple: an outcome that would "rarely" happen if a claim were true--is good evidence that the claim is NOT true." (p.363 top)
Suppose someone claims that the average height of Wells women over the years is 70" (5'10").  I take samples (151 classes) every year.  This year my sample has mean 65.67" (n = 20ish). Standard deviation for heights of women in population is supposed to be about 2.5" , so s.d. for means from samples of 20 is about 2.5/4.48= 0.56. IF the real mean is 70", my sample is astonishingly unusual (65.67-70)/0.56= -4.33 /0.56 = -7.73, 7.73 s.d's below the mean.  Conclude the claim is Not true.

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Extended Standard Normal Table--"Normal Tails" (also from Weblinks page, )
  z         P(Z < z)                         P(Z > z)   = same in scientific notation: E-03 = 10^-3
3.00   .9986501019683700    .0013498980316301    1.35E-03
4.00   .9999683287581670    .0000316712418331    3.17E-05
5.00   .9999997133484280    .0000002866515718    2.87E-07
6.00   .9999999990134120    .0000000009865877    9.87E-10
7.00   .9999999999987200    .0000000000012799    1.28E-12
8.00   .9999999999999990    .0000000000000007    6.66E-16  Below this, machine can't compute. If your assumptions lead you to a(n almost) impossible z value, question your assumptions!
(The basis of significance/hypothesis testing)

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Need machinery to analyze less "obvious" results--build in effect of standard deviation (if s.d. were 10" would my sample still be inconsistent with the claim?) and sample size (if n were only 4 would that change my result?) .
Do 15.2 p. 365 Are older students like traditionals, or higher (on average) on this measure?:  Normal, s.d. = 30.  Claim:  pop. mean = 115.  n = 25.   IF mean is really 115, Xbars are N(115, 6).   Sketch!
      xbar = 118.6 .  118.6-115=3.6.  This is 3.6/6 = 0.6 s.d.'s above the mean, a pretty typical kind of value.
      xbar = 125.8    125.8-115=10.8.   This is 10.8/6 = 1.8 s.d's above the mean, high enough to be pretty unusual (how unusual?) if the mean is really 115.
      xbar =  139     139-115=24. This is  24/6 =  4 s.d.'s above the mean, unreasonably high if the mean is really 115.
   So 125.8 or (more so!) 139 would be evidence that the mean for this group (older students) is NOT 115, is in fact higher.

Got to here Wednesday. Shoeboxes are outlined for HW above.
Shoeboxes (white and yellow slips): Take a sample of size 4 from each,  record, return numbers.
   I claim the mean value  for both shoeboxes is µ = 20.  Am I telling you the truth?  I can't remember for sure.  I do know that the distribution in the box is normal,  standard deviation is 4.
I do remember that if  µ is not 20, then it is greater than 20. µ > 20.
Take a sample of size 4, find xbar.   Once for each shoebox! (should have found xbar already)
How far from 20 is it?  far enough that I believe the mean is not 20??

<>Measure your xbar's distance from 20  in standard deviations of Xbar's. (That is, find z for xbar, assuming µ = 20. Note s.d. for sampling dist of xbar is 2 (why?) ).   Example:  If I got an xbar = 24,  z = (24-20)/2 = 2 s.d.'s above mean. 
Is this a far-out value of z? Look in the normal table TableA to see how much probability is in the tail to the right of it--gives a measure of far-out-ness independent of distribution ("P-value").   Prob. above 2 is about half of 5%, or .025, more exactly (1-.9772) = .0228.   IF the mean is really 20, I would see an xbar as high as 24 (or higher) about 2 to 3 in a hundred times.  So xbar = 24 is pretty strong evidence that real mean isn't 20.

The game:
Before taking data, define
H₀: "Null hypothesis" A claim or statement about the population we would like to show is NOT true.
   Stated usually as:  A parameter = a particular value. 
       H₀: µ =1000 hrs.  ("Average lightbulb life".)     H₀: µ =20 (shoebox mean=20)
H_a: "Alternative hypothesis" A claim or statement about the population we are trying to find evidence FOR.
    Stated usually as: The parameter  is >, or <, (one-tail tests) --
                       or NOT = the particular value. (two-tail)
    H_a:   µ  > 1000 hrs. (Suppose we have a New process that makes them burn longer. We hope.)   
    Other possible alternatives: H_a:   µ  < 1000 hrs.  (Want evidence that Mfr.'s claim is inflated)
             (two-sided=two-tail) H_a:   µ  Not = 1000 hrs.  (Want evidence that Assembly line process is"off")
    H_a: µ >20 (shoebox mean >20)

   Some authorities say you should always do two-sided tests.  Others say:  If you have a hope or suspicion; are only interested in one direction, then do it that way.  What's NOT OK is to look at your data and then decide your alternative hypothesis.

