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STA3032 Quiz 6 solutions - 29 1/2 13.2874 4.097 9.1904...

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STA3032 QUIZ 6 Solutions Spring 2011 Problem 13 0n page 423 (Section 6.4) The following Minitab output presents the results of a hypothesis test for a population mean, μ. Some of the numbers are missing. Fill them in, showing the formula (or reason) you have used to calculate them. You will get one point for the correct formula (or reason) and one point for the correct answer . One – Sample t: X Test of mu = 16 vs not 16 Variable n Mean StDev SE Mean 95% CI T p X 11 13.2874 (a) 1.8389 ( 29 ( ) , ( ) b c (d) 0.171 (a) ( ) 1.8389 , 1.8389 11 6.0989 S SE X SE Mean so n S = = = = =
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Unformatted text preview: ( 29 ( 1, /2) ( ), ( ) ( , ) , 13.2874 4.097 9.1904 ( ) 2.228 1.8389 4.097 n CI b c X ME X ME So b X ME because ME t SE X α-= =-+ =-=-= =% =• = (c) ( 29 ( 1, /2) ( ), ( ) ( , ) , 13.2874 4.097 17.3844 ( ) 2.228 1.8389 4.097 n CI b c X ME X ME So c X ME because ME t SE X-= =-+ = + = + = =% =• = (d) 13.2874 16 1.4751 ( ) 1.8389 cal X T T SE X μ--= = = -= (e) [Added] Is this a one-tailed or a two-tailed test? Why? Since Ho: μ = 16 vs. Ha: μ ≠ 16, we have a two-sided test....
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