Chapter Six: Two Independent Samples Methods 1/35

Transcription

1 Chapter Six: Two Independent Samples Methods 1/35

2 6.1 Introduction 2/35 Introduction It is not always practical to collect data in the paired samples configurations discussed previously. The majority of data collected in research contexts is unpaired. Methods designed for the analysis of paired data are not generally appropriate for the analysis of unpaired data. The slides in this section deal with hypothesis tests and confidence intervals that are designed to compare means or proportions of unpaired data.

3 6.2 Methods Related To Differences Between Means 3/35 Independent Samples t Test: Rationale Researchers often randomly assign subjects to treatment groups in order to avoid bias in the way the groups were formulated. In such circumstance, the question arises as to whether an observed difference between mean outcomes for two groups is attributable to the manner in which the groups were formulated or to a difference in the treatments afforded the subjects in the two groups. By employing the t distribution associated with the sampling distribution of x 1 x 2, the independent samples t test can assist in answering this question.

4 6.2 Methods Related To Differences Between Means 4/35 The Test: Null Hypothesis The null hypothesis to be tested by the independent samples t test is most often (but not always) H 0 : µ 1 = µ 2 or equivalently, H 0 : µ 1 µ 2 = 0

5 6.2 Methods Related To Differences Between Means 5/35 The Test: Two-Tailed Alternative The two-tailed alternative is most often (but not always) H A : µ 1 µ 2 or equivalently, H A : µ 1 µ 2 0

6 6.2 Methods Related To Differences Between Means 6/35 The Test: One-Tailed Alternative One-tailed alternatives are most often (but not always) of the form H A : µ 1 < µ 2 or equivalently, or or equivalently, H A : µ 1 µ 2 < 0 H A : µ 1 > µ 2 H A : µ 1 µ 2 > 0

7 The Test Statistic The test statistic is calculated by t = x 1 x 2 δ 0 ( ) n1 n2 s 2 P x 1 and x 2 represent the means of samples one and two, while n 1 and n 2 represent the number of observations in each of the two samples. The symbol δ 0 represents the hypothesized difference between µ 1 and µ 2 which is usually zero. However, as you will see in the discussion of equivalence tests based on the independent samples t test, this is not always the case. 6.2 Methods Related To Differences Between Means 7/35

8 6.2 Methods Related To Differences Between Means 8/35 The Test Statistic (continued) Degrees of freedom for the test are n 1 + n 2 2. sp 2 is an estimate of the population variance based on an averaging or pooling of the information in the two samples. You can see that the two sample variances are used in the estimate from the following. s 2 P = (n 1 1) s (n 2 1) s 2 2 n 1 + n 2 2 s1 2 and s2 2 represent the variances of the first and second samples respectively.

9 Calculation Of s 2 P A more direct computation of s 2 P is obtained by the following. s 2 P = ( ) ( x 2 1 (P x 1 ) 2 ) n 1 + x 2 2 (P x 2 ) 2 n 2 n 1 + n Methods Related To Differences Between Means 9/35

10 6.2 Methods Related To Differences Between Means 10/35 Example A group of workers exposed to a toxic mold in the ventilating system of the building in which they worked are administered a scale designed to assess symptoms experienced before discovery of the mold. The scale rates symptom experience from zero (no symptoms) to 40 (numerous symptoms at least some of which were severe). Researchers believe that men tend to minimize such experiences and therefore will score significantly lower on the scale than will women. Use the data in the following Table (slide #11) to perform a one-tailed independent samples t test at α =.05 to assess the researcher s theory. Begin by clearly stating the null and alternative hypotheses.

11 6.2 Methods Related To Differences Between Means 11/35 Example (continued) Table: Symptom scale scores of men and women exposed to toxic mold. Men Women Men Women

12 6.2 Methods Related To Differences Between Means 12/35 Solution If men are designated as group one and women as group two, then the null and alternative hypotheses can be stated as H 0 : µ 1 = µ 2 H A : µ 1 < µ 2 We note that X 1 = 147, X 2 1 = 2287, X 2 = 372 and X 2 2 = It follows that x 1 = and x 2 =

13 6.2 Methods Related To Differences Between Means 13/35 Solution (continued) We now calculate s 2 P = = ( ) ( x 2 1 (P x 1 ) 2 ) n 1 + x 2 2 (P x 2 ) 2 n 2 (2287 (147)2 11 = n 1 + n 2 2 ) (10222 (372)2 15 )

