1. Chapter 10 – Hypothesis Testing What is a hypothesis? A statement about a population that may or may not be true. What is hypothesis testing? A statistical test to prove or disprove a hypothesis. At the end of the test, either the hypothesis is rejected or not rejected.

2. Steps in hypothesis testing There are 5 steps in hypothesis testing: Step 1 – pattern of population distribution (normal) Step 2 – Formulation of hypothesis Null Hypothesis (H 0 ) Alternate Hypothesis (H 1 )

3. Steps in hypothesis testing Step 3 - Level of significance (α ) Alpha is the probability of rejecting a null hypothesis when it is true. Alpha is also known as the level of risk. It is described in terms of percent or decimals (5% or 0.05, 1% or 0.01, and so on.) Step 4 - Test statistic It is a quantity that is used to compare with critical Z from appendix D to determine if a null hypothesis is to be rejected or not.

4. Steps in hypothesis testing This quantity is the z-statistic _ Z = (x –  ) / (  n). Step 5 - Decision Rule: It is a rule by which a null hypothesis is rejected or not rejected. It always takes the form: Reject H 0 if Z-statistic is > Critical Z or < - Critical Z Critical Z-value is obtained from appendix D.

5. Steps in hypothesis testing Decision: Reject H 0 or Do not reject H 0 Hypothesis tests can be one-tailed or two- tailed depending on how the alternate hypothesis is written.

6. Let’s take an example Suppose: H : μ = 100 0 H : μ > 100 1 This is a one tailed test. Another example: H : μ = 100 0 H : μ < 100 1

7. An example of a two tailed test Suppose: H : μ = 100 0 H : μ ≠ 100 1 This is a two tailed test. Exercises from book: Problem 7(P age 239), Problem 8 (Page 240), Problem 9 (Page 240)

8. Exercises from Book Problem 7 (pg 283): a. ONE-TAILED b. ONE-TAILED c. ONE-TAILED d. ONE-TAILED e. TWO-TAILED Problem 8 (pg 283): a. Ho : μ = 500, H 1 : μ ≠ 500 b. Ho : μ = 500, H 1 : μ > 500 c. Ho : μ = 500, H 1 : μ < 500

9. Exercises from book Problem 9 (pg 283) Ho : μ = 60; H 1 : μ > 60 Problem 10 (pg 283) Ho : μ = 1000; H 1 : μ ≠ 1000 Ho : μ ≤ 1000; H 1 : μ > 1000

10. Testing for population mean Such testing can be made under 2 situations: (1) When population standard deviation is known and (2) when population standard deviation is unknown. This chapter discusses the first. Refer to example problem 10-2 (page 241) Two hypotheses are: H : μ = 10 0 H : μ < 10 1

11. Testing for population mean Alpha ( α ) is 5% (or 0.05). Critical Z from appendix D is –1.645 Reject H 0 if Z-statistic < - 1.645 Test statistic (Z-statistic) _ X – μ 8.8 - 10 Z = -------- = ---------- = - 2.0 σ / √ n 2.4 / 4

12. Testing for population mean Decision: Reject H 0 What does it mean? The new manufacturing process reduces tar content of cigarettes. Example problem 10-4.

13. Testing for population mean H : μ = 1000 0 H : μ > 1000 1 α = 0.025 Critical Z = 1.96 (From Appendix D) _ X – μ 1050 - 1000 Test Statistic Z = --------- = -------------- = 2.5 σ / n 100 / 5 Decision: Reject H 0 What does it mean? The new manufacturing process increases life of fuses.

14. This is a two tailed test Example Problem 10-7 (page 248) H : μ = 100 0 H : μ ≠ 100 1 Critical Z’s are – 2.575 and + 2.575 Decision Rule: Reject H 0 if Z-statistic > + 2.575 or < - 2.575 _ X - 100 Z-statistic = -------------- = 1.667 Decision: Do not reject H √ (225 / 25) 0 IQs of the school district students are no different from the national average

15. Example problem 13 Problem 13 (page 251) H : μ ≥ 30 0 H : μ < 30 1 Critical Z = - 2.326 Decision Rule: Reject H 0 if Z-statistic ≤ - 2.326 27 - 30 Z-Statistic = ------------- = - 5.0 6 / √ 100 Decision: Reject H 0 What does it mean? Vendor’s claim that mean weight of his chickens is at least 30 ounces is rejected.

