1.  What is Hypothesis Testing?  Testing for the population mean  One-tailed testing  Two-tailed testing  Tests Concerning Proportions  Types of Errors Hypothesis Testing

2. A Hypothesis is a statement about the value of a population parameter developed for the purpose of testing. Examples of hypotheses made about a population parameter are: – The mean monthly income for systems analysts is $3,625 (Statement on a population mean μ). – Twenty percent of all restaurant customers at return for another meal within a month (Statement on a population proportion π). Hypothesis Testing

3. Given or implied by the problem Decide if the z or the t distribution is to be used. Find the Critical Values and the accept/reject regions. Find the z or t value of the sample and check if it falls in an accept/reject region. Hypothesis testing is a procedure, based on sample evidence and probability theory, used to determine whether the hypothesis is a reasonable statement and should not be rejected, or is unreasonable and should be rejected.

4. o H 0 : null hypothesis and H 1 : alternate hypothesis o H 0 and H 1 are mutually exclusive and collectively exhaustive o H 0 is always presumed to be true o H 1 has the burden of proof Hypothesis Testing

5. In problem solving, look for key words and convert them into symbols. Some key words include: “improved, better than, as effective as, different from, has changed,… H 0 : μ = value H 1 : μ ≠ value H 0 : μ < value H 1 : μ > value H 0 : μ > value H 1 : μ < value Is there a Change? has not changed Possible Keywords:3 possible situations: is larger than is better than has improved is less than is less effective Hypothesis Testing

6. H 0 : μ = value H 1 : μ ≠ value Reject H 0 if : Hypothesis Testing (Two-Tailed) 0 Critical Value = -Z α/2 Accept H 0 α/2 Critical Value = Z α/2

7. H 0 : μ ≥ value H 1 : μ < value Reject H 0 if : Hypothesis Testing (One-Tailed) H 0 : μ ≤ value H 1 : μ > value 0 CV= Z α/2 Accept H 0 α 0 CV= -Z α/2 Accept H 0 α

8. Hypothesis Testing (for the mean μ)

9. Hypothesis Testing (for the proportion π)

10.  Example 26, page 356 50% of students change their major within the first year. A random sample of 100 students revealed that 48 students changed their major within the first year. Has there been a significant decrease in the number of students who changed their major? Test at a 0.05 level of significance. H 0 : π ≥ 0.5 H 1 : π < 0.5

11.  Example 26, page 356 p=48/100 = 0.48 α = 0.05 => z of CV = -1.65 => H 0 is rejected if z<-1.65 -0.4 > -1.65 => H o is not rejected. The proportion of students changing their major has not changed. H 0 : π ≥ 0.5 H 1 : π < 0.5

12. Researcher Null Accepts Rejects Hypothesis H o H o H o is true H o is false Correct decision Type I error  Type II Error  Correct decision Decisions and Consequences

13.  Is the probability that the null hypothesis is NOT rejected when it is actually false.  Example (page 356): At a plant manufacturing pins using steel, past experiences indicate that mean tensile strength of all incoming shipments µ 0 is 10,000 psi and standard deviation σ is 400 psi.  To make a decision, the manufacturer sets up the following rule to a quality control inspector: Take a sample of 100 steel bars.  At the.05 significance level, if the sample mean X-bar strength falls between 9,922 psi and 10,078 psi, accept a lot. Otherwise the lot is rejected.  Suppose that the unknown population mean of an incoming lot, designated by µ 1, is really 9,900 psi. What is the probability that the inspector will fail to reject the shipment (type II error)? Type II Error (β)

15. Then P(z>0.55)=.2088. So the probability of a type II error or β is 0.5-.2088=.2912. Always draw a picture of the normal curve and areas to solve! The z value of is:

16. Chapter 11 Two-Sample Tests of Hypothesis

17. Our Objectives  Conduct a test of hypothesis about the difference between two independent population means.  Conduct a test of a hypothesis about the difference between two population proportions.  Conduct a test of a hypothesis about the mean difference between paired or dependent observations.  Understand the difference between dependent and independent samples.

18. Two-Sample Tests of Hypothesis  We take samples from two populations and compare the population means.  In the one-sample test of hypothesis, we took a sample from a population and compared the sample statistic to the population parameter.  Example: Is there a difference in the mean value of residential real estate sold by male agents and female agents in a particular area?

19. Two-Sample Test of Hypothesis: Independent Samples  Example A financial accountant wishes to know whether there is a difference in the mean rate of return for high yield mutual funds and global mutual funds.  There are two independent populations: high yield mutual funds, and the global mutual funds.  If there is a difference between the population means, then we expect that there is a difference between the sample means.  If the size of the two samples is more than 30, we can reason that the distribution of the difference in the sample means is Normal.  Mean of the distribution of the differences:  If zero, we conclude that there is no difference in the two populations.  If positive or negative value, we conclude that two populations do not have the same mean.