14 6.2 Methods Related To Differences Between Means 14/35 Solution (continued) The t statistic is then t = x 1 x 2 δ 0 ( ) n1 n2 s 2 P = ( ) 15 = 3.886

15 6.2 Methods Related To Differences Between Means 15/35 Solution (continued) Appendix B shows that the critical value for a one-tailed t test with = 24 degrees of freedom conducted at α =.05 is so that the null hypothesis is rejected. We can conclude, therefore, that the men who experienced the toxic mold scored significantly lower on the scale than did the women. The design of this study did not permit the random assignment of subjects to groups. We must be particularly cautious as to how results are interpreted in the absence of random assignment. In this case, we cannot be sure that the significant difference found between the two groups is due to gender. It may be, for example, that men tended to be assigned jobs in areas of the building where there was less mold.

16 6.2 Methods Related To Differences Between Means 16/35 Equivalence Via The Independent Samples t Test: Rationale Rejection of a null hypothesis provides good evidence (though not proof positive) that the null hypothesis is false. Failure to reject a null hypothesis does not, generally, provide good evidence that the null hypothesis is true. In situations where you wish to establish the validity of the null hypothesis, you must employ an equivalence test to show that the null hypothesis is (approximately) true.

17 6.2 Methods Related To Differences Between Means 17/35 Two-Tailed Null And Alternative Hypotheses If we let EI U represent the upper end of EI and EI L the lower end, the equivalence null hypothesis for the (independent samples) two-tailed equivalence t test is The alternative is H 0E : µ 1 µ 2 EI L or µ 1 µ 2 EI U H AE : EI L < µ 1 µ 2 < EI U Notice that the null hypothesis states that the difference between population means is not in EI while the alternative states that this difference is in EI. The null hypothesis, then, is an assertion of non-equivalence while the alternative asserts equivalence.

18 6.2 Methods Related To Differences Between Means 18/35 Testing The Equivalence Null Hypothesis Testing the null hypothesis for a two-tailed equivalence test requires that two one-tailed tests be conducted. The null and alternative hypotheses for the two one-tailed tests for the difference between independent samples means are as follows Test One Test Two H 01 : µ 1 µ 2 = EI U H 02 : µ 1 µ 2 = EI L H A1 : µ 1 µ 2 < EI U H A2 : µ 1 µ 2 > EI L Both of these tests must be significant in order to reject the equivalence null hypothesis.

19 6.2 Methods Related To Differences Between Means 19/35 Example Use the data on the following slide (slide #20) to perform a two-tailed equivalence test at α =.05. Use equivalence interval EI L = 4 and EI U = 4. State the equivalence null and alternative hypotheses.

20 6.2 Methods Related To Differences Between Means 20/35 Example (continued) Table: Practice data for two-tailed equivalence test based on the independent samples t test. Group One Group Two Group One Group Two

21 6.2 Methods Related To Differences Between Means 21/35 Solution The null and alternative equivalence hypotheses are as follows. H 0E : µ 1 µ 2 4 or µ 1 µ 2 4 H AE : 4 < µ 1 µ 2 < 4 We note; X 1 = 108, x 2 1 = 800, X 2 = 102, and X 2 2 = 736.

22 6.2 Methods Related To Differences Between Means 22/35 Solution (continued) Using the previously calculated sums and sums of squares, P x 1 = x n = 108 P 16 = and x 2 = x n = = s 2 P = = ( ) ( x 2 1 (P x 1 ) 2 ) n 1 + x 2 2 (P x 2 ) 2 n 2 (800 (108)2 16 = n 1 + n 2 2 ) (736 (102)2 16 )

23 6.2 Methods Related To Differences Between Means 23/35 Solution (continued) t 1 = x 1 x 2 δ 0 ( ) n1 n2 s 2 P = ( ) = Appendix B shows that the critical value for a one-tailed t test with = 30 degrees of freedom conducted at α =.05 is It follows that the null hypothesis H 0 : µ 1 µ 2 = 4 is rejected in favor of the alternative H A : µ 1 µ 2 < 4.

24 6.2 Methods Related To Differences Between Means 24/35 Solution (continued) ( 4.0) t 2 = ( ) = = Because obtained t 2 is greater than critical t of 1.697, the null hypothesis H 0 : µ 1 µ 2 = 4 is rejected in favor of the alternative H A : µ 1 µ 2 > 4. Because both Test One and Test Two are significant, the equivalence null hypothesis is rejected in favor of the alternative. Equivalence, as defined by the equivalence interval, is thereby established.

