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Sample proportion
 phat = y/n
 y = success
 n = number of trials

Wilsonadjusted proportion
 ptilda = (y+2)/(n+4)
 to keep ptilda from 0 or 1

SE for phat
√(phat (1phat)/n)

CI for phat (95%)
 phat +/ z_{.025}√(phat(1phat)/n) =
 phat +/ 1.96√(phat(1phat)/n)
 set upper limit to 1 and lower limit to 0 if surpass
 unstable, sometimes over coverage, sometimes less (not always 95%)

Wilson adjusted 95% CI
 ptilda +/ 1.96√(ptilda(1ptilda)/(n+4))
 set upper limit to 1 and lower limit to 0
 Gives better coverage (closer to 95%)

One sided confidence interval
 (∞,ptilda + 1.65 * SE_{ptilda})
 (ptilda  1.65 * SE_{ptilda}, ∞)
 Still between 0 or 1

Wilson SE
√(ptilda(1  ptilda)/ (n + 4))

Χ^{2}^{ }for more than 2
 (Observed  Expected)^{2} / Expected + all values
 Use df
 All expected have to be greater than 5
 Can just say if they are different than expected
 Observedexpect^{2} will always be the same so

X^{2} for 2
 directional
 could use binomial
 H_{0}: p = .75
 H_{A}: p ≠ .75
 check with table
 For one sided, ts has to be >/< than 2*alpha ts AND on the right side of expected

Test for independence w/ contingency_{} tables
 p_{1} = (AB)
 p_{2} = (AC)
 H_{0}: p_{1} = p_{2}
 H_{A}: p_{1} = p_{2} (p_{1} >< p_{2})

Expected values in 2x2 tables
 row total * column total / Grand total
 Make sure each is at least 5

df in 2x2 tables
(# rows  1) * (# columns  1)

Directional test with X^{2} and 2x2
 X^{2} > X_{table}for 2alpha
 and
 Alternate hypothesis was satisfied
 Nondirectional don't double

Interpretation for X^{2}
 association not causal
 maybe causal in controlled study
 if one H_{0} is rejected, differently defined p will be also be rejected from same table

What is significance level
the likelyhood of making a type I error

CI for ptilda
 ptilda_{1}  ptilda_{2} +/ Z_{Apha/2} * SE_{p1p2}
 If it contains 0 no differences

SE for ptilda
 To keep it away from 0

Assumptions for ANOVA
 each population is normally distributed
 samples are independent
 samples are random

For several categories why not pairwise ttest?
 Because chance of committing type I error is large for whole test
 alpha for each pair
 1 (1alpha)^{# colums}
 Problem of multiple comparisons
 if lower alpha get higher type II

