1. Example: One-Sample T-Test Researchers are interested in whether the pulse rate of long-distance runners differs from that of other athletes They randomly sample 8 long-distance runners, measure their resting pulse, and obtain the following data: 45, 42, 64, 54, 58, 49, 48, 56 The average resting pulse of athletes in the general population is 60 beats per minute Test the null hypothesis at the 0.05 level of significance

2. Example: One-Sample T-Test H O : Pulse of long-distance runners = 60 H A : Pulse of long-distance runners differs from 60 Mean = 416/8 = 52 SS = 374; s 2 = 53.4; s = 7.3; SE = 2.6 df = 7 t = (52-60)/2.6 = 3 t crit (from table) at alpha of 0.05 = 2.365 Reject null hypothesis. There is a difference

3. Paired T-Test Used when two samples are not independent of each other Observations in one sample can be paired with observations in the other sample For example: –Before and after observations on the same subjects –A comparison of two different measurements or treatments on the same subjects

4. Paired T-Test: Procedure Calculate the difference (d i = y i − x i ) between the two observations on each pair, making sure you distinguish between positive and negative differences Calculate the mean difference Calculate the standard deviation of the differences (s d ) and use this to calculate the standard error of the mean difference SE = s d / n Calculate t = d / SE Degrees of freedom = n − 1 Use Table B.3 to obtain t crit

5. Example Four individuals with high levels of cholesterol went on a special diet, avoiding high cholesterol foods and taking special supplements. Using the.05 level of significance, was there a significant decrease in cholesterol level? Their total cholesterol levels before and after the diet were as follows: Before After 287 255 305 269 243 245 309 247

6. Example Before AfterDifferenced-d (d-d) 2 287 2553211 305 2693639 245 243231961 309 2476229841 132 (Total)1812 (SS) Mean difference = 132/4 = 33 Standard deviation = 1812/3 = 24.6 Standard error = 12.3 t = 33/12.3 = 2.683df = 3 t crit = 2.353 (one-sided test at α = 0.05) Reject null hypothesis

7. CI for Difference in Means Recall example of verbal skills in 8-year boys and girls t crit = 2.878 for α =0.01 99% CI = 37-31 ± 2.878(2) = 6 ± 5.756 = (0.244; 11.756) This interval does not include zero, therefore there is a difference between boys and girls C. I. = X 1 – X 2  t crit (S X1 – X2 )