Example a certain type of seed grows to a plant of an

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Example. A certain type of seed grows to a plant of an expected height of 8.5 ins. A sample of 49 seeds, grown under new conditions with a different fertiliser, produces plants with an average height of 8.8 ins. Using a 5% level of significance 2
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and making the assumption that the standard deviations of the heights is σ = 1 inch and that it is unaffected by the new conditions, you are asked to determine whether there has been a significant increase in the heights. Given ¯ x = 8 . 8, there is z = ¯ x μ 0 σ n = 8 . 8 85 1 / 7 = 2 . 1 . Under the null hypothesis, this is assumed to be a value sampled from a standard normal N (0 , 1) distribution. The critical value that isolates the upper 5% tail of the distribution is β = 1 . 645. This is exceeded by the value of z ; and so we are inclined to reject the null hypothesis that the fertiliser has had no effect. Type II errors. If there is a simple an a wholly specific alternative hypothesis, then we are able to calculate the probability of the Type II error. Let x N ( μ, σ 2 ) and let the null hypothesis be H 0 : μ = μ 0 and the alternative hypothesis be H q : μ = μ 1 . Then The probability of a Type I error is P ( d 1 | μ = μ 0 ) = P x > μ 0 + βσ/ n ), when ¯ x ( μ 0 , σ 2 /n ). The probability of a Type II error is P ( d 0 | μ = μ 1 ) = P x μ 1 + βσ/ n ), when ¯ x ( μ 1 , σ 2 /n ). 3
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  • Spring '12
  • D.S.G.Pollock
  • Normal Distribution, Null hypothesis, Statistical hypothesis testing, Type I and type II errors

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