1. When  is unknown  The sample standard deviation s provides an estimate of the population standard deviation .  Larger samples give more reliable estimates of . Population distribution Small sampleLarge sample

2. The t distributions We take 1 random sample of size n from a Normal population N(µ,σ):  When  is known, the sampling distribution of is Normal N(  /√n), and the statistic follows the standard Normal N(0,1).  When  is estimated from the sample standard deviation s, the statistic follows the t distribution t ( ,1) with n − 1 degrees of freedom.

3. When n is large, s is a good estimate of  and the t df n – 1 distribution is close to the standard Normal distribution. Standard Normal t distribution, df 4 t distribution, df 1 Standard Normal t distribution, df 100 t distribution, df 20

4. A medical study examined the effect of a new medication on the seated systolic blood pressure. The results, presented as mean ± SEM for 25 patients, are 113.5 ± 8.9. What is the standard deviation s of the sample data? Standard deviation versus standard error For a sample of size n, the sample standard deviation s is: n − 1 is the “degrees of freedom.” The value s/√n is called the standard error of the sample mean SEM. Scientists often present their sample results as the mean ± SEM. SEM = s/√n s = SEM*√n s = 8.9*√25=44.5

5. Table C When σ is known, we use the Normal distribution and z. When σ is unknown we use a t distribution with “n − 1” degrees of freedom (df). Table C shows the z-values and t-values corresponding to landmark P-values/ confidence levels.

6. The one-sample t test As before, a test of hypotheses requires a few steps: 1.Stating the null hypothesis (H 0 ) 2.Deciding on a one-sided or two-sided alternative (H a ) 3.Choosing a significance level  4.Calculating t and its degrees of freedom 5.Finding the area under the curve with Table C or software 6.Stating the P-value and concluding

7. We draw a random sample of size n from an N(µ, σ) population. When  is estimated from s, the distribution of the test statistic t is a t distribution with df = n – 1. This resulting t test is robust to deviations from Normality as long as the sample size is large enough.  1

8. One-sided (one-tailed) Two-sided (two-tailed) The P-value is the probability, if H 0 was true, of randomly drawing a sample like the one obtained or more extreme in the direction of H a.

9. Using Table C: 2.398 < t 2.7 < 2.821 so 0.02 > P-value > 0.01 For H a : μ > μ 0 if n = 10 and t = 2.70, then…

10. Variable N Mean SE Mean StDev s/√n s Weightchange 53 -2.404 0.720 5.243 Study Participants: 53 obese children ages 9 to 12 with a BMI above the 95 th percentile for age and gender Intervention: family counseling sessions on the stoplight diet (green/yellow/red approach to eating food) - after 8 weekly sessions and 3 follow-up sessions Assessment: Weight change at 15 weeks of intervention Was the intervention effective in helping obese children lose weight? H 0 :  = 0 versus H a :  < 0 (one-sided test)

11. MINITAB: Test of mu = 0 vs < 0 Variable N Mean StDev SE Mean T P Weightchange 53 -2.404 5.243 0.720 -3.34 0.001 There is a significant weight loss, on average, following intervention. df = 53-1=52 ≈ 50, 3.261 one-sided P >.0005 (software gives P = 0.0008 ≈ 0.001), highly significant.

12. Confidence intervals A C% confidence interval is a range of values (interval) that contains the true population parameter with confidence C. We have a set of data from a normal pop’n with both  and  unknown. We use x ̅ to estimate , and s to estimate  using a t dist’n (df n − 1). C t*t*−t*−t* m  C is the area between −t* and t*.  We find t* in the line of Table C.  The margin of error m is:

13. Data on the blood cholesterol levels (mg/dl) of 24 lab rats give a sample mean of 85 and a standard deviation of 12. We want a 95% confidence interval for the mean blood cholesterol of all lab rats. Assume normality of the data. We are 95% confident that the true mean blood cholesterol of all lab rats is between 79.9 and 90.1 mg/dl.

14. Matched pairs t procedures Sometimes we want to compare treatments or conditions at the individual level. The data sets produced this way are not independent. The individuals in one sample are related to those in the other sample.  Pre-test and post-test studies look at data collected on the same sample elements before and after some experiment is performed.  Twin studies often try to sort out the influence of genetic factors by comparing a variable between sets of twins.  Using people matched for age, sex, and education in social studies allows us to cancel out the effect of these potential lurking variables.

16. Variable N Mean SE Mean StDev Weightchange 53 -2.404 0.720 5.243 Study Participants: 53 obese children ages 9 to 12 with a BMI above the 95 th percentile for age and gender Intervention: family counseling sessions on the stoplight diet (green/yellow/red approach to eating food) - after 8 weekly sessions and 3 follow-up sessions Assessment: Weight change at 15 weeks of intervention Was the intervention effective in helping obese children lose weight? This is a pre-/post design. The weight change values are the difference in body weight before and after intervention for each participant.

17. Does lack of caffeine increase depression? Randomly selected caffeine-dependent individuals were deprived of all caffeine- rich foods and assigned to receive daily pills. At one time the pills contained caffeine and, at another time they were a placebo. Depression was assessed quantitatively (higher scores represent greater depression). (Assume normality.) This is a matched pairs design with 2 data points for each subject. We compute a new variable “Difference” Placebo minus Caffeine Caffeine

18. H 0 :  diff = 0 ; H a :  diff > 0 For df = 11-1=10, 3.169 P-value >.0025 (Software gives P = 0.0027.) Caffeine deprivation causes a significant increase in depression (P < 0.005, n = 11). (…) With 11 "difference" points, df = n – 1 = 10. We find: x ̅ diff = 7.36; s diff = 6.92; so SEM diff = s diff / √n = 6.92/√11 = 2.086 We test:

19. Robustness The t procedures are exactly correct when the population is exactly Normal. This is rare. The t procedures are robust to small deviations from Normality, but:  The sample must be a random sample from the population.  Outliers and skewness strongly influence the mean and therefore the t procedures. Their impact diminishes as the sample size gets larger because of the Central Limit Theorem. As a guideline:  When n < 15, the data must be close to Normal and without outliers.  When 15 < n < 40, mild skewness is acceptable, but not outliers.  When n > 40, the t statistic will be valid even with strong skewness.

20. Does oligofructose consumption stimulate calcium absorption? Healthy adolescent males took a pill for nine days and had their calcium absorption tested on the ninth day. The experiment was repeated three weeks later. Subjects received either an oligofructose pill first or a control sucrose pill first. The order was randomized and the experiment was double-blind. Fractional calcium absorption data (in percent of intake) for 11 subjects: Can we use a t inference procedure for this study? Discuss the assumptions.

21. Red wine, in moderation Does drinking red wine in moderation increase blood polyphenol levels, thus maybe protecting against heart attacks? Nine randomly selected healthy men were assigned to drink half a bottle of red wine daily for two weeks. The percent change in their blood polyphenol levels was assessed: 0.73.544.95.577.48.18.4 x ̅ = 5.5; s = 2.517; df = n − 1 = 8 Can we use a t inference procedure for this study? Discuss the assumptions.