Answer:
6907 people on that program must be surveyed if we want mean weight loss as 0.25 lb and standard deviation as 10.6 lb.
Step-by-step explanation:
Given that mean of weight loss in 0.25 lb and standard deviation is 10.6 lb.
For 95% confident level z-score must be 1.96.
That is [tex]Z_{c}[/tex]=1.96
Now required sample size is [tex](\frac{Z_{c} X(SE)}{m})^{2}[/tex]
Where SE means standard deviation=10.6 lb and mean m=0.25 lb
Hence sample size=[tex](\frac{1.96X10.6}{0.25})^{2}=6906.27[/tex]
Which we always should round up that is 6907.