Take data.  Calculate test statistic, usually based on one that estimates the parameter in the hypotheses.  For µ, test statistic is the z-score of xbar. A big z-score number means that xbar is far from µ.
    Is it an unlikely result if  H₀ is true?  Then that is evidence against H₀.

Measuring the strength of the evidence against H₀ (a common measuring stick for all distributions and parameters):
P-value of a test:  The probability, computed assuming that H₀ is true, that the observed outcome would take a value as extreme or more extreme than what we actually observed (if we could repeat taking-data again).  p. 368. 
    The smaller the P-value, the stronger the data's evidence against H₀ ( for H_a).

For a test of µ  , using xbar (sigma known), the P-value is
--the area of the tail beyond the observed xbar, in the direction of H_a (one tail)
(--or twice that area (two-tail).)
We usually calculate it by standardizing the observed xbar (assuming H₀ true) and looking in the normal table. (p. 369 on)
H₀: µ =20     H_a:   µ  > 20      How far from 20 is your xbar? Find z for xbar.   For xbar = 24, z = 2
Is this a far-out value of z? What is the probability of being farther out, i.e. being in the tail beyond this z?  That's the P-value.  P = .0228    Table A

<>Applet:  P-value of a test of significance automates this.  (Uses "raw" scale of xbars, rather than z-scores).  Use as check, guide. 
Applet How to: At top, put in H₀  value, choose direction of H_a,  put in sample size n, and s.d. of the population sigma.  Do Update (Reset sends back to 'opening" values).  The graph and scale axis show distribution of x-bars assuming H₀ is true.  Under the graph, put your "observed" x-bar value in the "I have data..." box and do Show P.  The P-value is the size of the tail, shown in gold.

For HW: draw the picture and label the axes both in "raw" and in z- values.  Show direction(s) of the alternative.  Mark xbar, z, and shade the area which is P-value. (And do the calculations of course.)

Example (one sided):  H₀: µ =1000 hrs.  (Average lightbulb life.)  Competing bulb:   Show it's better.
         H_a:   µ  > 1000 hrs.  (one-sided)
   Sample of size n = 25.  Population sigma = 150 hrs.  S.d. of xbars = 150/5 = 30.
       Get xbar = 1075 hrs.  Are these bulbs better than the "standard?"
       z =  (1075-1000) ÷ (150/5) = -75/30 = 2.5;
       P(Z > 2.5) =  .0062 = P-value.  More than  6 in a thousand and less than 7 in a thousand. More crudely, Less than 1% chance of getting a result this high if we did it again--if the real mean is 1000.

Example (one sided):  H₀: µ =1000 hrs.  (Average lightbulb life.)  Suspect company's cheating:   Show it's worse.
          H_a:   µ  < 1000 hrs.
    Sample of size n = 25.  Population sigma = 150 hrs.  S.d. of xbars = 150/5 = 30.
         Get xbar = 940 hrs.  Are these bulbs worse than claimed?
         z = (940-1000) ÷ (150/5) = -60/30 =  -2.
         P(Z < - 2) =  .0228 = P-value  More than  2% and less than 3% chance of getting a result this low (below1000) if we did it again--if the real mean is 1000.

Will do two-sided in more detail Now.
Example (two sided):     H₀: µ =1000 hrs.  (Average lightbulb life.)(Quality control on assembly line--find if it is "off" either way.)    
           H_a:   µ  Not = 1000 hrs. (two-sided)    
H_a: "Alternative hypothesis" A claim or statement about the population we are trying to find evidence FOR. 
              A value either much bigger than or much smaller than the H₀ value is evidence against H₀ & for H_a.  
   Sample of size n = 25.  Population sigma = 150 hrs.  S.d. of xbars = 150/5 = 30.
   Get xbar = 940 hrs.   Is the quality control "off?"
    z = (940-1000) ÷ (150/5) = -60/30 =  - 2;   P(Z < - 2) = .0228
P-value (two sided): We measure the probability of seeing something (again) as extreme as the observed value (or more so).
So you need to measure the P-value symmetrically both directions from the observed value--so the P value is double what it would be for a one-sided test.  P-value is approximately 5%; more precisely, 2·.0228 = .0456
So for a test of a mean, the P-value for one-sided is half that for two sided, IF the result is in the direction of evidence for the alternative.