1. Introduction to Hypothesis Testing: One Population Value Chapter 8 Handout

2. Chapter 8 Summary Hypothesis Testing for One Population Value: 1.Population Mean (  a.  (population standard deviation) is given (known):  Use z/standard normal/bell shaped distribution b.  (pop std dev) is not given but s (sample std dev) is given  Use student’s t distribution 2.Population proportion (  )  Use z/standard normal/bell shaped distribution 3.Population variance (     Use   (Chi-Square) distribution PS: population standard deviation = 

3. 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?

4. 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.

5. 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 or H A

6. 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

7. 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!

8. 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

9. 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, 

10. 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

11. 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

12. 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

13.   Reduce probability of one error and the other one goes up.  &  Have an Inverse Relationship

14. True Value of Population Parameter  Increases When Difference Between Hypothesized Parameter & True Value Decreases Significance Level   Increases When  Decreases Population Standard Deviation   Increases When   Increases Sample Size n  Increases When n Decreases Factors Affecting Type II Error,       n

15. 3 Methods for Hypotheses Tests Refer to Figure 8-6 (page 299) for a hypothesis test for means  with  (pop. std. dev.) is given: Method 1: Comparing X   X critical) with  X Method 2: Z test, i.e., comparing Z   critical) with Z (or Z statistics or Z calculated) Method 3: Comparing  significance level) with p-value You can modify those three methods for other cases. For example, if  is unknown, you must use student’s t distribution. If you would like to use Method 2, please compare t   t critical) with t (or t statistics or t calculated). Refer to Figure 8-8 (page 303).

16. You always get: Z   critical) from Z distribution t   t critical) from student’s t distribution.      critical) from   distribution You always get: Z or Z calculated or Z statistics from sample (page 299 and Figure 8-6) t or t calculated or t statistics from sample (Figure 8-8, page 299).   or   calculated or   statistics from sample (Figure 8-19, page 322)

17. 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

18. 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

19. 1.State H 0 H 0 :  3 2.State H 1 H 1 :  3.Choose   =.05 4.Choose n n = 100 5.Choose Method: Z Test (Method 2) Hypothesis Testing: Steps Test the Assumption that the true mean # of TV sets in US homes is at least 3.

20. 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)

21. 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)

22. 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

23. 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 _

24. Z.04.06 1.6.4495.4505.4515 1.7.4591.4599.4608 1.8.4671.4678.4686.4738.4750 Z0  Z = 1 1.645.50 -.05.45. 05 1.9.4744 Standardized Normal Probability Table (Portion) What Is Z Given  = 0.05?  =.05 Finding Critical Values: One Tail Critical Value = 1.645

25.  = 0.025 n = 25 Critical Value: 1.645 Test Statistic: Decision: Conclusion: Do Not Reject Ho 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

26. 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

27. 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

28. 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

29.  = 0.05 n = 25 Critical Value: ±1.96 Test Statistic: Decision: Conclusion: Do Not Reject Ho 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

30. Two tail hypotheses tests = 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 Ho. _

31. Assumptions  Population is normally distributed  If not normal, only slightly skewed & a large sample taken Parametric test procedure t test statistic t-Test:  Unknown

32. 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  0.01 level. 368 gm. H 0 :  368 H 1 :  368  is not given,

33.  = 0.01 n = 36, df = 35 Critical Value: 2.4377 Test Statistic: Decision: Conclusion: Do Not Reject Ho at  =.01 No Evidence that True Mean Is More than 368 Z 0 2.4377.01 Reject Example Solution: One Tail H 0 :  368 H 1 :  368

34. Involves categorical variables Fraction or % of population in a category If two categorical outcomes, binomial distribution  Either possesses or doesn’t possess the characteristic Sample proportion (p) Proportions

35. 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 n = 500 were surveyed with x = 25 responses. Solution: Test at the  =.05 significance level.

36.  =.05 n = 500, x = 25 p = x/n = 25/500 = 0.05 Do not reject Ho at Do not reject Ho at  =.05 Z Test for Proportion: Solution H 0 :  .04 H 1 :  .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-   (1 -  ) n =.05-.04.04 (1 -.04) 500 = 1.14 Z0 Reject.025