HW: See Web Project proposal due next Thursday. See web for more detail.

1. HW: See Web • Project proposal due next Thursday. See web for more detail.

2. Inference from Small SamplesChapter 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. Fabric Data:Tried to light 4 samples of 4 different (unoccupied!) pajama fabrics on fire. 18 Higher #meanslessflamable Mean=16.85std dev=0.94 17 16 15 e m i 14 T n 13 r u B 12 Mean=10.95std dev=1.237 Mean=11.00std dev=1.299 11 Mean=10.50std dev=1.137 10 9 1 2 3 4 Fabric

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

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

6. What are degrees of freedom? • Think of them as a parameter • t-distribution has one parameter: df • Normal distribution has 2 parameters: mean and variance

7. 0.4 0.3 0.2 density 0.1 0.0 -4 -2 0 2 4 x “Student” t-distribution (like a normal distribution, but w/ “heavier tails”) Idea: estimating stddev leads to “morevariability”. Morevariability = higher chance of “extreme”observation Normal dist’n t dist’t with 3df As df increases, tn-1 becomes the normal dist’n. Indistinguishable for n > 30 or so.

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

9. Back to burn time example x s t0.025,3 95% CI Fabric 1 16.85 0.940 3.182 (15.35,18.35) Fabric 2 10.95 1.237 3.182 (8.98, 12.91) Fabric 3 10.50 1.137 3.182 (8.69, 12.31) Fabric 4 11.00 1.299 3.182 (8.93, 13.07)

10. t-based Hypothesis test for a single mean • Mechanics: replace za/2 cutoff with ta/2,n-1ex: fabric 1 burn time data H0: mean is 15HA: mean isn’t 15Test stat: |(16.85-15)/(0.94/sqrt(4))| = 3.94 Reject at a=5% since 3.94>t0.025,3=3.182P-value = 2*Pr(T>3.94) where T~t3. This is between 2% and 5% since t0.025,3=3.182 and t0.01,3=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?

11. 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: H0: mean of fabric 2 = 10.5HA: mean of fabric 2 doesn’t = 10.5 • What’s the appropriate hypothesis test? x for fabric 3

12. H0: mean fab 2 – mean fab 3 = 0 HA : mean fab 2 – mean fab 3 not = 0 • Let’s do this w/ a confidence interval (a=0.05). • 95% Large sample CI would be: (x2 – x3) +/- z0.05/2sqrt[s22/n2 + s23/n3] • Can’t use this because it will be “too narrow” (i.e. claim 95% CI but actually it’s an 89%...)

13. 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 za/2 with ta/2,n2+n3-2df = n2+n3-2 = (n2-1)+(n3-1)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<(S23/S22)<3 Read section 10.4 More conservative

14. Minitab: Stat: Basic statistics: 2 sample t 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

15. Hypothesis test: comparison of 2 means • As in the 1 mean case, replace za/2 with the appropriate “t based” cutoff value. • When s21 approximately = s22 then test statistic is t=|(x1–x2)+/-sqrt(s21/n1+s22/n2)| Reject if t > ta/2,n1+n2-2 Pvalue = 2*Pr(T > t) where T~tn1+n2-2 For unequal variances, software adjusts df on cutoff.

16. “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

17. 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 = city = country 180 160 City mean = 110.47 140 IQ 120 100 Country mean = 106.33 80 60 2 4 6 8 10 12 14 The twins are“linked” by these #s Number

18. country - city 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 5 0 IQ -5 Mean difference = -4.14(country – city) -10 -15 2 4 6 8 10 12 14 Index Of course, Mean difference = mean( country ) - mean( city ) If we want to test “difference = 0”, we need variance of differences.

19. 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, estimate var(differences) with sample variance of differences. • This is not the same as var( city ) + var( country) • 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.

20. “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

21. Compare “Paired T-test” vs “2 sample 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 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

22. Compare “Paired T-test” vs “2 sample t-test” • 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 (df) too: • t0.025,13 for paired • t0.025,28 for 2 sample

23. Compare “Paired T-test” vs “2 sample t-test” • 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 (df) too: • t0.025,13 for paired • t0.025,28 for 2 sample

24. Where we’ve been We can use data to address following questions: 1. Question: Is a mean = some number a. Answer: If n>30, large sample “Z” test and confidence interval for means (chapters 8 and 9) b. Answer: If n<=30 and data is approximately normal, then “t” test and confidence intervals for means (chapter 10) 2. Question: Is a proportion = some percentage Answer: If n>30, large sample “Z” test and confidence interval for proportions (chapters 8 and 9) If n<=30, t-test is not appropriate

25. Where we’ve been 3. Question: Is a difference between two means = some # a. Answer: If n>30 and samples are independent (not paired), large sample “Z” test and confidence interval for means (chapters 8 and 9)b. Answer: If n<=30 and samples are independent (not paired), large sample “t” test and confidence interval for means (chapter 10)c. Answer: If samples are dependent, paired t-test (chap 10) 4. Question: Is a difference between two proportions = some % Answer: If n>30 and samples are independent, “Z” test for proportions (chapters 8 and 9) (no t-test…)