1. 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

2. 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)

3. 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) = , s = 1.674

4. 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

5. Lower-tail test ((μ1 - μ2 < 0)
Draw the picture:
Solution:
Decision:
Conclusion:
tcalc = (( )-0)/(1.674*sqrt(1/4 – 1/5)) = -0.70
Approach 1:
df = 7, t0.5,7 =  tcrit =
Approach 2:
=TDIST(0.7,7,1) =
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.

6. Upper-tail test (μ2 – μ1 > 0)
Draw the picture:
Solution:
Decision:
Conclusion:
tcalc = (( )-0)/(1.674*sqrt(1/4 – 1/5)) = 0.70
Approach 1:
df = 7, t0.5,7 =  tcrit = 1.895
Approach 2:
=TDIST(0.7,7,1) =
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

7. Two-tailed test (μ1 - μ2 ≠ 0)
Draw the picture:
Solution:
Decision:
Conclusion:
tcalc = (( )-0)/(1.674*sqrt(1/4 – 1/5)) = 0.70
Approach 1:
df = 7, t0.025,7 =  tcrit = , 2.365
Approach 2:
=TDIST(0.7,7,2) =
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.

8. 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 = s12 = 11
n2 = 10 x2 = s22 = 23
tcalc = ______________________
ν = ________________ (using equation in table 10.2)
tcalc = =( )/(SQRT(11/15+23/10)) =
df from equation in table 10.2, pg. 313
ν = ≈ 15

9. 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) =
2-sided p-value =TDIST(1.895,15,2) =