# © 1999 Prentice-Hall, Inc. Chap. 8 - 1 Chapter Topics Hypothesis Testing Methodology Z Test for the Mean (  Known) p-Value Approach to Hypothesis Testing.

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© 1999 Prentice-Hall, Inc. Chap. 8 - 1 Chapter Topics Hypothesis Testing Methodology Z Test for the Mean (  Known) p-Value Approach to Hypothesis Testing Connection to Confidence Interval Estimation One Tail Test t Test of Hypothesis for the Mean Z Test of Hypothesis for the Proportion

© 1999 Prentice-Hall, Inc. Chap. 8 - 2 A hypothesis is an assumption about the population parameter.  A parameter is a Population mean or proportion  The parameter must be identified before analysis. I assume the mean GPA of this class is 3.5! © 1984-1994 T/Maker Co. What is a Hypothesis?

© 1999 Prentice-Hall, Inc. Chap. 8 - 3 States the Assumption (numerical) to be tested e.g. The average # TV sets in US homes is at least 3 (H 0 :  3) Begin with the assumption that the null hypothesis is TRUE. (Similar to the notion of innocent until proven guilty) The Null Hypothesis, H 0 Refers to the Status Quo Always contains the ‘ = ‘ sign The Null Hypothesis may or may not be rejected.

© 1999 Prentice-Hall, Inc. Chap. 8 - 4 Is the opposite of the null hypothesis e.g. The average # TV sets in US homes is less than 3 (H 1 :  < 3) Challenges the Status Quo Never contains the ‘=‘ sign The Alternative Hypothesis may or may not be accepted The Alternative Hypothesis, H 1

© 1999 Prentice-Hall, Inc. Chap. 8 - 5 Steps:  State the Null Hypothesis (H 0 :   3)  State its opposite, the Alternative Hypothesis (H 1 :  < 3) Hypotheses are mutually exclusive & exhaustive Sometimes it is easier to form the alternative hypothesis first. Identify the Problem

© 1999 Prentice-Hall, Inc. Chap. 8 - 6 Population Assume the population mean age is 50. (Null Hypothesis) REJECT The Sample Mean Is 20 Sample Null Hypothesis Hypothesis Testing Process No, not likely!

© 1999 Prentice-Hall, Inc. Chap. 8 - 7 Sample Mean  = 50 Sampling Distribution It is unlikely that we would get a sample mean of this value...... if in fact this were the population mean.... Therefore, we reject the null hypothesis that  = 50. 20 H0H0 Reason for Rejecting H 0

© 1999 Prentice-Hall, Inc. Chap. 8 - 8 Defines Unlikely Values of Sample Statistic if Null Hypothesis Is True  Called Rejection Region of Sampling Distribution Designated  (alpha)  Typical values are 0.01, 0.05, 0.10 Selected by the Researcher at the Start Provides the Critical Value(s) of the Test Level of Significance, 

© 1999 Prentice-Hall, Inc. Chap. 8 - 9 Level of Significance,  and the Rejection Region H 0 :   3 H 1 :  < 3 0 0 0 H 0 :   3 H 1 :  > 3 H 0 :   3 H 1 :   3    /2 Critical Value(s) Rejection Regions

© 1999 Prentice-Hall, Inc. Chap. 8 - 10 Type I Error  Reject True Null Hypothesis  Has Serious Consequences  Probability of Type I Error Is  Called Level of Significance Type II Error  Do Not Reject False Null Hypothesis  Probability of Type II Error Is  (Beta) Errors in Making Decisions

© 1999 Prentice-Hall, Inc. Chap. 8 - 11 H 0 : Innocent Jury Trial Hypothesis Test Actual Situation Verdict InnocentGuilty Decision H 0 TrueH 0 False Innocent CorrectError Do Not Reject H 0 1 -  Type II Error (  ) Guilty Error Correct Reject H 0 Type I Error (  ) Power (1 -  ) Result Possibilities

© 1999 Prentice-Hall, Inc. Chap. 8 - 12   Reduce probability of one error and the other one goes up.  &  Have an Inverse Relationship

© 1999 Prentice-Hall, Inc. Chap. 8 - 13 Convert Sample Statistic (e.g., ) to Standardized Z Variable Compare to Critical Z Value(s)  If Z test Statistic falls in Critical Region, Reject H 0 ; Otherwise Do Not Reject H 0 Z-Test Statistics (  Known) Test Statistic X

© 1999 Prentice-Hall, Inc. Chap. 8 - 14 Probability of Obtaining a Test Statistic More Extreme  or  ) than Actual Sample Value Given H 0 Is True Called Observed Level of Significance  Smallest Value of a H 0 Can Be Rejected Used to Make Rejection Decision  If p value  Do Not Reject H 0  If p value < , Reject H 0 p Value Test

© 1999 Prentice-Hall, Inc. Chap. 8 - 15 1.State H 0 H 0 :  3 2.State H 1 H 1 :  3.Choose   =.05 4.Choose n n = 100 5.Choose Test: Z Test (or p Value) Hypothesis Testing: Steps Test the Assumption that the true mean # of TV sets in US homes is at least 3.

