glossary_stat    physics    chemistry   
Estimate a population mean, mods 7,8
"Estimate a population mean". modules 7 and 8  TODO. ref p115 top
Estimating a population mean, mod 18
Estimating a population mean. fm p115 top. module 18
Distribution of Sample Means
"Inference for Means". typical problem, p112, _m17, 4of4, 'Distribution of Sample Means'

problem:
    popuMean= __ ; popuSigma = __ ; curve is normal
    Is it surprising to find a sample, of sampleSize 'n',
        with a sampleMean >= surpriseX

formulation:  compute (P(sampleMean) > surpriseX)

process:
    # since curve is normal, no need that sample size >= 30
    # popuMean will be used to calculate surpriseZ
    sampleSigma = popuSigma/sqrt(sampleSize)
    # convert surpriseX to surpriseZ
    surpriseZ = (surpriseX - popuMean) / sampleSigma
    surpriseP = p4z(surpriseZ) # convert Z into its probability P
Inference for one proportion summary
Inference for one proportion summary. m16
p0 = from some published survey.
Propose that other conditions have changed the related probability
H0 = "null hypothesis", based on the original article and its data
Ha = "alternate hypothesis"
alpha = (significance level). typically 0.05 could be diff

determine 'n' for your sample
    bNormal # n*p0 > 10 etc forms the minimum.

collect sample data; extract pHat (the sample mean)
stdErr = sqrt(p0 * (1-p0) / n)
zScore = (pHat - p0) / stdErr)
pValue = p4z(zScore)  # careful, need both tails if Ha used 'notEqual'
if pValue < alpha, conclude that Ha is correct
if pValue > alpha, no conclusion, data doesn't resolve this issue.
hypothesize
    ##
    # relation = one of 'LT','NE','GT'
    def hypothesize(self,p0,pHat,relation,n,alpha)
        # bNormal(self, n,p0)  normality test.
        # print H0, Ha
        # calc stderr
        stndErr = stdError(self, n,p0) : #
        zValue  = zScore(self, stndErr,p,pHat) #  = (pHat - p0) / stndErr

        write something to return area for zValue. from table. include x's as well as y's

        # get area
        # trim/fix area value
        # apply
30% of smokers
Learn by Doing, p77.8 // claim: 30% of smokers started before age 16

The National Health Survey uses household interviews to describe the health-related habits of U.S. adults. From these interviews they estimate population parameters associated with behaviors such as alcohol consumption, cigarette smoking, and hours of sleep for all U.S. adults.

In the 2005-2007 report, they estimated that 30% of all current smokers started smoking before the age of 16. Imagine that we want to verify this estimate. So we randomly select a sample of 100 smokers and calculate the proportion who started smoking before the age of 16. How much error do we expect in the sample proportions if the 30% is correct for the population overall? Use the applet and a give an error based on 2 standard deviations. (WHY 2 SDs? dont we want 'typical' ?)

Applet: set these values
p = 0.3  // population probability that smokers start before age 16
n = 100  // set sample size to 100, as instructed.
pHat: slider bar  // red line and pHat not needed for this answer

"Sampling Distribution: mean = 0.2999 standard deviation = 0.0458

answer is 0.0458 x 2 = 0.09

We are instructed to use 2 standard deviations for the margin of error. So proportions that vary less than 2 x SD

Skittles: Is 6 out of 20 (30%) a surprise...
Suppose that we purchase a small bag of Skittles. Assume that this size bag always has 20 candies. In this particular bag, 6 are green: 6 out of 20 is 30%. Is this a surprising result?

Use the applet to conduct a simulation. Which option gives the right answer and the best explanation?

set p=0.2, n=20, pHat=0.2 histogram'ish plot shows 0.3 just beyond 1 sd. mean 0.2019, std dev = 0.0906 "so error of 0.10 is not unusual...

Skittles: 100
same skittles problem except the bag now has 100 candies.
n=100  only change needed
now the std dev is 0.04 and an x value of 0.3 (vs mean of 0.2) is
2.5 std devs away from the mean, therefore 'unusual'=='surprising'
surprised if sample 35% when P=40%
Learn by Doing, p77.85   // surprised if exit poll sample 35% when P=40%
'surprised' implies 'unusual' (to me). think that works for Them too.
n=100 (again)
p=0.4
ran the Applet again. mean about .4, std dev is 0.049 or about 0.05
So unusally (low) would be mean - 2 x 0.05 or .4 - .1 = .3 and the
value given is .35 so it's not unusual (surprising).

ALL wrong. A _local_ exit poll is not a random sample. So no such conclusions!