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

2. The Problem  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 µ 1 and variance σ 1 2. –Population 2 has mean µ 2 and variance σ 2 2. –A sample of size n 1 will be taken from population 1. –A sample of size n 2 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  X 1 -  X 2 normal  1 and  2 are unknownSince X 1 and X 2 are both assumed to be normal, or the sample sizes, n 1 and n 2 are assumed to be large, then because  1 and  2 are unknown, the random variable  X 1 -  X 2 has a: –Distributiont –Distribution -- t –Mean  1 -  2 –Mean =  1 -  2 –Standard deviation –Standard deviation that depends on whether or not the standard deviations of X 1 and X 2 (although unknown) can be assumed to be equal –Degrees of freedom –Degrees of freedom that also depends on whether or not the standard deviations of X 1 and X 2 can be assumed to be equal

5. Appropriate Standard Deviation For  X 1 -  X 2 When  ’s Are Known Recall the appropriate standard deviation for  X 1 -  X 2 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 weighted average of s 1 2 and s 2 2If we can assume that the populations have equal variances, then the variance of  X 1 -  X 2 is the weighted average of s 1 2 and s 2 2, weighted by: DEGREES OF FREEDOM There are n 1 - 1 degrees of freedom from the first sample and n 2 -1 degrees of freedom from the second sample, so Total Degrees of Freedom = n 1 + n 2 -2Total Degrees of Freedom for the hypothesis test or confidence interval = (n 1 -1) + (n 2 -1) = n 1 + n 2 -2

7. The Appropriate Standard Deviation For  X 1 -  X 2 When Are  ’s Unknown, but Can Be Assumed to Be Equal The best estimate for  2 then is the pooled variance, s p 2 : Thus the best estimates for the variance and standard deviation of  X 1 -  X 2 are:

8. t-Statistic t-Statistic and t-Confidence Interval Assuming Equal Variances Degrees of Freedom = n 1 + n 2 -2 Confidence Interval

9. The Appropriate Standard Deviation For  X 1 -  X 2 When Are  ’s Unknown, And Cannot Be Assumed to Be Equal If we cannot assume that the populations have equal variances, then the best estimate for  1 2 is s 1 2 and the best estimate for  2 2 is s 2 2. Thus the best estimates for the variance and standard deviation of  X 1 -  X 2 are:

10. t-Statistic and t-Confidence Interval Assuming Unequal Variances t-Statistic Confidence Interval 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: H 0 :        They are equal) H A :        They are different) This is an F-test! If the larger of s 1 2 and s 2 2 is put in the numerator, then the test is: Reject H 0 if F = s    s   > F  DF1, DF2

14. Hypothesis Test/Confidence Interval Approach With Unknown  ’s Take a sample of size n 1 from population 1 –Calculate  x 1 and s 1 2 Take a sample of size n 2 from population 2 –Calculate  x 2 and s 2 2 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, –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. Example 1 – F-test F-test Do an F-test to determine if variances can be assumed to be equal. H 0 :  W 2 /  M 2 = 1 (Equal Variances) H A :  W 2 /  M 2  1 (Unequal Variances) Select α =.05. Reject H 0 (Accept H A ) if Larger s 2 /Smaller s 2 > F.025,DF(Larger s 2 ),DF(Smaller s 2 ) = F.025,31,29 = 2.09 * Calculation: s W 2 / s M 2 = (13.92) 2 /(11.79) 2 = 1.39 Cannot Since 1.39 < 2.09, Cannot conclude unequal variances.

17. Example 1 Example 1 The Equal Variance t-Test Example 1 H 0 :  W -  M = 0 H A :  W -  M > 0 Select α =.05. Reject H 0 (Accept H A ) if t > t.05,60 = 1.658 cannot Since t =.608, p-value = 0.273 > 0.05, we cannot conclude that women average better than men on the tests. Example 1 Example 1 The Equal Variance t-Test Example 1

18. Example 1 Example 1 95% Confidence Interval 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. normal distributions 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. = 1014.97 LA Lakers Number sampled = 11 Sample Average = 12,009 Sample St’d Dev. = 3276.73 LA Clippers

20. Example 2 – F-test F-test Do an F-test to determine if variances can be assumed to be equal. H 0 :  C 2 /  L 2 = 1 (Equal Variances) H A :  C 2 /  L 2  1 (Unequal Variances) Note: Clipper variance is the larger sample variance Choose α =.05. Reject H 0 (Accept H A ) if Larger s 2 /Smaller s 2 > F.025,DF(Larger variance),DF(Smaller variance) = F.025,10,12 = 3.37 Calculation: s C 2 / s L 2 = (3276.73) 2 /(1014.97) 2 = 10.42 Can Since 10.42 > 3.37, Can conclude unequal variances. Unequal Variance t-test. 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: = 11.626 = 12 This rounded to 12 degrees of freedom.

22. Proceed to the hypothesis test for the difference in means with unequal variances: H 0 :  L -  C = 2000 H A :  L -  C > 2000 Select α =.05. Reject H 0 (Accept H A ) if t > t.05,12 = 1.782 can Since t = 2.595, p-value = 0.0117 < 0.05, we can conclude that the Lakers average more than 2000 per game more than the Clippers. Example 2 – the t-Test the t-Testthe t-Test

23. Example 2 95% Confidence Interval 95% Confidence Interval 4666 ± 2238.47 2427.53  6904.47

24. Excel Approach Data Analysis.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 H 0 and H A 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: –H A 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) –H A is a “< test” and the t-statistic is negative This is the usual case If t>0, the true p-value is 1 – (p-value printed by Excel)

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

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

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 (.371671) Cannot conclude Unequal Variances Use Equal Variance t-test

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

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

33. Example 1 – 95% Confidence Interval

34. Excel For Example 2 – F-Test Example 2 Example 2 Go Data 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) Enter =2*F9 to give the p-value for the two-tailed test p-value for one-tail test Low p-value (.000352) – Can conclude Unequal Variances Use Unequal Variance t-test

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

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

39. Example 2 – t-test (Cont’d) Low p-value for 1-tail test (compared to α =.05)! Can 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: –Variances are assumed equal –Variances are not assumed equal Hypothesis Tests/ Confidence Intervals for Differences in Means (Assuming Equal or Unequal Variances) –Summary Data - By Formula –Detailed Data - By Data Analysis Tool