# Chapter 9, Part B Hypothesis Tests. Learning objectives

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1 Chater 9, Part B Hyothesis Tests Slide 1 Learning objectives 1. Able to do hyothesis test about Poulation Proortion 2. Calculatethe Probability of Tye II Errors 3. Understand ower of the test 4. Determinethe Samle Size for Hyothesis Tests About a Poulation Mean Slide 2 1

2 A Summary of Forms for Null and Alternative Hyotheses About a Poulation Proortion The equality art of the hyotheses always aears in the null hyothesis. In general, a hyothesis test about the value of a oulation roortion must take one of the following three forms (where 0 is the hyothesized value of the oulation roortion). H : H 0 0 : a < 0 H H 0 : 0 H 0 : = 0 : a > 0 H : a 0 One-tailed (lower tail) One-tailed (uer tail) Two-tailed Slide 3 Tests About a Poulation Proortion Test Statistic z = 0 σ where: 0 ( 1 0 ) σ = n assuming n > 5 and n(1 ) > 5 Slide 4 2

3 Tests About a Poulation Proortion Rejection Rule: Value Aroach Reject H 0 if value < α Rejection Rule: Critical Value Aroach H 0 : < 0 Reject H 0 if z > z α H 0 : > 0 H 0 : = 0 Reject H 0 if z < -z α Reject H 0 if z < -z α/2 or z > z α/2 Slide 5 Two-Tailed Test About a Poulation Proortion Examle: National Safety Council For a Christmas and New Year s week, the National Safety Council estimated that 500 eole would be killed and 25,000 injured on the nation s roads. The NSC claimed that 50% of the accidents would be caused by drunk driving. Slide 6 3

4 A samle of 120 accidents showed that 67 were caused by drunk driving. Use these data to test the NSC s claim with α =.05. Two-Tailed Test About a Poulation Proortion Examle: National Safety Council Slide 7 Two-Tailed Test About a Poulation Proortion Value and Critical Value Aroaches 1. Determine the hyotheses. 2. Secify the level of significance. H 0: =.5 H a:.5 α = Comute the value of the test statistic. a common error is using in this formula σ (1 ).5(1.5) = = = n = = = 0 (67/120).5 z = = = σ Slide 8 4

5 Two-Tailed Test About a Poulation Proortion Value Aroach 4. Comute the -value. For z = 1.28, cumulative robability =.8997 value = 2(1.8997) = Determine whether to reject H 0. Because value =.2006 > α =.05, we cannot reject H 0. Slide 9 Two-Tailed Test About a Poulation Proortion Critical Value Aroach 4. Determine the critical value and rejection rule. For α/2 =.05/2 =.025, z.025 = 1.96 Reject H 0 if z < or z > Determine whether to reject H 0. Because > and < 1.96, we cannot reject H 0. Slide 10 5

6 In-class exercise #35 (.368) (two-tail test) #36 (.368) (one-tail test) Slide 11 Statistical ower Tye II error As said earlier (Part a.), the conclusion that the research hyothesis is truecomes from the test that contradict (or reject) the nullhyothesis. When a test fails to reject a null hyothesis (i.e., accets H 0 ), tye II error (β) needs to be checked. A Tye II error is acceting H 0 when it is false. When a test reject H 0, ossible error is Tye I error that can be controlled easily. Article by Sawyer and Ball (1981) Slide 12 6

7 Calculating the Probability of a Tye II Error in Hyothesis Tests About a Poulation Mean 1. Formulate the null and alternative hyotheses. 2. Using the critical value aroach, use the level of significance α to determine the critical value and the rejection rule for the test. 3. Using the rejection rule, solve for the value of the samle mean corresonding to the critical value of the test statistic. Slide 13 Calculating the Probability of a Tye II Error in Hyothesis Tests About a Poulation Mean 4. Use the results from ste 3 to state the values of the samle mean that lead to the accetance of H 0 ; this defines the accetance region. 5. Using the samling distribution of x for a value of µ satisfying the alternative hyothesis, and the accetance region from ste 4, comute the robability that the samle mean will be in the accetance region. (This is the robability of making a Tye II error at the chosen level of µ.) Slide 14 7

