1. HW: –Due Next Thursday (4/18): 10.15 10.28 10.39 –Project proposal due next Thursday (4/18).

2. Inference from Small Samples Chapter 10 Data from a manufacturer of child’s pajamas Want to develop materials that take longer before they burn. Run an experiment to compare four types of fabrics. (They considered other factors too, but we’ll only consider the fabrics. Source: Matt Wand)

3. 4321 18 17 16 15 14 13 12 11 10 9 Fabric B u r n T i m e 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

4. Confidence Intervals? Suppose we want to make confidence intervals of mean “burn time” for each fabric type. Can I use: x +/- z  /2 s/sqrt(n) for each one? Why or why not?

5. Answer: t n-1 is the “t distribution” with n-1 degrees of freedom (df) Sample size (n=4) is too small to justify central limit theorem based normal approximation. More precisely: –If x i is normal, then (x –  )/[  /sqrt(n)] is normal for any n. –x i is normal, then (x –  )/[s/sqrt(n)] is normal for n > 30. –New: Suppose x i is approximately normal (and an independent sample). Then (x –  )/[s/sqrt(n)] ~ t n-1 Parameter: (number of data points used to estimate s) - 1

6. “Student” t-distribution (like a normal distribution, but w/ “heavier tails”) t dist’t with 3df Normal dist’n As df increases, t n-1 becomes the normal dist’n. Indistinguishable for n > 30 or so. Idea: estimating std dev leads to “more variability”. More variability = higher chance of “extreme” observation

7. t-based confidence intervals 1-  level confidence interval for a mean: x +/- t  /2,n-1 s/sqrt(n) where t  /2,n-1 is a number such that Pr(T > t  /2,n-1 ) =  /2 and T~t n-1 (see table opposite normal table inside of book cover…)

8. Back to burn time example xst 0.025,3 95% CI Fabric 116.850.9403.182(15.35,18.35) Fabric 210.951.2373.182(8.98, 12.91) Fabric 310.501.1373.182(8.69, 12.31) Fabric 411.001.2993.182(8.93, 13.07)

9. t-based Hypothesis test for a single mean Mechanics: replace z  /2 cutoff with t  /2,n-1 ex: fabric 1 burn time data H 0 : mean is 15 H A : mean isn’t 15 Test stat: |(16.85-15)/(0.94/sqrt(4))| = 3.94 Reject at  =5% since 3.94>t 0.025,3 =3.182 P-value = 2*Pr(T>3.94) where T~t 3. This is between 2% and 5% since t 0.025,3 =3.182 and t 0.01,3 =4.541. (pvalue=2*0.0146) from software) See minitab: basis statistics: 1 sample t test Idea: t-based tests are harder to pass than large sample normal based test. Why does that make sense?