Statistics Homework Solutions 161

Statistics Homework Solutions 161 - 161 SECTION 6.15 (c)...

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SECTION 6.15 161 (c) Greater than 0.9. The power is greater than 0.9 against a difference of 3, so it will be greater than 0.9 against any difference greater than 3. Section 6.14 1. (a) There are six tests, so the Bonferroni-adjusted P -values are found by multiplying the original P -values by 6. For the setting whose original P -value is 0.002, the Bonferroni-adjusted P -value is therefore 0.012. Since this value is small, we can conclude that this setting reduces the proportion of defective parts. (b) The Bonferroni-adjusted P -value is 6(0.03) = 0.18. Since this value is not so small, we cannot conclude that this setting reduces the proportion of defective parts. 3. The original P -value must be 0 05 20 0 0025. 5. (a) No. Let X represent the number of times in 200 days that H 0 is rejected. If the mean burn-out amperage is equal to 15 A every day, the probability of rejecting H 0 is 0.05 each day, so X Bin 200 0 05 . The probability of rejecting
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This note was uploaded on 12/20/2011 for the course STA 3163 taught by Professor Mattgilg during the Fall '11 term at UNF.

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