8 Calculating the Probability of a Tye II Error Examle: Metro EMS (revisited) Recall that the resonse times for a random samle of 40 medical emergencies were tabulated. The samle mean is minutes. The oulation standard deviation is believed to be 3.2 minutes. The EMS director wants to erform a hyothesis test, with a.05 level of significance, to determine whether or not the service goal of 12 minutes or less is being achieved. Slide 15 Calculating the Probability of a Tye II Error 1. Hyotheses are: H 0 : µ < 12 and H a : µ > Rejection rule is: Reject H 0 if z > Value of the samle mean that identifies the rejection region: x 12 z = / x = We will accet H 0 when x < Slide 16 8

9 Calculating the Probability of a Tye II Error 5. Probabilities that the samle mean will be in the accetance region: µ z = Values of µ 3.2/ 40 β 1-β Slide 17 Calculating the Probability of a Tye II Error Calculating the Probability of a Tye II Error Observations about the receding table: When the true oulation mean µ is close to the null hyothesis value of 12, there is a high robability that we will make a Tye II error. Examle: µ = , β =.9500 When the true oulation mean µ is far above the null hyothesis value of 12, there is a low robability that we will make a Tye II error. Examle: µ = 14.0, β =.0104 Slide 18 9

10 Power of the Test The robability of correctly rejecting H 0 when it is false is called the ower of the test. For any articular value of µ, the ower is 1 β. We can show grahically the ower associated with each value of µ; such a grah is called a ower curve. (See next slide.) Slide 19 Power Curve Probability of Correctly Rejecting Null Hyothesis H 0 False µ Slide 20 10

11 In-class exercise #46 (.374) #50 (.375) Slide 21 Determining the Samle Size for a Hyothesis Test About a Poulation Mean The secified level of significance determines the robability of making a Tye I error. By controlling the samle size, the robability of making a Tye II error is controlled. Slide 22 11

12 Determining the Samle Size for a Hyothesis Test About a Poulation Mean Samling distribution of x when H 0 is true and µ = µ 0 c α Reject H 0 H 0 : µ < µ 0 H a : µ > µ 0 x Note: σ = x σ n µ 0 β Samling distribution of x when H 0 is false and µ a > µ 0 c µ a x Slide 23 Determining the Samle Size for a Hyothesis Test About a Poulation Mean where 2 2 α β σ 2 0 a ( z + z ) n = ( µ µ ) z α = z value roviding an area of α in the tail z β = z value roviding an area of β in the tail σ = oulation standard deviation µ 0 = value of the oulation mean in H 0 µ a = value of the oulation mean used for the Tye II error Note: In a two-tailed hyothesis test, use z α /2 not z α Slide 24 12

13 Relationshi Among α, β, and n Once two of the three values are known, the other can be comuted. For a given level of significance α, increasing the samle size n will reduce β. For a given samle size n, decreasing α will increase β, whereas increasing α will decrease b. Slide 25 Determining the Samle Size for a Hyothesis Test About a Poulation Mean Let s assume that the director of medical services makes the following statements about the allowable robabilities for the Tye I and Tye II errors: If the mean resonse time is µ = 12 minutes, I am willing to risk an α =.05 robability of rejecting H 0. If the mean resonse time is 0.75 minutes over the secification (µ = 12.75), I am willing to risk a β =.10 robability of not rejecting H 0. Slide 26 13

14 Determining the Samle Size for a Hyothesis Test About a Poulation Mean α =.05, β =.10 z α = 1.645, z β = 1.28 µ 0 = 12, µ a = σ = 3.2 ( z ) ( )(3.2) n = = = ( ) ( ) α + z β σ µ 0 µ a Slide 27 In-class exercise #54 (.379) #55 (.379) Slide 28 14

15 End of Chater 9, Part B Slide 29 15

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