Where we’ve been & where we’re going We can use data to address following questions: 1.Question:Is a mean = some number? Large sample z-test and CI Small sample t-test and CI 2.Question:Is a proportion = some %? Proportion version of large sample z-test and CI

Where we’ve been & where we’re going 3.Question:Is a diff between two means = some # Independent samples: large sample z test and CI small sample t test and CI paired samples: small sample paired t test and CI 4.Question:Is diff between 2 proportions = some % Proportion version of large sample z test and CI

Topics to be covered in remaining 8 classes (including today) Analysis of Variance and Linear Regression (Chapters 11, 12 and 13) “response =  0 +  1 covariate 1 + … +  p covariate p + error” Categorical Data / Contingency Tables “when response is discrete…”

Fabric B u r n T i m e Back to Fabric Data: Tried to light 4 samples of 4 different (unoccupied!) pajama fabrics on fire. Higher # means less flamable Mean=16.85 std dev=0.94 Mean=10.95 std dev=1.237 Mean=10.50 std dev=1.137 Mean=11.00 std dev=1.299

Suppose we want to test: H 0 :         H A : at least one mean is not equal. at level  = Note that this is the probability of making a false claim (if they are all equal). First idea for how to do this: do four tests at level  (         etc  and reject H 0 if at least one is rejected.

Test 1 H 0 :     H A : not equal Level  Test 2 H 0 :     H A : not equal Level  Test 3 H 0 :     H A : not equal Level  Test 4 H 0 :     H A : not equal Level  Reject all means equal if at least one test fails. This will give you a decision, but what’s the overall probability of making a false claim (if all means are equal) (  level) for this procedure? >,<, or equal to  ?

Overall  = Pr(Falsely reject H 0 :         ) =Pr( at least one test falsely rejects) =1-Pr(none falsely reject) =1-Pr( test 1 doesn’t and … and test 4 doesn’t) =1-(0.95^4) = 0.19 (last step uses independence…) Point: We thought we were doing a level 0.05 test, but it’s actually level 0.18! That’s a problem! Name for this problem: multiple testing problem. What’s one solution?

Solution 1: Do the 4 tests each at a level less than  Many methods to do this: Bonferroni and Tukey are some common ones. We won’t go into much mathematical detail, but these test are often conservative. (True  is smaller than the planned  and power is lower than planned.) For instance, divide  by # of tests you do: 1-(1-(  /4)) 4 = 1-(1-0.05/4) 4 = …

Solution 2: Analysis of Variance! Idea: 1.Variability in the fabric data occurs at two levels: within fabric type and across fabric types. 2.If across fabric type variability is “large” relative to variability within each fabric type, then the means are not equal.

Fabric B u r n T i m e Vertical spread of data points within each oval is one type of variability. Vertical spread of the ovals is another type of variability.

Suppose  1 2 =  2 2 If  1 2 is estimated by s 1 2 from n 1 data points and  2 2 is estimated with s 2 2 from n 2 data points (and the data are normal and independent), then s 2 2 / s 1 2 ~ F n2-1,n1-1 To use the idea to test, we need a fact about variances: Another distribution. The F distribution. n2-1 = numerator df n1-1 = denominator df

Use the test to define “large” H 0 :       H A :       Level  test: reject H 0 at level  if s 2 2 / s 1 2 > F 1- ,n2-1,n1-1

Test for fabric: Formally: –At least one of the means is different if: –Variance among fabric types is greater than the variance within fabric types –Variance among fabric types / Variance within fabric types > F 1- ,3-1,16-3 When one does the test, one uses software that produce: Analysis of variance or ANOVA tables.

Suppose there are k treatments and n data points. ANOVA table: Source Sum ofMean of VariationdfSquaresSquare F P Treatmentk-1SSTMST=SST/(k-1)MST/MSE Errorn-kSSEMSE=SSE/(n-k) Totaln-1total SS ESTIMATE OF “WITHIN FABRIC TYPE” VARIABILITY ESTIMATE OF “AMONG FABRIC TYPE” VARIABILITY “SUM OF SQUARES” IS WHAT GOES INTO NUMERATOR OF s 2 : “(X 1 -X) 2 + … + (X n -X) 2” P-VALUE FOR TEST. (REJECT IF LESS THAN  )

One-way ANOVA: Burn Time versus Fabric Analysis of Variance for Burn Time Source DF SS MS F P Fabric Error Total Explaining why ANOVA is an analysis of variance: MST = / 3 = Sqrt(MST) describes standard deviation among the fabrics. MSE = / 12 = 1.35 Sqrt(MSE) describes standard deviation of burn time within each fabric type. (MSE is estimate of variance of each burn time.) F = MST / MSE = It makes sense that this is large and p-value = Pr(F 4-1,16-4 > 27.15) = 0 is small because the variance “among treatments” is much larger than variance within the units that get each treatment. (Note that the F test assumes the burn times are independent and normal with the same variance.)