Question

Americans spend over $70 billion per year trying to lose welght. Thus, it would seem to be important to see which diets are better than others. Diet number 1 (D1) involves sticking to a special diet and requirad exercise. Diet number 2 (02) invoves sticking to the special diet but no e ense. Di tician lane Jay randomly selected m-63 subjects who stick to Dit D1 and n-45 sub ects who stick to Diet #2 D2 The means weight loss or the D1 subjects was 25 pounds while the mean weight loss for the D2 su ects was y-14 pounds. rt is recognize that the true stan ard dev ations for a DI d t subject is σχ 9.7 poundsand the true standard deviation for a D2 diat subject is 7y 4.9 pounds. The true unknown mean weight loss for DI d t subject is while thaunknawn) maan weight loss for a D2 diat subjact is My. Wa would like to examinep a) what is the standard deviatian of the distribution afY b) What is the standard deviation of the distribution of x-Y c) Create a 96% confidence interval for x- y ? d) What is the length of the contidence interval in part c) a) Copy your R script far tha ahova into tha text box hera.

Answer 1

e) The R script for the above parts is as follows:

> #----Given-------
> m=63; n=45; Xbar=25; Ybar=14 ; Sx=9.7; Sy=4.9;
>
> #____________a)_______________________
> SdYbar=Sy/sqrt(n); SdYbar
[1] 0.7304489
>
> #____________b)_______________________
> SdXbar_Ybar=sqrt((Sx^2/m)+(Sy^2/n)); SdXbar_Ybar
[1] 1.423744
>
> #____________c) To find 96% CI for Mx-My_______________________
> Upper_Limit=Xbar-Ybar+abs(qnorm(0.02))*SdXbar_Ybar
> Lower_Limit=Xbar-Ybar-abs(qnorm(0.02))*SdXbar_Ybar
> CI=c(Lower_Limit, Upper_Limit) ; CI
[1] 8.075987 13.924013
> #____________d)_______________________
> Lenght=Upper_Limit-Lower_Limit; Lenght
[1] 5.848026