Question

You want to estimate the mean weight loss of people one year after using a popular weight-loss program being advertised on TV. How many people on that program must be surveyed if we want to be 95% confident that the sample mean weight loss is within 0.25 lb of the true population mean? Assume that the population standard deviation is known to be 10.6 lb.

Answer 1

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.

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
`Z_(c)`=1.96

Now required sample size is
`((Z_(c) X(SE))/(m))^(2)`

Where SE means standard deviation=10.6 lb and mean m=0.25 lb

Hence sample size=
`((1.96X10.6)/(0.25))^(2)=6906.27`

Which we always should round up that is 6907.