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Unformatted text preview: C HAPTER 11 T WO S AMPLE T ESTS OF H YPOTHESIS 1. a. Twotailed test b. Reject H o if z <  2.05 or z > 2.05 c. 2.59, found by 2 2 102 99 5 6 40 50 z = + d. Reject H o and accept H 1 e. p = 0.0096, found by 2(0.5000 – 0.4952) This pvalue shows there is a very small chance of finding a z statistic this extreme when the null hypothesis is true. 2. a. Onetailed test b. Reject H o and accept H 1 if z > 1.41 c. 0.607 found by 2 2 2.67 2.59 (0.75) (0.66) 65 50 z = + d. Fail to reject H o e. p = 0.2709, found by 0.5000 – 0.2291 This pvalue shows there is a very large chance of finding a z statistic this extreme when the null hypothesis is true. 3. Step 1: H o : μ 1 ≥ μ 2 H 1 : μ 1 < μ 2 Step 2: The 0.05 significance level was chosen Step 3: Reject H o and accept H 1 if z <  1.65 Step 4: 0.94, found by 2 2 7.6 8.1 (2.3) (2.9) 40 55 z = + Step 5: Fail to reject H o . Babies using the Gibbs brand did not gain less weight. p = 0.1736 found by 0.5000 – 0.3264 4. Step 1: H o : μ c = μ p H 1 : μ c ≠ μ p Step 2: The 0.05 significance level was chosen Step 3: Reject H o and accept H 1 if z less than  1.96 or greater than 1.96 Step 4: 1.53, found by 2 2 370 380 (30) (26) 35 40 z = + Step 5: Fail to reject H o . There is no difference in the mean number of miles traveled per month. The pvalue is 0.126 found by 2(0.5000 – 0.4370) Chapter 11 98 5. Twotailed test. Because we are trying to show a difference exists between two means. Reject H o if z <  2.58 or z > 2.58 2.66, found by 2 2 31.4 34.9 (5.1) (6.7) 32 49 z = + Reject H o at the 0.01 level. There is a difference in the mean turnover rate. The pvalue is 0.0078 found by 2(0.5000 – 0.4961). 6. H o : μ 1 ≤ μ 2 H 1 : μ 1 > μ 2 If z > 2.05, reject H o . 2 2 20.75 19.80 2.09 (2.25) (1.90) 40 45 z = = + Reject the null. It is reasonable to conclude union nurses earn more. The pvalue is 0.0183. This pvalue shows there is a moderate chance of finding a z statistic this extreme when the null hypothesis is true. 7. a. H o is rejected if z > 1.65 b. 0.64, found by 70 90 100 150 c p + = + c. 1.61, found by 0.70 0.60 (0.64)(0.36) (0.64)(0.36) 100 150 z = + d. H o is not rejected 8. a. H o is rejected if z <  1.96 or z > 1.96 b. 0.80, found by 170 110 200 150 c p + = + c. 2.70, found by 0.85 0.7333 (0.80)(0.20) (0.80)(0.20) 200 150 z = + d. H o is rejected 9. a. H o : π 1 = π 2 H 1 : π 1 ≠ π 2 b. H o is rejected if z <  1.96 or z > 1.96 c. 24 40 0.08 400 400 c p + = = + d. 2.09, found by 0.06 0.10 (0.08)(0.92) (0.08)(0.92) 400 400 z = + e. H o is rejected. The proportion infested is not the same in the two fields. Chapter 11 99 10. a. H o : π 1 ≥ π 2 H 1 : π 1 < π 2 b. H o is rejected if z <  1.65 1530 2010 0.59 3000 3000 c p + = = + c. 12.60, found by 0.51 0.67 (0.59)(0.41) (0.59)(0.41) 3000 3000 z = + d. H o is rejected. The proportion of women who think men are thoughtful has declined....
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This note was uploaded on 04/25/2008 for the course ECON 211 taught by Professor Haley during the Spring '08 term at Wisc Oshkosh.
 Spring '08
 HALEY

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