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Unformatted text preview: 9.11 a In order to make sure that the average weight was one pound, you would test a H : 1 versus H : 1 µ µ = ≠ b-c The test statistic is 1.01 1 .33 .18 35 x x z n s n µ µ σ − − − = ≈ = = with ( ) ( )-value .33 2 .3707 .7414 p P z = > = = . Since the p-value is greater than .05, the null hypothesis should not be rejected. The manager should report that there is insufficient evidence to indicate that the mean is different from 1. 9.12 a The average density for hogweed in its native area is 5 plants/m 2 . To see if the average density is different in the invaded area, you should test a H : 5 versus H : 5 µ µ = ≠ with the test statistic 11.17 5 11.19 3.9 50 x x z n s n µ µ σ − − − = ≈ = = Since this is a two-tailed test, the rejection region with .05 α = is set in the both tails of the z distribution as .025 | | 1.96 z z > = (similar to Exercise 9.3 b ). Since the observed value 11.19 z = falls in the rejection region, H is rejected. There is evidence that the average density in the invaded area is different from...
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- Spring '10
- Chemical Engineering, Null hypothesis, Statistical hypothesis testing, computer science.