PENGUJIAN HIPOTESIS 1 Pertemuan 9 Matakuliah : I0272 - STATISTIK PROBABILITAS Tahun : 2009 PENGUJIAN HIPOTESIS 1 Pertemuan 9

Uji hipotesis nilai tengah Uji hipotesis beda dua nilai tengah Materi Uji hipotesis nilai tengah Uji hipotesis beda dua nilai tengah Bina Nusantara University

Hypothesis Testing Developing Null and Alternative Hypotheses Type I and Type II Errors One-Tailed Tests About a Population Mean: Large-Sample Case Two-Tailed Tests About a Population Mean: Tests About a Population Mean: Small-Sample Case Bina Nusantara University 4

Developing Null and Alternative Hypotheses Hypothesis testing can be used to determine whether a statement about the value of a population parameter should or should not be rejected. The null hypothesis, denoted by H0 , is a tentative assumption about a population parameter. The alternative hypothesis, denoted by Ha, is the opposite of what is stated in the null hypothesis. Hypothesis testing is similar to a criminal trial. The hypotheses are: H0: The defendant is innocent Ha: The defendant is guilty Bina Nusantara University 5 5

Testing Research Hypotheses The research hypothesis should be expressed as the alternative hypothesis. The conclusion that the research hypothesis is true comes from sample data that contradict the null hypothesis. Bina Nusantara University 6 6

A Summary of Forms for Null and Alternative Hypotheses about a Population Mean The equality part of the hypotheses always appears in the null hypothesis. In general, a hypothesis test about the value of a population mean  must take one of the following three forms (where 0 is the hypothesized value of the population mean). H0:  > 0 H0:  < 0 H0:  = 0 Ha:  < 0 Ha:  > 0 Ha:  ≠ 0 Bina Nusantara University 7 7

Type I and Type II Errors Contoh Soal: Metro EMS Population Condition H0 True Ha True Conclusion ( ) ( ) Accept H0 Correct Type II (Conclude  Conclusion Error Reject H0 Type I Correct (Conclude  rror Conclusion Bina Nusantara University 8 8

The Steps of Hypothesis Testing Determine the appropriate hypotheses. Select the test statistic for deciding whether or not to reject the null hypothesis. Specify the level of significance  for the test. Use to develop the rule for rejecting H0. Collect the sample data and compute the value of the test statistic. a) Compare the test statistic to the critical value(s) in the rejection rule, or b) Compute the p-value based on the test statistic and compare it to to determine whether or not to reject H0. Bina Nusantara University 9 9

One-Tailed Tests about a Population Mean: Large-Sample Case (n > 30) Hypotheses H0:   or H0:  Ha: Ha: Test Statistic  Known  Unknown Rejection Rule Reject H0 if z > zReject H0 if z < -z Bina Nusantara University 10

Two-Tailed Tests about a Population Mean: Large-Sample Case (n > 30) Hypotheses H0: =  Ha:  Test Statistic  Known  Unknown Rejection Rule Reject H0 if |z| > z Bina Nusantara University 11

Example: One Tail Test H0: m £ 368 H1: m > 368 Q. Does an average box of cereal contain more than 368 grams of cereal? A random sample of 25 boxes showed = 372.5. The company has specified s to be 15 grams. Test at the a = 0.05 level. 368 gm. H0: m £ 368 H1: m > 368 Bina Nusantara University 12

Finding Critical Value: One Tail What is Z given a = 0.05? Standardized Cumulative Normal Distribution Table (Portion) .05 Z .04 .06 1.6 .9495 .9505 .9515 .95 a = .05 1.7 .9591 .9599 .9608 1.8 .9671 .9678 .9686 1.645 Z Critical Value = 1.645 1.9 .9738 .9744 .9750 Bina Nusantara University

Example Solution: One Tail Test H0: m £ 368 H1: m > 368 Test Statistic: Decision: Conclusion: a = 0.5 n = 25 Critical Value: 1.645 Reject Do Not Reject at a = .05 .05 No evidence that true mean is more than 368 1.645 Z Bina Nusantara University 14 1.50

p -Value Solution p-Value is P(Z ³ 1.50) = 0.0668 Z 1.50 Use the alternative hypothesis to find the direction of the rejection region. P-Value =.0668 1.0000 - .9332 .0668 Z 1.50 From Z Table: Lookup 1.50 to Obtain .9332 Z Value of Sample Statistic Bina Nusantara University 15