20. H 0 : μ 1 = μ 2 H 1 : μ 1 ≠ μ 2 Two-Sample Test of Hypothesis: Independent Samples with known population SD σ H 0 : μ 1 - μ 2 =0 H 1 : μ 1 - μ 2 ≠ 0

21. Standardize the distribution of the differences. The test statistic for the difference between two means is: The variance of the distribution of differences in sample means is: Two-Sample Test of Hypothesis: Independent Samples with known population SD σ

22. Given or implied by the problem Decide if the z or the t distribution is to be used. Find the Critical Values and the accept/reject regions. Find the z or t value of the sample and check if it falls in an accept/reject region.

23.  Example 2, page 374 First population: A sample of 65 observations is selected. Population SD = 0.75. Sample mean = 2.67. Second population: A sample of 50 observations is selected. Population SD = 0.66. Sample mean = 2.59. Use a 0.08 significance level. H 0 : μ 1 ≤ μ 2 H 1 : μ 1 > μ 2

24.  Example 2 a) Is this a one-tailed or a two-tailed test? b) State the decision rule. c) Compute the value of the test statistic d) What is your decision regarding H 0 ? e) What is the p-value? a) It is a one-tailed test. b) For α = 0.08 and a one tailed test, then we reject H 0 if z>1.41 (CV=1.41). c)

25.  Example 2 d) 0.607 < 1.41, We fail to reject H 0 e) p-value of the sample is

26. Standardize the distribution of the differences. The test statistic for the difference between two proportions is: Two-Sample Test of Hypothesis (Proportions)

27. p 1 is the first sample proportion (p 1 =x 1 /n 1 ) p 2 is the second sample proportion (p 2 =x 2 /n 2 ) p c is the pooled proportion Two-Sample Test of Hypothesis (Proportions)  The pooled estimate of the population proportion is computed using the formula:

28.  Example 12, page 378 Single People: A sample of 400 people is selected. 120 had at least one accident in the past three years. Married People: A sample of 600 people is selected. 150 had at least one accident in the past three years. Use a 0.05 significance level. Is there a significant difference in the proportion of single and married people having accidents?

29.  Example 12 0.05 significance level, z of Critical Values: z = 1.96 and z=-1.96 (two tailed test). Accept Region is between 1.96 and -1.96. H 0 : π m = π s H 1 : π m ≠ π s

30.  Example 12  H o is not rejected. There is no difference in the proportion of married and single drivers who have accidents.

31. o Use the following t distribution if: o Independent Samples o Both samples have unknown but equal population SD o At least one of the samples is less than 30 Two-Sample Test of Hypothesis: Independent Samples with unknown population SD σ and at least one of the samples is less than 30.

32.  We use the t statistic. We compute the t value using the formula: S p squared is pooled estimate of population variance. We use n 1 +n 2 -2 degrees of freedom. So to find the value of t, 3 steps are performed: 1.compute s 1 and s 2 2.compute s p 3.determine t

33.  Pooled variance is computed using the formula: Where s 1 squared is the variance of the 1 st sample; s 2 squared is the variance of the 2 nd sample.

34.  Example 15, page 384 Men Examination Scores: 72 69 98 66 85 76 79 80 77 Women Examination Scores: 81 67 90 78 81 80 76 Is it reasonable to conclude that women score higher than men? Use the 0.01 significance level.

35.  Example 15, page 384 H o :  f   m H 1 :  f >  m Use Appendix B2 to obtain Critical Values. For significance level 0.01, one tailed test and df=n 1 +n 2 -2=14 we obtain a t=2.624 for the Critical Value. Accept H 0 if the sample t is less than 2.624. S f =6.88 S m =9.49

36. t < 2.624 so we accept H 0

37. o Use the following t distribution if: o Independent Samples o Both samples have unknown but can not assume equal population SD o At least one of the samples is less than 30 Two-Sample Test of Hypothesis: Independent Samples with unknown population SD σ and at least one of the samples is less than 30.

38.  We use the t statistic. We compute the t value using the formula:  Use the following for degree of freedom (round down if not an integer):

39.  Example 22, page 388 Klein Models: 5.0, 4.5, 3.4, 3.4, 6.0. 3.3, 4.5, 4.6, 3.5, 5.2, 4.8, 4.4, 4.6, 3.6, 5.0 Clairborne Models: 3.1, 3.7, 3.6, 4.0, 3.8, 3.8, 5.9, 4.9, 3.6, 3.6, 2.3, 4.0 Is it reasonable to conclude that Clairborne Models earn more? Use the 0.05 significance level and assume the population standard deviations are not the same.

40.  Example 15, page 384 H o :  k   c H 1 :  k >  c Use Appendix B2 to obtain Critical Values. For significance level 0.05, one tailed test and df=22.

41. we obtain a t=1.717 for the Critical Value. Accept H 0 if the sample t is less than 2.624. t < 1.717 so we fail to reject the null hypothesis