Please refer to the following information for Question 15 Th

Please refer to the following information for Question 15.

The tête-à-tête between MSF and FAP broke down, as anyone would have anticipated. They are going to court.

To prepare for the upcoming court case, FAP managed to get hold of the same 10 teens randomly selected by MSF, and performed its own testing on them. Their corresponding findings on the teens, in the same order as listed by MSF, are listed below.

2.45 2.85 2.87 2.32 1.98 2.51 1.75 1.98 2.03 2.89

It is understandable that FAP would have findings with a higher level of cobalamine in the teens. But it is important to authenticate FAP\'s findings.

It is imperative to make sure that FAP actually have the same group of teens tested by MSF. His Honor, Judge Ig Sushi is presiding over this landmark case. Unexpectedly, Professor Au is deemed to be able to act impartially, and appointed by His Honor to be the amicus curiæ in this pending court case. (As a reward for his faithful service, Professor Au will be offered lifelong supply of bananas from the republic.)

After a few tasty bananas, Professor Au suggested to His Honor that the court should perform a statistical test to see if there is any difference in the means of the MSF group and the FAP group based on a court chosen significance level. Furthermore, Professor Au informed His Honor that it would be unlikely that either MSF or FAP actually knew the population standard deviation at all.

15. (10 points) Perform the statistical test recommended by Professor Au. What significance levels do you think MSF would prefer? On the other hand, what significance levels you think FAP would like to set?

Solution

The important \"Please refer to the following information for Question 15.\" is not available to the TUTOR. So, expect an approximate answer.

> t.test(c(2.45,2.85,2.87,2.32,1.98,2.51,1.75,1.98,2.03,2.89),alternative = \"less\")

   One Sample t-test

data: c(2.45, 2.85, 2.87, 2.32, 1.98, 2.51, 1.75, 1.98, 2.03, 2.89)
t = 17.817, df = 9, p-value = 1
alternative hypothesis: true mean is less than 0
95 percent confidence interval:
-Inf 2.606114
sample estimates:
mean of x
2.363