Answer to Question #204478 in Statistics and Probability for charm

Question

A weight-reducing clinic claims that after completion of its advanced course, the weight of
participants is decreased by an average of 15%. A random sample of 25 graduates of this
course found an average 14.14 % the sample standard deviation was 3.61%. Test the clinic’s
claim at 5% level of significance.

Answer 1

One sample t-test.

$H_0:\mu=15.\\ H_a:\mu<15.$

Degrees of freedom: $df=25-1=24.$

Test statistic: $t=\frac{14.14-15}{\frac{3.61}{\sqrt{25}}}=-1.19.$

P-value: $p=P(T<-1.19)=0.1228.$

Since the P-value is greater than 0.05, fail to reject the null hypothesis.

There is no sufficient evidence to reject the clinic's claim.