1. Hypothesis Testing and Confidence Intervals
Two Population Means
Hypothesis Testing and Confidence Intervals
With Unknown
Standard Deviations

2. The Problem OBJECTIVES 1 or 2 are unknown
1 and 2 are not known (the usual case)
OBJECTIVES
Test whether 1 > 2 (by a certain amount)
or whether 1  2
Determine a confidence interval for the difference in the means: 1 - 2

3. KEY ASSUMPTIONS Sampling is done from two populations.
Population 1 has mean µ1 and variance σ12.
Population 2 has mean µ2 and variance σ22.
A sample of size n1 will be taken from population 1.
A sample of size n2 will be taken from population 2.
Sampling is random and both samples are drawn independently.
Either the sample sizes will be large or the populations are assumed to be normally distribution.

4. Distribution of X1 - X2
Since X1 and X2 are both assumed to be normal, or the sample sizes, n1 and n2 are assumed to be large, then because 1 and 2 are unknown, the random variable X1 -X2 has a:
Distribution -- t
Mean = 1 - 2
Standard deviation that depends on whether or not the standard deviations of X1 and X2 (although unknown) can be assumed to be equal
Degrees of freedom that also depends on whether or not the standard deviations of X1 and X2 can be assumed to be equal

5. Appropriate Standard Deviation For X1 -X2 When s’s Are Known
Recall the appropriate standard deviation for X1 - X2 is:
Now if 1 = 2 we can simply call it  and write it as:
So if the standard deviations are unknown, we need an estimate for the common variance, 2.

6. Estimating 2 Degrees of Freedom
If we can assume that the populations have equal variances, then the variance of X1 - X2 is the weighted average of s12 and s22, weighted by:
DEGREES OF FREEDOM
There are n1- 1 degrees of freedom from the first sample and n2-1 degrees of freedom from the second sample, so
Total Degrees of Freedom for the hypothesis test or confidence interval = (n1 -1) + (n2 -1) = n1 + n2 -2

7. The Appropriate Standard Deviation For X1 - X2 When Are s’s Unknown, but Can Be Assumed to Be Equal
The best estimate for 2 then is the pooled variance, sp2:
Thus the best estimates for the variance and standard deviation of X1 - X2 are:

8. t-Statistic and t-Confidence Interval Assuming Equal Variances
Degrees of Freedom = n1 + n2 -2

9. The Appropriate Standard Deviation For X1 - X2 When Are s’s Unknown, And Cannot Be Assumed to Be Equal
If we cannot assume that the populations have equal variances, then the best estimate for 12 is s12 and the best estimate for 22 is s22.
Thus the best estimates for the variance and standard deviation of X1 - X2 are:

10. t-Statistic and t-Confidence Interval Assuming Unequal Variances
Total Degrees of Freedom
Round the resulting value.

11. s are known z-distribution
Standard Error

12. s are unknown t-distribution
1 = 2
1 ≠ 2
Standard
Error
Degree Of
Freedom
n1 + n2 -2

13. Testing whether the Variances Can Be Assumed to Be Equal
The following hypothesis test tests whether or not equal variances can be assumed:
H0: s12/s22 = 1 (They are equal)
HA: s12/s22  1 (They are different)
This is an F-test!
If the larger of s12 and s22 is put in the numerator,
then the test is:
Reject H0 if F = s12/s22 > Fa/2, DF1, DF2

14. Hypothesis Test/Confidence Interval Approach With Unknown ’s
Take a sample of size n1 from population 1
Calculate x1 and s12
Take a sample of size n2 from population 2
Calculate x2 and s22
Perform an F-test to determine if the variances can be assumed to be equal
Perform the Appropriate Hypothesis Test or Construct the Appropriate Confidence Interval

15. Example 1 Based on the following two random samples, Women Men
Can we conclude that women on the average score better than men on civil service tests?
Construct a 95% for the difference in average scores between women and men on civil service tests.
Because the sample sizes are large, we do not have to assume that test scores have a normal distribution to perform our analyses.
Number sampled = 32
Sample Average = 75
Sample St’d Dev. = 13.92
Women
Number sampled = 30
Sample Average = 73
Sample St’d Dev. = 11.79
Men

16. H0: W2/M2 = 1 (Equal Variances) HA: W2/M2  1 (Unequal Variances)
Example 1 – F-test
Do an F-test to determine if variances can be assumed to be equal.
H0: W2/M2 = (Equal Variances)
HA: W2/M2  1 (Unequal Variances)
Select α = .05.
Reject H0 (Accept HA) if Larger s2/Smaller s2 > F.025,DF(Larger s2),DF(Smaller s2) = F.025,31,29 = 2.09 *
Calculation: sW2/ sM2 = (13.92)2/(11.79)2 = 1.39
Since 1.39 < 2.09, Cannot conclude unequal variances.

17. Example 1 The Equal Variance t-Test
H0: W - M = 0
HA: W - M > 0
Select α = .05.
Reject H0 (Accept HA) if t > t.05,60 = 1.658
Since .608 < 1.658, we cannot conclude that
women average better than men on the tests.

