Question

Scientists want to estimate the mean weight of mice after they have been fed a special diet. From previous studies, it is known that the weight is normally distributed with standard deviation 5 grams. How many mice must be weighed so that a 95% confidence interval will have margin of error of 0.6 grams?

Answer 1

Answer: n = 267

Step-by-step explanation: Margin of Error shows the percentage that will differ the result you get from the real population value or, in other words, is the range of values in a confidence interval.

It can be calculated as

margin of error = $z\frac{s}{\sqrt{n} }$

in which

z is z-score related to the confidence interval, which is this case is 1.96;

s is standard deviation;

n is the number in a sample;

So, the number of mice must be:

margin of error = $z\frac{s}{\sqrt{n} }$

$0.6=1.96\frac{5}{\sqrt{n} }$

$\sqrt{n}=\frac{1.96*5}{0.6}$

$\sqrt{n}=16.33$

n = 267

For the margin of error with 95% confidence interval be 0.6, it is needed 267 mice.