Class Notes, Chapter 10, 10.1 - 10.3

One- and Two-Sample Tests of Hypotheses

Reading: 10.1 - 10.3

Estimation

We started with:
Point estimates and estimators
Methods for finding estimators
Properties of good estimators
Continued with:
Interval estimates
Confidence intervals
CI picture: move Parent distr. back and forth, to some criterion
Needed to find:
Critical values of test statistics like t or chi-squared
These are what's in the tables
Look into the tables based on:
alpha level
confidence coefficient = 1 - alpha

Confidence intervals for:
I.      Mean of a normal distribution with "known" variance
II.     Mean of a normal distribution with unknown variance
III.    Difference of means from normal distributions, variances "known"
IV.    Difference of means from normal distributions, variances unknown, and equal
V.     Difference of means, normal distributions, variances unknown, and unequal: *** Use equal n ***
VI.    Paired observations: mean difference: *** Very different model ***
VII.  Estimating a proportion: Use the normal approximation and Z * Obviously, large sample *
VIII. Difference between two proportions: Use the normal approximation and Z
IX.    Estimating the variance from a normal distribution: * Assumption of normality more important *
X.     Estimating the ratio of variances from two normal distributions


Hypothesis testing - finally, Decision Theory is put into the mix

One- and Two-Sample Tests of Hypotheses

Statistics theory
Probability theory
Expectation
Estimation
Decision theory

Koosh ball demo revisited

You have a bag with magenta and yellow koosh balls in it, but you don't know the actual numbers, except that there are ten total.

Your friend Yama thinks it would make sense that the bag holds half magenta and half yellow koosh balls. Gustavo heard a rumor that there's only one yellow, and the rest magenta. They want you to help them decide if either of these hypotheses is reasonable.
Suppose your sampling scheme is to take 5 koosh balls, with replacement, and count the number of magentas. (n = 5, iid RVs)
Your sample turns out to be 3 magentas and 2 yellows.
If there are 5 magenta and 5 yellow in the bag, there is a 31% chance of getting 3 and 2. What do you think? If there are 9 magenta and 1 yellow, there is a 7% chance of getting 3 and 2. What do you think?
The bag (bag = parent distribution) has 10 koosh balls in it, so it could have held any one of the following (M = magenta, Y = yellow):
M = 0, Y = 10
M = 1, Y = 9
M = 2, Y = 8
M = 3, Y = 7
M = 4, Y = 6
M = 5, Y = 5
M = 6, Y = 4
M = 7, Y = 3
M = 8, Y = 2
M = 9, Y = 1
M = 10, Y = 0
Point estimation
Maximum likelihood estimation
Calculate the likelihood (probability) of this sample, based on each parent distribution (bag contents), and plot:
M = 0, Y = 10 0
M = 1, Y = 9 .0082
M = 2, Y = 8 .0512
M = 3, Y = 7 .1323
M = 4, Y = 6 .2304
M = 5, Y = 5 .3125
M = 6, Y = 4 .3456
M = 7, Y = 3 .3087
M = 8, Y = 2 .2048
M = 9, Y = 1 .0729
M = 10, Y = 0 0
Maximum likelihood estimate is 6 Magenta, 4 Yellow for the population values.

Interval estimation
Calculate the probability of this sample (3 Magenta) or a more extreme one (4 or 5 Magentas), for each of the possible bag contents (parent distributions). This involves three calculations, summed, for each bag.

Example: Suppose the bag (population) contains 2 Magenta, 8 Yellow. The sampling distribution has as possible outcomes 0, 1, 2, 3, 4 or 5 Magenta, with probabilities (this is a binomial distribution) plotted:

 
The probability of 3, 4 or 5 Magenta would be about .05 + a little more. Somewhat borderline.

The probability of r = 3, 4 or 5 given each bag:

M = 10, Y = 0           Can't produce this result.
M = 9, Y = 1 .9914 - could have produced this or a more extreme result
M = 8, Y = 2 .9421 - ditto
M = 7, Y = 3 .8369 - ditto
M = 6, Y = 4 .6826 - ditto
M = 5, Y = 5 .5000 - ditto
M = 5, Y = 5 .5000 - ditto
M = 4, Y = 6 .3174 - ditto
M = 3, Y = 7 .1631 - ditto
M = 2, Y = 8 .0579 - borderline
M = 1, Y = 9 .0086 - highly unlikely to have produced this or a more extreme result
M = 0, Y = 10           Can't produce this result.
What do you think of Yama's hypothesis? What do you think of Gustavo's hypothesis?


Statistical hypothesis testing

Statistical hypothesis
Set up population distribution
Make some necessary assumptions about the population distribution
If necessary, use sample values for estimation of parent distribution parameters
Make a hypothesis, a statement about population values
Set up sampling distribution for the test statistic
Use sampling scheme you have chosen
Use parent distribution that has been set up
Generate expectations for test statistic
Comes from the sampling distribution of that test statistic
Use sample data to get a value for the test statistic
Compare test statistic value from data to expectation from sampling distribution
Accept (retain, fail to reject) or reject
based on probabilities
too extreme to be due to chance
could have happened by chance
Null hypothesis, alternative hypothesis
Decision criterion
Test statistic
sampling distribution of the test statistic
decision criterion in the sampling distribution of the test statistic
Critical value(s), critical region
test statistic values
probability values (p-values) in sampling dist. of test statistic
alpha - the level of significance
effects on beta of changing alpha
effects on power of changing sample size
one- and two-tailed
p-value approach

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