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11.2 One-Tailed Tests about a Population Mean: Large Sample Case Example (continue): Suppose the population variance is 2 0.18 2 . Also, suppose a random sample of 36 cans of coffee is selected. Intuitively, if H 0 : 3 is true, have large value 2 ( X N , ). n For example, as x should 7 0.18 2 X N 7, 36 . Then, it is very likely that the sample mean x , should have value close to 7 and therefore should be much larger than 3. On the other hand, if x is much larger than 3, it seems to indicate that H 0 : 3 is a sensible hypothesis. That is, not rejection of H 0 : 3 might be a sensible choice. Thus, a sensible statistical procedure would be where c reject H 0 : x 3 c not reject H 0 : x 3 c is some constant. Next Question: how to determine the value of c ? Answer: control (the probability of making type I error) to determine the value of c . Suppose H0 0.01 is the predetermined Type I error (the level of significance). As is true, for ease of explanation, suppose 0.18 2 3, X N 3, . 36 1 Then, 0.01 the probabilit y that wro ngly reject H 0 P( H 0 is true but is rejected) P 3; X 3 c X 3 c P 0.18 0.18 36 36 c 0.18 36 c P Z 0.18 36 z 0.01 2.33 c 2.33 0.03 0.0699 . Thus, a sensible statistical procedure would be reject H 0 : x 3 0.0699 not reject H 0 : x 3 0.0699 with P 3; reject H 0 PX 3 0.0699; 3 0.01 . Based on the above statistical procedure, we can use sample mean x to test the hypothesis. For example, if x 2 x 3 2 3 1 0.0699 reject H 0 if x 2.92 x 3 2.92 3 0.08 0.0699 reject H 0 if x 2.97 x 3 2.97 3 0.03 0.0699 not reject H 0 if x 4 x 3 4 3 1 0.0699 not reject H 0 . Note: Even as the true mean 3 , not 3 , the Type I error is still smaller than 0.01, i.e., PX 3 0.0699; 3 0.01 . 2 That is, the above statistical procedure can control the Type I error u 3. less than or equal to 0.01 for any Note: The hypothesis in this example is H 0 : 0 3 vs. H a : 0 3 . The resulting statistical procedure is reject H 0 : x 3 0.0699 not reject H 0 : x 3 0.0699 reject H 0 : not reject H 0 : x 3 x 0 z z0.01 n 0.18 36 x 0 z0.01 n 0.0699 0.18 36 General Case: as n 30 and level of significance As is known, let z x 0 X x 0 . n (I): H 0 : 0 vs. H a : 0 Then, reject H 0 : z z not reject H 0 : z z (II): H 0 : 0 vs. 3 H a : 0 Then, reject H 0 : z z not reject H 0 : z z As is unknown, z x 0 x 0 sX s . n (I): H 0 : 0 vs. H a : 0 Then, reject H 0 : z z not reject H 0 : z z (II): H 0 : 0 vs. H a : 0 Then, reject H 0 : z z not reject H 0 : z z Example 1: H 0 : 600 Suppose vs. H a : 600 . n 36, x 605, s 12 . What is the conclusion as 0.05 ? [solution:] z x 0 605 600 5 2.5 z z0.05 1.64 . s 12 2 n 36 reject H0 4 Example 2: n 50 A sample with provides a sample mean of 36 and sample standard deviation of 12. Consider the following hypothesis test H 0 : 30 vs. H a : 30 . Using a critical value, test the hypothesis at the 10% level of significance (classical approach). [solution:] n 50, 0.1, x 36, s 12, 0 30, z Therefore, we reject x 0 36 30 3.535 z 0.1 1.28 s 12 50 n H0 . Note: z in (I) or z in (II) are called critical value. Online Exercise: Exercise 11.2.1 5