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Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 1 Understandable Statistics S eventh Edition By Brase and Brase Prepared by: Lynn Smith Gloucester County College Chapter Nine Part 5 (Section 9.7) Hypothesis Testing

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 2 Independent Sampling Distributions There is no relationship whatsoever between specific values of the two distributions. Example: comparison of the incomes of individuals from two groups.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 3 Testing the Differences of Means for Large, Independent Samples Let x 1 and x 2 have normal distributions with means  1 and  2 and standard deviations  1 and  2 respectively. Take independent random samples of size n 1 and n 2 from each distribution.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 4

5 If both n 1 and n 2 are 30 or larger The Central Limit Theorem can be applied even if the original distributions are not normal.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 6 When testing the difference of means The null hypothesis usually indicates that there is no difference between the means. H 0 :  1 –  2 = 0 or, equivalently H 0 :  1 =  2

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 7 Left-Tailed Test H 0 :  1 =  2 H 1 :  1 <  2

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 8 Right-Tailed Test H 0 :  1 =  2 H 1 :  1 >  2

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 9 Two-Tailed Test H 0 :  1 =  2 H 1 :  1   2

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 10 Use the Critical Values for the Normal Distribution

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 11

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 12 A college professor wishes to determine (at 0.05 level of significance) if there is a difference in the final exam results between the two groups of students. One group studies Calculus in the traditional classroom setting; the other is taught via distance learning.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 13 Sample exam results:

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 14 We wish to determine if the observed difference in exam results (88 versus 85) is significant. H 0 :  1 =  2 H1:1  2H1:1  2 Using a two-tailed test, the critical z values are  1.96 for  = 0.05.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 15 Compute the sample test statistic

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 16 Convert the test statistic to a z value

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 17 Since 3.14 > 1.96 We reject the null hypothesis - the statement that there was no difference between the two groups.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 18 We conclude that the exam results for the distance learning class are significantly different from those for the traditional group at 5% level of significance.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 19 P Value Approach To find the P value for the sample statistic of the difference in the means, use the corresponding z value, z = Since we are working with a two-tailed test, add the area to the right of z = 3.14 to the area to the left of z = – P value = 2(0.0008) =

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 20 P value = We would reject H 0 for any   We, therefore reject H 0 for  = 0.05.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 21 Inferences about the Differences of Two Means of Small Independent Samples

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 22 Assumptions Independent random samples are drawn from two populations with means  1 and  2. The parent populations have normal distributions. The standard deviations for the populations (  1 and  2 ) are equal.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 23 Best Estimate of the Common or Pooled Standard Deviation for Two Populations

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 24 When testing the difference of means The null hypothesis usually indicates that there is no difference between the means. H 0 :  1 –  2 = 0 or, equivalently H 0 :  1 =  2

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 25 Left-Tailed Test H 0 :  1 =  2 H 1 :  1 <  2

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 26 Right-Tailed Test H 0 :  1 =  2 H 1 :  1 >  2

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 27 Two-Tailed Test H 0 :  1 =  2 H 1 :  1   2

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 28 To Test the Hypothesis Use the t Statistic

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 29 The Pooled Standard Deviation

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 30 Test H 0 :  1 =  2 Against H 1:  1 >  2 Use a 1% Level of Significance.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 31 Sample Data

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 32 Degrees of Freedom d.f. = n 1 + n 2 – 2 = – 2 = 15

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 33 For a Right-Tailed Test at 1% Level of Significance and d.f. = 15 Use the column headed by  = The critical t value is t =

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 34

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 35 Pooled Standard Deviation

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 36

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 37 Critical t Value =

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 38 Test Statistic  Do not reject H 0.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 39 P Value Approach To test H 0 :  1 =  2 Against H 1 :  1 >  2 t value = 1.593, d.f. = 15 For a right-tailed test, use the column headed by . For d.f. = 15, t = falls between t = and t = We conclude that < P <

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 40 Conclusion: < P < We reject H 0 only for   P. We could not reject H 0 for  = 0.01.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 41 Tests for Differences of Proportions We will draw independent random samples of size n 1 and n 2 from two binomial distributions. For large values of n 1 and n 2,, the distribution of sample differences in proportions of successes will be approximately normally distributed.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 42 Symbols Used r 1 and r 2 = number of successes from the first and second samples p 1 and p 2 = probability of success on each trial from the first and second binomial distributions

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 43 For large values of n 1 and n 2 :

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 44 The Null Hypothesis p 1 = p 2

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 45 The proportions p 1 and p 2 are unknown and must be estimated

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 46 Critical Values for Tests Involving a Difference of Two Proportions (Large Samples)

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 47 “Large” Samples All of the following are larger than five:

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 48 Sample Test Statistic

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 49 A college is attempting to determine whether a reminder phone call encourages students to participate in early registration. A group of 1200 students was divided into two groups. One received a reminder phone call; the other did not.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 50 The data:

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 51 Use a 5% level of significance to test the claim that the reminder phone calls increased participation in early registration. Let p 1 = proportion from the first group (reminder) who registered early. Let p 2 = proportion from the second group (no reminder) who registered early.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 52 The null hypothesis is that there is no difference in proportions. H 0 :p 1 = p 2

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 53 The alternate hypothesis is that the reminders improve participation rate. H 1 :p 1 > p 2

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 54 H 1 :p 1 > p 2 Use a right-tailed test. For 5% level of significance, the critical z value =

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 55 For the Reminder group, r 1 = 475, n 1 = 600.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 56 For the No Reminder group, r 1 = 452, n 1 = 600.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 57 The Sample Statistic

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 58 The pooled estimates of proportion

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 59 Convert the test statistic to a z value:

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 60 Critical z Value =

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 61 Test statistic = z = Do not reject the null hypothesis.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 62 Conclusion At 5% level of significance, we cannot reject the claim that the proportions (of students who register early) are equal. We conclude that the reminder phone calls do not make a significant difference in the number of students who participate in early registration.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 63 P Value Approach For a right-tailed test, the value of P is the area in the right tail of the distribution, larger than z = From Table 5a, we find the P value. P = – = We would reject H 0 only for   We cannot reject H 0 for  = 0.05.