1. Homework and project proposal due next Thursday. Please read chapter 10 too…

2. 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

3. 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)

4. Comparison of 2 means: Example: –Is mean burn time of fabric 2 different from mean burn time of fabric 3? –Why can’t we answer this w/ the hypothesis test: H 0 : mean of fabric 2 = 10.5 H A : mean of fabric 2 doesn’t = 10.5 –What’s the appropriate hypothesis test? x for fabric 3

5. H 0 : mean fab 2 – mean fab 3 = 0 H A : mean fab 2 – mean fab 3 not = 0 Let’s do this w/ a confidence interval. Large sample CI: (x 2 – x 3 ) +/- z  /2 sqrt[s 2 2 /n 2 + s 2 3 /n 3 ]

6. CI is based on small sample distribution of difference between means. That distribution is different depending on whether the variances of the two means are approximately equal equal or not Small sample CI: –If var(fabric 2) is approximately = var(fabric 3), then just replace z  /2 with t ,n2+n3-2 This is called “pooling” the variances. –If not, then use software. (Software adjusts the degrees of freedom for an “appoximate” confidence interval.) Rule of thumb: OK if 1/3<(S 2 3 /S 2 2 )<3 More conservative Read section 10.4

7. Two-sample T for f2 vs f3 N Mean StDev SE Mean f2 4 10.95 1.24 0.62 f3 4 10.50 1.14 0.57 Difference = mu f2 - mu f3 Estimate for difference: 0.450 95% CI for difference: (-1.606, 2.506) T-Test of difference = 0 (vs not =): T-Value = 0.54 P-Value = 0.611 DF = 6 Both use Pooled StDev = 1.19 Minitab: Stat: Basic statistics: 2 sample t

8. Hypothesis test: comparison of 2 means As in the 1 mean case, replace z  /2 with the appropriate “t based” cutoff value. When  2 1 approximately =  2 2 then test statistic is t=|(x 1 –x 2 )+/-sqrt(s 2 1 /n 1 +s 2 2 /n 2 )| Reject if t > t  /2,n1+n2-2 Pvalue = 2*Pr(T > t) where T~t n1+n2-2 For unequal variances, software adjusts df on cutoff.

9. “Paired T-test” In previous comparison of two means, the data from sample 1 and sample 2 were unrelated. (Fabric 2 and Fabric 3 observations are independent.) Consider following experiment: –“separated identical twins” (adoption) experiments. 15 sets of twins 1 twin raised in city and 1 raised in country measure IQ of each twin want to compare average IQ of people raised in cities versus people raised in the country since twins share common genetic make up, IQs within a pair of twins probably are not independent

10. Data: One Way of Looking At it country city [1,] 117 118 [2,] 153 156 [3,] 73 71 [4,] 64 65 [5,] 95 109 [6,] 120 123 [7,] 94 88 [8,] 106 121 [9,] 90 95 [10,] 96 110 [11,] 67 66 [12,] 102 112 [13,] 111 110 [14,] 127 133 [15,] 180 180 Number IQ 2468101214 60 80 100 120 140 160 180 = city = country City mean = 110.47 Country mean = 106.33

11. country city diff [1,] 117 118 -1 [2,] 153 156 -3 [3,] 73 71 2 [4,] 64 65 -1 [5,] 95 109 -14 [6,] 120 123 -3 [7,] 94 88 6 [8,] 106 121 -15 [9,] 90 95 -5 [10,] 96 110 -14 [11,] 67 66 1 [12,] 102 112 -10 [13,] 111 110 1 [14,] 127 133 -6 [15,] 180 180 0 country - city Index IQ 2468101214 -15 -10 -5 0 5 Mean difference = -4.14 (country – city) Of course, Mean difference = mean( country ) - mean( city ) If we want to test “difference = 0”, we need variance of differences too.

12. Paired t-test One twin’s observation is dependent on the other twin’s observation, but the differences are independent across twins. As a result, we can do an ordinary one sample t- test on the differences. This is called a “paired t-test”. When data naturally come in pairs and the pairs are related, a “paired t-test” is appropriate.

13. “Paired T-test” Minitab: basic statistics: paired t-test: Paired T for Country - City N Mean StDev SE Mean Country 15 106.33 31.03 8.01 City 15 110.47 31.73 8.19 Difference 15 -4.13 6.46 1.67 95% CI for mean difference: (-7.71, -0.56) T-Test of mean difference = 0 (vs not = 0): T-Value = -2.48 P-Value = 0.027 Compare this to a 2-sample t-test

14. Two-sample T for Country vs City N Mean StDev SE Mean Country 15 106.3 31.0 8.0 City 15 110.5 31.7 8.2 Difference = mu Country - mu City Estimate for difference: -4.1 95% CI for difference: (-27.6, 19.3) T-Test of difference = 0 (vs not =): T-Value = -0.36 P-Value = 0.721 DF = 28 Both use Pooled StDev = 31.4 Estimate of difference is the same, but the variance estimate is very different: –Paired: std dev(difference) = 1.67 –2 sample: sqrt[ (31.0^2 /15) + (31.7^2/15) ] = 11.46 –“Cutoff” is different too: t 0.025,13 for paired t 0.025,28 for 2 sample