18. Example 1 95% Confidence Interval
2 ± 6.57
-4.57  8.57

19. Example 2
Based on the following random samples of basketball attendances at the Staples Center,
Can we conclude that the Lakers average attendance is more than 2000 more than the Clippers average attendance at the Staples Center?
Construct a 95% for the difference in average attendance between Lakers and Clippers games at the Staples Center.
Since sample sizes are small, we must assume that attendance at Lakers and Clipper games have normal distributions to perform the analyses.
Number sampled = 13
Sample Average = 16,675
Sample St’d Dev. =
LA Lakers
Number sampled = 11
Sample Average = 12,009
Sample St’d Dev. =
LA Clippers

20. Example 2 – F-test
Do an F-test to determine if variances can be assumed to be equal.
H0: C2/L2 = (Equal Variances)
HA: C2/L2  1 (Unequal Variances)
Note: Clipper variance is the larger sample variance
Choose α = .05.
Reject H0 (Accept HA) if Larger s2/Smaller s2 > F.025,DF(Larger variance),DF(Smaller variance) = F.025,10,12 = 3.37
Calculation: sC2/ sL2 = ( )2/( )2 = 10.42
Since > 3.37, Can conclude unequal variances.
Do Unequal Variance t-test.

21. Degrees of Freedom for the Unequal Variance t-Test
The degrees of freedom for this test is given by: = =
This rounded to 12 degrees of freedom.

22. Example 2 – the t-Test
Proceed to the hypothesis test for the difference in means with unequal variances:
H0: L - C = 2000
HA: L - C > 2000
Select α = .05.
Reject H0 (Accept HA) if t > t.05,12 = 1.782
Since t = > 1.782, we can conclude that the Lakers average more than 2000 per game more than the Clippers at the Staples Center.

23. Example 2 95% Confidence Interval
4666 ± 

24. Excel Approach
F-test, t-test Assuming Equal Variances, t-test Assuming Unequal Variances are all found in Data Analysis.
Excel only performs a one-tail F-test.
Multiply this 1-tail p-value by 2 to get the p-value for the 2-tail F-test.
Formulas must be entered for the LCL and UCL of the confidence intervals.
All values for these formulas can be found in the Equal or Unequal Variance t-test Output.

25. Inputting/Interpreting Results From Hypotheses Tests
Express H0 and HA so that the number on the right side is positive (or 0)
The p-value returned for the two-tailed test will always be correct.
The p-value returned for the one-tail test is usually correct. It is correct if:
HA is a “> test” and the t-statistic is positive
This is the usual case
If t < 0, the true p-value is 1 – (p-value printed by Excel)
HA is a “< test” and the t-statistic is negative
If t>0, the true p-value is 1 – (p-value printed by Excel)

26. Excel For Example 1 – F-Test
Go Tools
Select Data Analysis
Select F-Test Two-Sample For Variances

27. Example 1 – F-Test (Cont’d)
Use Women (Column A) for Variable Range 1
Use Men (Column B) for Variable Range 2
Check
Labels
Designate first cell
for output.

28. Example 1 – F-Test (Cont’d)
p-value for
one-tail test

29. Example 1 – F-Test (Cont’d)
p-value for
one-tail test
=2*D9
Multiply the one-tail p-value
by 2 to get the 2-tail p-value.
High p-value ( )
Cannot conclude Unequal Variances
Use Equal Variance t-test

30. Select t-Test: Two-Sample Assuming Equal Variances
Example 1 – t-Test
Go Tools
Select Data Analysis
Select t-Test: Two-Sample Assuming Equal Variances

31. Example 1 – t-Test (Cont’d)
Since HA is W - M > 0, enter
Column A for Range 1
Column B for Range 2
0 for Hypothesized Mean Difference
Check
Labels
Designate first cell
for output.

32. Example 1 – t-test (Cont’d)
High p-value for 1-tail test!
Cannot conclude average women’s score > average men’s score
p-value for
the one-tail “>” test
p-value for at
two-tail “” test

33. Example 1 – 95% Confidence Interval

34. Excel For Example 2 – F-Test
Go Tools
Select Data Analysis
Select F-Test Two-Sample For Variances

35. Example 2 – F-Test (Cont’d)
Use Lakers (Column B) for Variable Range 1
Use Clippers (Column D) for Variable Range 2
Check
Labels
Designate first cell
for output.

36. Example 2 – F-Test (Cont’d)
p-value for
one-tail test
Enter =2*F9
to give the p-value
for the two-tailed test
Low p-value ( ) – Can conclude Unequal Variances
Use Unequal Variance t-test

37. t-Test: Two Sample Assuming Unequal Variances
Example 2 – t-Test
Go Tools
Select Data Analysis
Select
t-Test: Two Sample Assuming Unequal Variances

38. Example 2 – t-Test (Cont’d)
Since HA is L - C > 2000, enter
Column B for Range 1
Column D for Range 2
2000 for Hypothesized Mean Difference
Check
Labels
Designate first cell
for output.

39. Example 2 – t-test (Cont’d)
Low p-value for 1-tail test
(compared to α = .05)!
Can conclude the Lakers average more than 2000 more people per game than the Clippers.
p-value for
the one-tail “>” test
p-value for at
two-tail “” test

40. Example 2 – 95% Confidence Interval
=(F15-G15)-TINV(.05,F19)*SQRT(F16/F17+G16/G17) *
Highlight Cell I14
Add $ Signs Using F4 key
Drag to cell I15
Change “-” to “+”

41. Review Standard Errors and Degrees of Freedom when:
Variances are assumed equal
Variances are not assumed equal
F-statistic to determine if variances differ
t-statistic and confidence interval when:
Hypothesis Tests/ Confidence Intervals for Differences in Means (Assuming Equal or Unequal Variances)
Summary Data - By Formula
Detailed Data - By Data Analysis Tool