© 1999 Prentice-Hall, Inc. Chap. 8 - 16 6. Set Up Critical Value(s) Z = -1.645 7. Collect Data 100 households surveyed 8. Compute Test Statistic Computed Test Stat.= -2 9. Make Statistical Decision Reject Null Hypothesis 10. Express Decision The true mean # of TV set is less than 3 in the US households. Hypothesis Testing: Steps Test the Assumption that the average # of TV sets in US homes is at least 3. (continued)

© 1999 Prentice-Hall, Inc. Chap. 8 - 17 Assumptions  Population Is Normally Distributed  If Not Normal, use large samples  Null Hypothesis Has  or  Sign Only Z Test Statistic: One-Tail Z Test for Mean (  Known)

© 1999 Prentice-Hall, Inc. Chap. 8 - 18 Z 0  Reject H 0 Z 0 0  H 0 :  H 1 :  < 0 H 0 :  0 H 1 :  > 0 Must Be Significantly Below  = 0 Small values don’t contradict H 0 Don’t Reject H 0 ! Rejection Region

© 1999 Prentice-Hall, Inc. Chap. 8 - 19 Does an average box of cereal contain more than 368 grams of cereal? A random sample of 25 boxes showed X = 372.5. The company has specified  to be 15 grams. Test at the  0.05 level. 368 gm. Example: One Tail Test H 0 :  368 H 1 :  > 368 _

© 1999 Prentice-Hall, Inc. Chap. 8 - 20 Z.04.06 1.6.5495.5505.5515 1.7.5591.5599.5608 1.8.5671.5678.5686.5738.5750 Z0  Z = 1 1.645.50 -.05.45. 05 1.9.5744 Standardized Normal Probability Table (Portion) What Is Z Given  = 0.05?  =.05 Finding Critical Values: One Tail Critical Value = 1.645

© 1999 Prentice-Hall, Inc. Chap. 8 - 21  = 0.025 n = 25 Critical Value: 1.645 Test Statistic: Decision: Conclusion: Do Not Reject at  =.05 No Evidence True Mean Is More than 368 Z0 1.645.05 Reject Example Solution: One Tail H 0 :  368 H 1 :  > 368

© 1999 Prentice-Hall, Inc. Chap. 8 - 22 Z0 1.50 p Value. 0668 Z Value of Sample Statistic From Z Table: Lookup 1.50.9332 Use the alternative hypothesis to find the direction of the test. 1.0000 -.9332.0668 p Value is P(Z  1.50) = 0.0668 p Value Solution

© 1999 Prentice-Hall, Inc. Chap. 8 - 23 0 1.50 Z Reject (p Value = 0.0668)  (  = 0.05). Do Not Reject. p Value = 0.0668  = 0.05 Test Statistic Is In the Do Not Reject Region p Value Solution

© 1999 Prentice-Hall, Inc. Chap. 8 - 24 Does an average box of cereal contains 368 grams of cereal? A random sample of 25 boxes showed X = 372.5. The company has specified  to be 15 grams. Test at the  0.05 level. 368 gm. Example: Two Tail Test H 0 :  368 H 1 :  368

© 1999 Prentice-Hall, Inc. Chap. 8 - 25  = 0.05 n = 25 Critical Value: ±1.96 Test Statistic: Decision: Conclusion: Do Not Reject at  =.05 No Evidence that True Mean Is Not 368 Z 0 1.96.025 Reject Example Solution: Two Tail -1.96.025 H 0 :  386 H 1 :  386

© 1999 Prentice-Hall, Inc. Chap. 8 - 26 Connection to Confidence Intervals For X = 372.5oz,  = 15 and n = 25, The 95% Confidence Interval is: 372.5 - (1.96) 15/ 25 to 372.5 + (1.96) 15/ 25 or 366.62    378.38 If this interval contains the Hypothesized mean (368), we do not reject the null hypothesis. It does. Do not reject. _

© 1999 Prentice-Hall, Inc. Chap. 8 - 27 Assumptions  Population is normally distributed  If not normal, only slightly skewed & a large sample taken Parametric test procedure t test statistic t-Test:  Unknown

© 1999 Prentice-Hall, Inc. Chap. 8 - 28 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   15. Test at the  0.01 level. 368 gm. H 0 :  368 H 1 :  368  is not given,

© 1999 Prentice-Hall, Inc. Chap. 8 - 29 Example:Z Test for Proportion Problem: A marketing company claims that it receives 4% responses from its Mailing. Approach: To test this claim, a random sample of 500 were surveyed with 25 responses. Solution: Test at the  =.05 significance level.

© 1999 Prentice-Hall, Inc. Chap. 8 - 30  =.05 n = 500 Do not reject at Do not reject at  =.05 Z Test for Proportion: Solution H 0 : p .04 H 1 : p .04 Critical Values:  1.96 Test Statistic: Decision: Conclusion: We do not have sufficient evidence to reject the company’s claim of 4% response rate. Z  p - p p (1 - p) n s =.04 -.05.04 (1 -.04) 500 = 1.14 Z0 Reject.025

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