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Test and power

# Test and power - Exercise The department of Transportation...

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Exercise: The department of Transportation reported Amtrack trains had a 78% on-time arrival record over the previous 12 months. Assume that in 2004, a study found 330 of 400 Amtrack trains arrived on time. Does the sample indicate the Amtrack arrival rate has change? Test 0 : 0.78 : 0.78 a Hp at 0.05 (a) What is the sample proportion of Amtrack trains arriving on time? (b) Compute the value of the test statistic ( z ) (c) What is the p value? (d) What is your conclusion?

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Solution: (a) 330/ 400 0.825 p  0 00 330 400(0.78) (1 ) 400(0.78)(0.22) 18 2.17 8.28 x np z np p p value = 2(0.5-0.4850)=0.03 (b) z value of 2.17 > 1.96 critical value, (c) Reject H 0
Power calculation Power=1-β=1-P(Type II error)=1-P(not reject | H α is true) Steps. The following are the steps for computing the β for large sample test with σ unknown. Power of other types of tests can be
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• Spring '11