To evaluate the effect of a treatment, a sample of n=9 isobtained from a populat
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Question
To evaluate the effect of a treatment, a sample of n=9 isobtained from a population with a mean of =40, and thetreatment is administered to the individuals in the sample. Aftertreatment, the sample mean is found to be M= 33. a. If the sammple has a standard deviation of s=9, are thedata sufficient to conclude that the treatment has a significanteffect using a two tailed test with =0.05? df= s2= Sm= t= Conclusion b. If the sample has a standard deviation of s=15, arethe data sufficient to conclude that the treatment has asignificant effect using a two-tailed test with =0.050? df= S2= Sm= t= Conclusion c. Comparing you answer for parts a and b, how does thevariability of the scores in the sample inluence the outcome of ahypothesis test? df= S2 Sm= t= Conclusion To evaluate the effect of a treatment, a sample of n=9 isobtained from a population with a mean of =40, and thetreatment is administered to the individuals in the sample. Aftertreatment, the sample mean is found to be M= 33. a. If the sammple has a standard deviation of s=9, are thedata sufficient to conclude that the treatment has a significanteffect using a two tailed test with =0.05? df= s2= Sm= t= Conclusion b. If the sample has a standard deviation of s=15, arethe data sufficient to conclude that the treatment has asignificant effect using a two-tailed test with =0.050? df= S2= Sm= t= Conclusion c. Comparing you answer for parts a and b, how does thevariability of the scores in the sample inluence the outcome of ahypothesis test? df= S2 Sm= t= ConclusionExplanation / Answer
To evaluate the effect of a treatment, a sample of n=9 isobtained from a population with a mean of =40, and thetreatment is administered to the individuals in the sample. Aftertreatment, the sample mean is found to be M= 33. a. If the sammple has a standard deviation of s=9, are thedata sufficient to conclude that the treatment has a significanteffect using a two tailed test with =0.05? df= 8 s2= Sm= t= (33-40)/(9/9)=-2.33 2.306 from the table with 8 degrees of freedom, two tail test Conclusion treatment is not effectiveb. If the sample has a standard deviation of s=15, arethe data sufficient to conclude that the treatment has asignificant effect using a two-tailed test with =0.050? df= 8 S2= Sm= t= (33-40)/(15/ 9)=-1.4 2.306 again Conclusionis effective
c. Comparing you answer for parts a and b, how does thevariability of the scores in the sample inluence the outcome of ahypothesis test? df= S2 Sm= t= Conclusion yes, larger s, more spread.
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