Ten individuals have participated in a diet-modification program to stimulate we

Question

Ten individuals have participated in a diet-modification program to stimulate weight loss. Their weight both before and after participation in the program is shown in the following list. Calculate the difference for each subject (d = After - Before). You may just add another column in the table. Calculate the sample mean (d) and sample standard deviation (s_d) of the differences (d_j). Is there evidence to support the claim that this particular diet-modification program is effective in producing a mean weight reduction? Use alpha = 0.05.

Explanation / Answer

there are n=10 subjects

ley X denotes the before weight and Y denotes the after weight

assumption is that X and Y jointly follows a normal distribution.

hence d=Y-X follows a normal distribution.

let d~N(u,sigma2)

a) after calculating d's the table as follows

before   after       d
195           187   -8
213            195   -18
247      221   -26
201      190   -11
187            175   -12
210            197   -13
215            199   -16
246      221   -25
294            278   -16
310            285   -25

b) the sample mean is dbar=(-8-18-26-11-12-13-16-25-16-25)/10=-17 [answer]

the sample standard deviation is sd=[(-8+17)2+(-18+17)2+(-26+17)2+(-11+17)2+(-12+17)2+(-13+17)2+(-16+17)2+(-25+17)2+(-16+17)2+(-25+17)2]/(10-1)=6.41   [answer]

c) we want to test the claim that this particular diet modification program is effective in producing a mean weight reduction

that tis the claim is E[Y]<E[X]

now u=E[d]=E[Y-X]=E[Y]-E[X]<0   under the claim

hence we want to test H0: u=0 vs alternative hypothesis H1: u<0

to test this the test statistic is given by

T=(dbar-0)*sqrt(n)/sd which under H0 follows a t distribution with df n-1

[since sigma is unknown it was replaced by sd]

since the alternative hypothesis is less than type hence the test is left tailed

so H0 is rejected iff t<-talpha;n-1 where t is the observed value of T and talpha;n-1 is the alpha percentage point of a t distribution with df n-1

given that dbar=-17 sd=6.41   n=10 alpha=0.05   t0.05;9=1.83

now t=-17*sqrt(10)/6.41=-8.3866958

so t<-1.83=-t0.05;9

hence at 5% level of significance H0 is rejected and the conclusion is that u<0 that is the claim that this particular diet modification program is effective in producing a mean weight reduction is true [answer]

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.