16. Testing for difference between 2 means These tests are performed when one wants to compare 2 groups of populations. For example, is these a difference between average GPAs of students of two universities? Or, are average monthly incomes the same between two groups of people? The testing procedure is similar to what has been presented in the previous section.

17. Testing for difference between 2 means The null hypothesis is always of the form: H : μ = μ 0 1 2 The alternative hypothesis can be either of the 3 forms: μ > μ or μ < μ or μ ≠ μ 1 2 1 2 1 2

18. Testing for difference between 2 means Formula 10-1, p. 252 The decision rule is the same as before Example problem 10-8 (page 252-253) H : μ = μ H : μ ≠ μ 0 1 2 1 1 2

19. Testing for difference between 2 means 0.8 – 1.0 Z- statistic = --------------------------- = - 1.0 √ (0.36/25) + (0.64/25) At α = 0.05, critical Z = -1.96 and +1.96 Decision: Do not reject H 0 Interpretation: The avg. nicotine contents of two brands of cigarettes are equal.

20. Problem 4, pg 255 Is average hourly output of male workers less than that of female workers? This is a test involving difference of means between two population groups. Data are given in the table Ho: μ 1 = μ 2 ; H 1 : μ 1 < μ 2 (μ 1 indicates output of males)

21. Z-statistic = (150-153)/√[(70/36) + (74/36)] = -1.5 At α = 0.05, critical Z is -1.645 Decision: Do not reject Ho Interpretation: Average outputs of male and female workers are the same.

22. Testing for population proportion In some cases, we are interested in population proportions rather than population means. For example, are more than 50% of CSULB students females? Or, is a new medicine more than 60% effective? Two hypothesis are: H : p = a given value 0 H : p > a given value OR 1 p < a given value OR p ≠ a given value Example 10-11 (Page 257-258)

23. The test statistic Z-statistic = (X – n.p)/√[(n.p(1-p)] The decision rule is the same as before.

24. Example problem 10-12 (pg 259) A TV program attracts 50% audience. A new anchorperson has been hired. Does the new hire increase audience level? Use 5% level of significance. Ho: p = 0.5; H 1 : p > 0.5 Given n = 100; critical Z = 1.645 P = 55/100

25. Problem 10-12, cont. Z-statistic = (55-50)/√[(100)(0.5)(0.5)] = 1 Decision: Do not reject Ho Interpretation: –The audience level has remained the same (at 50%) after hiring the new anchorperson.

26. Problem 6, pg 261 An auto manufacturer claims that 20% of customers prefer his products. In a sample of 100 customers, 15 indicate their preferences for his products. At 5% level of significance, can we support his claim? Ho: p = 0.2; H 1 : p ≠ 0.2

27. Problem 6, cont. Z-statistic = (15-20)/√[(20)(0.08)] = -1.25 Decision: Do not reject Ho. Interpretation –The auto manufacturer’s claim cannot be disputed.

28. Hypothesis Testing Strength of rejection (p-value) –When we reject a null hypothesis, there is a quantity that describes the strength of rejection. This quantity is called the p-value. –The lower the p-value, the greater the strength of refection. In other words, the lower the p-value, the greater the strength in the rejection.

29. What is p-value? It is the area to right right of the calculated value of Z-statistic when the area of rejection is on the right side of the normal curve. When the area of rejection is on the left side of the normal curve, the p-value is the area to the left of the calculated value of the Z-statistic. If there are two area of rejection, add the area to the right of Z- statistic and the area to the left of Z- statistic.

30. How do we find the P-value? Let’s take an example problem: Test if the mean waiting time is less than 3 minutes. α = 0.05 μ = 3, σ = 1, n=50, X=2.75 Ho: μ = 3; H 1 : μ < 3 Decision rule: Reject Ho if z-statistic < critical Z Z-statistic = (2.75 - 3)/(1/√50) = -1.768 Critical Z - -1.645 so we reject Ho. Interpretation: Mean waiting time is < 3 minutes.

31. What is the p-value? Find the area to the left of -1.768. This area is the p-value. P-value = 1.0 – 0.9616 = 0.0384