2011_SS1_Homework 06 - Solutions

7 tx tx t x 14 sx 5 5 9 for alpha 005

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Unformatted text preview: tudy, imagine that we calculate the sample variance to be 225 instead. Recalculate and report the value of the appropriate test statistic used to determine if there is evidence of an effect of treatment. sX = s2 , sX = n 225 , 9 s X = 25 , s X = 5 tX = X!µ 33 ! 40 !7 , tX = , tX = , t X = !1.4 sX 5 5 9) For alpha = 0.05, the critical values for the above test are t critical = ± 2.306 10) Given the test statistic and critical values you found above, what should we conclude about the effect of the treatment? 1. Reject the null hypothesis. We conclude that there really is an effect of treatment. 2. Fail to reject the null hypothesis. We conclude that there is not sufficient evidence to say that there is an effect of this treatment. |tobserved| < |tcritical|, | ­1.4| < |2.306|, Fail to Reject Null Hypothesis Notice: The sample variance has a direct effect on the hypothesis test; a larger sample variance leads to a less extreme t value, which makes rejecting the null less likely 11) Some...
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