uzdx=whlevine-0007&quiz-actions; takeuz&quiz-probGuid-QNAPCOA80; 10 10000003dd58
ID: 3062726 • Letter: U
Question
uzdx=whlevine-0007&quiz-actions; takeuz&quiz-probGuid-QNAPCOA80; 10 10000003dd581 A random sample of n-12 individuals is selected from a population with -70, and a treatments administered to each individual in the sample. After treatment, the sample mean is found to be M- 74.5 with SS 297. Use the Distributions tool to help answer the questions that follow. t Distribution Degrees of Freedom 21 5000 5000 3.0 -2.0 0.0 1.0 2.0 3.0 How much difference is there between the mean for the treated sample and the mean for the original population? (Note: In a hypothesis test, this value forms the numerator of the t statistic.) If there is no treatment offect, how much difference is expected just by chance between the sample mean and its population mean? That is, find the standard error for M. (Note: In a hypothesis test, this value is the denominator of the t statistic) Based on tha sample data, does the treatment have a signficant effect? Usq two-tailed test with three decimal places.) .05. (Show t-critical . The resuits indicateExplanation / Answer
Ans:
1)diffrence in the mean of treated sample and original population=74.5-70=4.5
2)standard error of M=sqrt(297/(12-1))=5.196
3)n=12
df=12-1=11
critical t value=tinv(0.05,11)=+/-2.201
Test statistic:
t=(74.5-70)/5.196
t=0.866
As,t=0.866 does not lie in the critical region of rejection(i.e. t<-2.201 or t>2.201),so we fail to reject null hypothesis and there is not a significant treatment effect.
last option is correct.
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