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stats 363 HW-Section-10-3

stats 363 HW-Section-10-3 - Section 10.3 Exercise 10.1(a...

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Section 10.3 Exercise 10.1 . ( a ) 2 : 015 ( b ) 2 : 306 ( c ) 1 : 330 ( d ) 1 : 960 Exercise 10.2 . ( a ) A two-tailed test with ° = 0 : 01 and 12 df gives t °= 2 = t 0 : 005 = 3 : 055. The rejection region is j t j > 3 : 055; that is, either t < ° 3 : 055 or t > 3 : 055. ( b ) A right-tailed test with ° = 0 : 05 and 16 df gives t ° = t 0 : 05 = 1 : 746. The rejection region is t > 1 : 746. ( c ) A two-tailed test with ° = 0 : 05 and 25 df gives t °= 2 = t 0 : 025 = 2 : 060. The rejection region is j t j > 2 : 060; that is, either t < ° 2 : 060 or t > 2 : 060. ( d ) A left-tailed test with ° = 0 : 01 and 7 df gives t ° = t 0 : 01 = 2 : 998. The rejection region is t < ° 2 : 998. Exercise 10.3 . ( a ) A two-tailed test with t = 2 : 43 and 12 df : The p -value is given by p -value = P f t < ° 2 : 43 g + P f t > 2 : 43 g = 2 P f t > 2 : 43 g : Then it is clear that 1 2 ( p -value) = P f t > 2 : 43 g : Now we read from Table 4 that P f t > 2 : 179 g = 0 : 025 P f t > 2 : 43 g = 1 2 ( p -value) P f t > 2 : 681 g = 0 : 010 : It follows that 0 : 01 < 1 2 ( p -value) < 0 : 025 : Hence, 0 : 02 < p -value < 0 : 05 : ( b ) A right-tailed test with t = 3 : 21 and 16 df : The p -value is given by p -value = P f t > 3 : 21 g : Now we read from Table 4 that P f t > 2 : 583 g = 0 : 010 P f t > 2 : 921 g = 0 : 005 P f t > 3 : 21 g = p -value.
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It follows that p -value < 0 : 005 : ( c )
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