25 6.2 Methods Related To Differences Between Means 25/35 One-Tailed Null And Alternative Hypotheses The null and alternative hypotheses for the one-tailed equivalence test are one of the following: H 0E : µ 1 µ 2 EI U H AE : µ 1 µ 2 < EI U OR H 0E : µ 1 µ 2 EI L H AE : µ 1 µ 2 > EI L The first one-tailed equivalence hypothesis given above is tested by Test One with the second being carried out by means of Test Two.

26 6.2 Methods Related To Differences Between Means 26/35 Example Suppose the data in the next slide (#27) was collected in connection with an equivalence study designed to show that the drug given group one is not more effective than the drug given group two insofar as control of hypertension is concerned. It is decided that drug one will be declared not more effective than drug two if the mean level of blood pressure achieved by drug one is less than five units below than that achieved by drug two. Use the independent samples t test to conduct a one-tailed equivalence test on this data at α =.05. State the equivalence null and alternative hypotheses before conducting the test.

27 6.2 Methods Related To Differences Between Means 27/35 Example (continued) Table: Practice data for two-tailed equivalence test based on the independent samples t test. Group One Group Two Group One Group Two

28 6.2 Methods Related To Differences Between Means 28/35 Solution The equivalence null hypothesis takes the form The alternative takes the form H 0E : µ 1 µ 2 5 H AE : µ 1 µ 2 > 5 The test is carried out via Test Two with null and alternative hypotheses H 02 : µ 1 µ 2 = 5 and H A2 : µ 1 µ 2 > 5

29 6.2 Methods Related To Differences Between Means 29/35 Solution (continued) The values of x 1, x 2 and sp 2 are 131.8, 135.2, and respectively. Obtained t is then t 2 = x 1 x 2 δ 0 ( ) n1 n2 s 2 P ( 5) = ( ) = = Appendix B shows that the critical value for a one-tailed t test with = 28 degrees of freedom conducted at α =.05 is Because obtained t of.380 is less than critical t of 1.701, the null hypothesis is not rejected. We have, therefore, not been able to demonstrate equivalence.

30 6.2 Methods Related To Differences Between Means 30/35 CI For The Difference Between Means: Rationale The independent samples t test attempts to determine whether there is a difference between the means of two populations (or whether the difference is of some specified value). A related and usually more informative question is, How large is the difference between the population means? This difference can be estimated with a confidence interval.

31 6.2 Methods Related To Differences Between Means 31/35 Construction Of The Interval The confidence interval for the difference between means based on independent samples has the following forms for L and U. L = ( x 1 x 2 ) t U = ( x 1 x 2 ) + t s 2 P s 2 P ( 1 n n 2 ( 1 n n 2 where x 1, x 2 and sp 2 are as previously defined for the independent samples t test, and t is the appropriate t value with n 1 + n 2 2 degrees of freedom. ) )

32 6.2 Methods Related To Differences Between Means 32/35 Example Use the data on slide #27 to form a two-sided 95% confidence interval for the estimation of µ 1 µ 2. Interpret the result.

33 6.2 Methods Related To Differences Between Means 33/35 Solution We note that, x 1 = 131.8, x 2 = 135.2, and sp 2 Equations 6.3 and 6.4 = Then, by L = ( x 1 x 2 ) t s 2 P ( 1 = ( ) = n 1 n 2 ) ( ) 15

34 6.2 Methods Related To Differences Between Means 34/35 Solution (continued) U = ( x 1 x 2 ) + t s 2 P ( 1 = ( ) = n 1 n 2 ) ( ) 15

35 6.2 Methods Related To Differences Between Means 35/35 Solution (continued) A strictly statistical interpretation of this interval would maintain that one can assert with 95% confidence that the difference µ 1 µ 2 is between and From a research point of view, this interval maintains, with 95% confidence, that the average blood pressure level realized by patients treated with drug one minus the average blood pressure level of patients treated with drug two is between and This is an unsatisfactory result for a researcher trying to evaluate the relative effectiveness of the two drugs since the difference might be negative, indicating an advantage for drug one, positive, thereby indicating an advantage for drug two or zero indicating no difference.