Math 151 , Spring 2005, Day 40 Mon. May 9 Hit reload 

What is the significance to Statistics of the Guinness Stout Bottle ?
Day 40 (Mon. May 9): Reading: D&V Ch. 23 thru p. 462, then 465-9.  You will not need to compute a two-sample t procedure by hand, but you will need to know how to identify the situation, to use SPSS, and understand the results. This is the end of the course material.

Hand in  p. 471, #1, 3 CPMP 6 a,b only Pulse rates 7 Cereal (SPSS: Data is in labs at Math151 D&V\spss data files D&V\dv01_24_07.sav ) 8 Egyptians (SPSS: Data is in labs at Math151 D&V\spss data files D&V\dv01_24_08.sav, BUT it's in the wrong form!  It's in 2 columns as if it were Paired but it's not paired data.  You can highlight the 30 items in one column, copy and paste at the bottom of the other.  Then make a grouping variable to distinguish the two groups.) 17 Job satisfaction (What should you do? (Don't do it...)) 12 a,b (c optional) Memory 11 Hurricanes. Do a back-to-back tally of the two sets. Don't do the test, just think about appropriateness.  The answer in the back was a little misled by the inappropriate boxplot into thinking there are outliers, tho there aren't, there's just "granularity" (small whole numbers here).

Read,   to  discuss

Optional

Homework questions? Day 39
Chapter 24, Comparing two means"Two-sample tests". Chapter 24  Two random samples,  independent of each other, from distinct  populations. (Populations are normally distributed)  p. 454-5
Often--comparing means from an experiment with two treatments (usually control and "treatment").
                /--- Group 1, n₁---- Treatment 1---\
              /                                    \
 Random asst.                                       Compare results
              \                                    /
               \--- Group 2, n₂---- Treatment 2---/
To examine  the difference of the  two means, µ₁ - µ₂:
We need fairly normal populations; no extreme outliers.  Back to back stemplots are good; boxplots will do.
(Above 40, Central Limit Th. helps:  15 to 40, a little skewness ok.  p. 455)
We use the difference of the two y-bars,  diff =  ybar₁ - ybar₂ = .
We need the Standard Error of the difference  ybar₁ - ybar₂ , and then we can proceed as before, more or less.
The Standard Error is calculated like the hypotenuse of a right triangle (Pythagorean Theorem),  from the individual standard errors.
 SE(diff) = SE( ybar₁ - ybar₂ )= sqrt[SE(ybar₁)² + SE(ybar₂)² ] 
P. 453 has another way of writing the same thing:

This almost fits the  t-model. Degrees of freedom are weird.(p. 454)

(For doing by hand, if you must: df = smaller of (n₁- 1) and (n₂- 1).)
Will give a "conservative" result--slightly wider C.I., slightly less significance, than a "sharper" value.  If your results hinge on the difference between this result and the computer result, they're too close for comfort anyway.

From a computer:  df = complicated formula on p. 494 bottom.  Produces non-integer degrees of freedom.  Very good approximation to the exact distribution, if both sample sizes are at least 5.   Always between "smaller of (n₁- 1) and (n₂- 1)" and [(n₁- 1) + (n₂- 1)].   Unsuitable for doing by hand.

Once we have (ybar₁ - ybar₂) , SE(diff) ,  and the df, our formulas pattern on the earlier ones. Optional Example by hand
CI :  estimate + t* ^. SE(estimate)
    CI for µ₁ - µ₂, difference of means,  is 
Test:  H₀: µ₁ - µ₂ = 0 same as µ₁ = µ₂ , "no difference" "always"
        H_a: µ₁ - µ₂ > 0 same as µ₁ > µ₂Be careful with these, that you know which direction you want.
    or H_a: µ₁ - µ₂ < 0 same as µ₁ < µ₂ Often we label our variables "1" and "2" so that we expect µ₁ > µ₂
    or H_a: µ₁ - µ₂ 0 same as µ₁  µ₂  (not equal)
        Calculate  find P-value

SPSS will do our computations when we are given raw data.  See handout.   Datasets
Analyze>Compare means> Independent-samples t. We use the Equal-variances-not-assumed line of the results.
  Does same example as Optional Example by hand: twosampexample.htm

Optional: Tukey's Quick test (p. 465) for two independent samples.  Doesn't need Normal!!
 (Not well known;  but worth knowing!)  Put data in order.  One group must have the highest value and the other group the lowest to use this.  How much do they not overlap?
Count the number of items in the "Higher" set that are bigger than all the items in the "Lower"set. Plus all the items in the "Lower" set that are smaller than all the items in the "Higher" set. (A tie at the edge = 1/2.)
"7, 10, 13"  7 or more? (2-sided) Sig. at .05.  10 or more? (2-sided)Sig. at .01.  13 or more? (2-sided)Sig. at .001.