(p-Value = 0.0668) ³ (a = 0.05) Do Not Reject. p -Value Solution (continued) (p-Value = 0.0668) ³ (a = 0.05) Do Not Reject. p Value = 0.0668 Reject a = 0.05 Z 1.645 1.50 Test Statistic 1.50 is in the Do Not Reject Region Bina Nusantara University 16

Example: Two-Tail Test Q. Does an average box of cereal contain 368 grams of cereal? A random sample of 25 boxes showed = 372.5. The company has specified s to be 15 grams. Test at the a = 0.05 level. 368 gm. H0: m = 368 H1: m ¹ 368 Bina Nusantara University 17

Example Solution: Two-Tail Test H0: m = 368 H1: m ¹ 368 Test Statistic: Decision: Conclusion: a = 0.05 n = 25 Critical Value: ±1.96 Reject Do Not Reject at a = .05 .025 .025 No Evidence that True Mean is Not 368 -1.96 1.96 Z Bina Nusantara University 18 1.50

(p Value = 0.1336) ³ (a = 0.05) Do Not Reject. p-Value Solution (p Value = 0.1336) ³ (a = 0.05) Do Not Reject. p Value = 2 x 0.0668 Reject Reject Test Statistic 1.50 is in the Do Not Reject Region a = 0.05 Z 1.50 1.96 Bina Nusantara University 19

Tests about a Population Mean: Small-Sample Case (n < 30) Test Statistic  Known  Unknown This test statistic has a t distribution with n - 1 degrees of freedom. Rejection Rule One-Tailed Two-Tailed Ha:  Reject H0 if t > t Ha:  Reject H0 if t < -t Ha:   Reject H0 if |t| > t Bina Nusantara University 20 20

Example: One-Tail t Test Does an average box of cereal contain more than 368 grams of cereal? A random sample of 36 boxes showed X = 372.5, and s = 15. Test at the a = 0.01 level. 368 gm. H0: m £ 368 H1: m > 368 s is not given Bina Nusantara University 21

Example Solution: One-Tail H0: m £ 368 H1: m > 368 Test Statistic: Decision: Conclusion: a = 0.01 n = 36, df = 35 Critical Value: 2.4377 Reject Do Not Reject at a = .01 .01 No evidence that true mean is more than 368 2.4377 t35 Bina Nusantara University 1.80

(p Value is between .025 and .05) ³ (a = 0.01). Do Not Reject. p -Value Solution (p Value is between .025 and .05) ³ (a = 0.01). Do Not Reject. p Value = [.025, .05] Reject a = 0.01 t35 1.80 2.4377 Test Statistic 1.80 is in the Do Not Reject Region Bina Nusantara University 23

Hypothesis Tests About the Difference Between the Means of Two Populations: Independent Samples Hypotheses H0: 1 - 2 < 0 H0: 1 - 2 > 0 H0: 1 - 2 = 0 Ha: 1 - 2 > 0 Ha: 1 - 2 < 0 Ha: 1 - 2  0 Test Statistic Large-Sample Small-Sample where, Bina Nusantara University 24

Critical Value untuk statistik uji z: Ha wilayah kritik (tolak H0) 1 - 2 > d0 z > z 1 - 2 < d0 z < -z 1 - 2  d0 Bina Nusantara University 25

Critical Value untuk statistik uji t: Ha wilayah kritik 1 - 2 > d0 t > t 1 - 2 < d0 t < -t 1 - 2  d0 v = n1 – n2 - 2 derajat bebas Bina Nusantara University 26

Contoh Soal: Specific Motors Hypothesis Tests About the Difference Between the Means of Two Populations: Small-Sample Case Rejection Rule Reject H0 if t > 1.734 (a = 0.05, d.f. = 18) Test Statistic where: Bina Nusantara University 27 27