If the p value is below a given significance level

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If the p -value is below a given significance level , then we say that the result is statistically significant at the level . For the previous example, we could not have rejected H 0 at the = 0 . 05 significance level. Indeed, we could not have rejected H 0 at any level below the p -value, 0.0770786. On the other hand, we would reject H 0 for any significance level above this value. Many statistical software packages (including R , see the example below) do not need to have the significance level in order to perform a test procedure. This is especially important to note when setting up a hypothesis test for the purpose of deciding whether or not to reject H 0 . In these circumstances, the significance level of a test is a value that should be decided before the data are viewed. After the test is performed, a report of the p -value adds information beyond simply saying that the results were or were not significant. It is tempting to associate the p -value to a statement about the probability of the null or alternative hypothesis being true. Such a statement would have to be based on knowing which value of the parameter is the true state of nature. Assessing whether of not this parameter value is in 0 is the reason for the testing procedure and the p -value was computed in knowledge of the data and our choice of 0 . In the example above, the test is based on having a test statistic S ( x ) (namely - ¯ x ) exceed a level k 0 , i.e., we have decision reject H 0 if and only if S (( x )) k 0 . This choice of k 0 is based on the choice of significance level and the choice of 0 2 0 so that ( 0 ) = P 0 { S ( X ) k 0 } = , the lowest value for the power function under the null hypothesis. If the observed data x takes the value S ( x ) = k , then the p -value equals P 0 { S ( X ) k } . This is the lowest value for the significance level that would result in rejection of the null hypothesis if we had chosen it in advance of seeing the data . Example 18.9. Returning to the example on the proportion of hives that survive the winter, the appropriate composite hypothesis test to see if more that the usual normal of hives survive is 286
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Introduction to the Science of Statistics Composite Hypotheses H 0 : p 0 . 7 versus H 1 : p > 0 . 7 . The R output shows a p -value of 3%. > prop.test(88,112,0.7,alternative="greater") 1-sample proportions test with continuity correction data: 88 out of 112, null probability 0.7 X-squared = 3.5208, df = 1, p-value = 0.0303 alternative hypothesis: true p is greater than 0.7 95 percent confidence interval: 0.7107807 1.0000000 sample estimates: p 0.7857143 Exercise 18.10. Is the hypothesis test above signfiicant at the 5% level? the 1% level? 18.4 Answers to Selected Exercises 18.2. In this case the critical regions is C = { x ; ¯ x k ( μ 0 ) } for some value k ( μ 0 ) . To find this value, note that P μ 0 { Z  - z } = P μ 0 { ¯ X  - σ 0 p n z + μ 0 } and k ( μ 0 ) = - ( σ 0 / p n ) z + μ 0 . The power function ( μ ) = P μ { ¯ X  - σ 0 p n z + μ 0 } = P μ { ¯ X - μ  - σ 0 p n z - ( μ - μ 0 ) } = P μ ¯ X - μ σ 0 / p n  - z - μ - μ 0 σ 0 / p n = Φ - z - μ - μ 0 σ 0 / p n .
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  • Spring '17
  • KASMIS MISGANEW
  • Null hypothesis, Statistical hypothesis testing, Type I and type II errors, power function, Science of Statistics

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