Chapter8.1

# Chapter8.1 - value From Table A we find 8106 88 = ≤ Z P...

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Chapter 8: Inference for Means 8.1 Inference for Proportions Overview In this section we consider inference about a population proportion p from an SRS of size n based on the sample proportion n X p = ˆ where X is the number of successes in the sample. Large-Sample Significance test for a Population Proportion Draw an SRS of size n from a large population with unknown proportion p of successes. To test the hypothesis 0 0 : p p H = , compute the z-statistic n p p p p z ) 1 ( ˆ 0 0 0 - - =

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In terms of a standard normal random variable Z, the appropriate P-value for a test of 0 H against 0 : p p H a is P( z Z ) 0 : p p H a < is P( z Z ) 0 : p p H a is 2P( | | z Z ) Example 1. The French naturalist Count Buffon once tossed a coin 4040 times and obtained 2048 heads. This is a binomial experiment with n=4040. The sample proportion is 5069 . 0 4040 2048 ˆ = = p If Buffon’s coin was balanced, then the probability of obtaining heads on any toss is 0.5. To assess whether the data provide evidence that the coin was not balanced, we test 5 . 0 : 0 = p H 5 . 0 : p H a
The test statistic is 88 . 0 4040 ) 5 . 0 1 ( 5 . 0 5 . 0 5069 . 0 4040 ) 5 . 0 1 ( 5 . 0 5 . 0 ˆ = - - = - - = p z Figure 1. illustrates the calculation of the P-

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Unformatted text preview: value. From Table A we find 8106 . ) 88 . ( = ≤ Z P . The probability in each tail is 1-0.8106 = 0.1894, and the P-value is 38 . 1894 . 2 = × = P . Since P-value is larger then 05 . = α , we do not reject 5 . : = p H at the level 05 . = . Figure 1. The P-value for Example 1. Example 2. A coin was tossed n=4040 times and we observed X=1992 tails. We want to test the null hypothesis that the coin is fair- that is, that the probability of tails is 0.5. So p is the probability that the coin comes up tails and we test 5 . : = p H 5 . : ≠ p H a The test statistic is 88 . 4040 ) 5 . 1 ( 5 . 5 . 4931 . 4040 ) 5 . 1 ( 5 . 5 . ˆ-=--=--= p z Using Table A, we find that 38 . 1894 . 2 = × = P . Since P-value is larger then 05 . = α , we do not reject 5 . : = p H at the level 05 . = ....
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Chapter8.1 - value From Table A we find 8106 88 = ≤ Z P...

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