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.

At least 6907 people.

Step-by-step explanation:

Population std deviation = sigma= 10.6

Since population std deviation is known, we can use normal probability table to get sample size from confidence interval.

The sample mean weight loss is within 0.25 lb of the true population mean.

Hence  margin of error < 0.25

Margin of error = z critical (std dev/n) where n = sample size

Z critical for 95% = 1.96

Hence 0.25 >1.96(10.6)/sq rt n

Simplify to get

sq rt n > 1.96(10.6)/0.25 = 83.104

Square both the sides to get

n > 83.104 square = 6906.27

i.e. sample size should be atleast 6907.

answer: 42

use a calculator to find that 845/73 = 11.575

the whole part of the answer is 11. multiply this by 73 to get 11*73 = 803.

subtract that result from 845 to get 845 - 803 = 42