2 1 1 1 2 Example Two Sample Pooled TestOwen Lawn Care Inc manufactures and

2 1 1 1 2 example two sample pooled testowen lawn

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𝑝 2 1 𝑛 1 1 𝑛 2
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Example: Two-Sample Pooled Test Owen Lawn Care Inc manufactures and assembles lawnmowers that are shipped to dealers throughout US and Canada. Two different procedures have been proposed, the Welles (W) method and the Atkins(A) method. The question is: Is there a difference in the mean time to mount the engines on the frames of the lawnmowers? To evaluate, a time and motion study was conducted. A sample of five employees is timed using the Welles method and six using the Atkins method. Is there a difference in the mean mounting times? Use the .10 significance level.
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Example: Two-Sample Pooled Test Step : State ? 0 and ? 1 Step : Select level of significance 𝛼 = 0.1 (given) Step : Select test statistics ? distribution since 𝜎 is unknown ? 0 : 𝜇 𝑊 = 𝜇 𝐴 ? 1 : 𝜇 𝑊 ≠ 𝜇 𝐴 Two-tailed test
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Example: Two-Sample Pooled Test Step : Formulate decision rule Two tailed, 𝛼 = 0.10 𝑑𝑓 = 𝑛 1 + 𝑛 2 − 2 = 5 + 6 − 2 = 9 From the table, critical ? = ±1.833 Reject ? 0 if computed t > 1.833 or if computed t < −1.8333
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Example: Two-Sample Pooled Test Step : Make a decision ? 𝑝 2 = 𝑛 𝑊 −1 𝑠 𝑊 2 + 𝑛 𝐴 −1 𝑠 𝐴 2 𝑛 𝑊 +𝑛 𝐴 −2 = 5−1 2.9155 2 6−1 2.0976 2 5+6−2 = 6.2222 ҧ𝑥 𝑊 = 20 5 = 4 ҧ𝑥 𝐴 = 30 6 = 5 ? 𝑊 = 34 5 − 1 = 2.9155 ? 𝐴 = 22 6 − 1 = 2.0976
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Example: Two-Sample Pooled Test Step : Make a decision ҧ𝑥 𝑊 = 20 5 = 4 ҧ𝑥 𝐴 = 30 6 = 5 ? 𝑊 = 34 5 − 1 = 2.9155 ? 𝐴 = 22 6 − 1 = 2.0976 ? = ҧ𝑥 𝑊 ҧ𝑥 𝐴 ? 𝑝 2 1 𝑛 𝑊 1 𝑛 𝐴 = 4 − 5 6.222 1 5 + 1 6 = −0.662 As −0.662 is between −1.833 and 1.833 , the decision is not to reject the null hypothesis. Reject ? 0 if computed t > 1.833 or if computed t < −1.8333
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Example: Two-Sample Pooled Test Step : Interpret result The conclusion is that the sample data failed to show a difference between the mean assembly times of the two methods. What is the 𝑝 -value? We need the probability value of ? to be closest to 0.622, with the 𝑑𝑓 value of 9. This value is 1.383, corresponding to the significance level of .20. ∴ 𝑝 -value > 0.20 Therefore, even if the significance level used is 20%, we would not have rejected the ? 0
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The null and alternate hypotheses are: ? 0 : 𝜇 1 = 𝜇 2 ? 1 : 𝜇 1 ≠ 𝜇 2 A random sample of 15 observations from the first population revealed a sample mean of 350 and a sample standard deviation of 12. A random sample of 17 observations from the second population revealed a sample mean of 342 and a sample standard deviation of 15. At the .10 significance level, is there a difference in the population means? Try This ....
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