Two-Sample Hypothesis Testing Example: A professor has designed an experiment to test the effect of reading the textbook before attempting to complete a homework assignment. Four students who read the textbook before attempting the homework recorded the following times (in hours) to complete the assignment: 3.1, 2.8, 0.5, 1.9 hours Five students who did not read the textbook before attempting the homework recorded the following times to complete the assignment: 0.9, 1.4, 2.1, 5.3, 4.6 hours

Two-Sample Hypothesis Testing Define the difference in the two means as: μ1 - μ2 = d0 What are the Hypotheses? H0: _______________ H1: _______________ or NOTE: d0 is often 0 (there is, statistically speaking, no difference in the means) H0: μ1 - μ2 = 0 H1: μ1 - μ2 < 0 (note: compare lower to higher for lower-tail test) H1: μ1 - μ2 ≠ 0 H1: μ1 – μ2 > 0 (note: compare higher to lower for upper-tail test)

Our Example Reading: n1 = 4 x1 = 2.075 s12 = 1.363 No reading: If we assume the population variances are “equal”, we can calculate sp2 and conduct a __________. = __________________ t-test sp2 = (3(1.363)+4(3.883))/(4+5-2) = 2.803 , s = 1.674

Your turn … Lower-tail test ((μ1 - μ2 < 0) “Fixed α” approach (“Approach 1”) at α = 0.05 level. “p-value” approach (“Approach 2”) Upper-tail test (μ2 – μ1 > 0) “Fixed α” approach at α = 0.05 level. “p-value” approach Two-tailed test (μ1 - μ2 ≠ 0) Recall  Note that we have to compare higher – lower mean to conduct an upper-tail test

Lower-tail test ((μ1 - μ2 < 0) Draw the picture: Solution: Decision: Conclusion: tcalc = ((2.075-2.860)-0)/(1.674*sqrt(1/4 – 1/5)) = -0.70 Approach 1: df = 7, t0.5,7 = 1.895  tcrit = -1.895 Approach 2: =TDIST(0.7,7,1) = 0.253259 Decision: fail to reject H0 Conclusion: the data do not support the hypothesis that the mean time to complete homework is less for students who read the textbook or There is no statistically significant difference in the time required to complete the homework for the people who read the text ahead of time vs those who did not. The data do not support the hypothesis that the mean completion time is less for readers than for non-readers.

Upper-tail test (μ2 – μ1 > 0) Draw the picture: Solution: Decision: Conclusion: tcalc = ((2.860-2.075)-0)/(1.674*sqrt(1/4 – 1/5)) = 0.70 Approach 1: df = 7, t0.5,7 = 1.895  tcrit = 1.895 Approach 2: =TDIST(0.7,7,1) = 0.253259 Decision: fail to reject H0 Conclusion: the data do not support the hypothesis that the mean time to complete homework is more for students who do not read the textbook

Two-tailed test (μ1 - μ2 ≠ 0) Draw the picture: Solution: Decision: Conclusion: tcalc = ((2.860-2.075)-0)/(1.674*sqrt(1/4 – 1/5)) = 0.70 Approach 1: df = 7, t0.025,7 = 2.365  tcrit = -2.365, 2.365 Approach 2: =TDIST(0.7,7,2) = 0.506518 Decision: fail to reject H0 Conclusion: the data do not support the hypothesis that the mean time to complete homework different when a student reads the text versus when the student does not read the text.

Another Example Suppose we want to test the difference in carbohydrate content between two “low-carb” meals. Random samples of the two meals are tested in the lab and the carbohydrate content per serving (in grams) is recorded, with the following results: n1 = 15 x1 = 27.2 s12 = 11 n2 = 10 x2 = 23.9 s22 = 23 tcalc = ______________________ ν = ________________ (using equation in table 10.2) tcalc = =(27.2-23.9)/(SQRT(11/15+23/10)) = 1.894759 df from equation in table 10.2, pg. 313 ν = 14.69 ≈ 15

Example (cont.) What are our options for hypotheses? At an α level of 0.05, One-tailed test, t0.05, 15 = ________ Two-tailed test, t0.025, 15 = ________ How are our conclusions affected? H0: μ1 - μ2 = 0 vs H0: μ1 - μ2 = 0 H1: μ1 - μ2 > 0 H1: μ1 - μ2 ≠ 0 t0.05, 15 = 1.753 t0.025, 15 = 2.131 Our data don’t support a conclusion that the two meals are different at an alpha level of .05 Our data do support a conclusion that meal 1 has more carbs than meal 2 at an alpha level of .05 note, 1-sided p-value =TDIST(1.895,15,1) = 0.038766 2-sided p-value =TDIST(1.895,15,2) = 0.077533