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 3 grams. How many mice must be weighed so that a 99% confidence interval will have a margin of error of 0.5 gram?

Answer 1

Solution :-

Given -

Margin of error ( E ) = 0.5

Standard Deviation ( ) = 3 grams

At 99 % Confidence level ' z ' is,

= 1 - 99 %

= 1 - 0.99

= 0.01

/2 = 0.01 / 2

Z /2 = 2.576

Sample Size ( n ) = ( Z /2 * ) ^ 2 / E ^ 2

Sample Size ( n ) = ( 2.576 * 3 )^2 / 0.5 ^ 2

Sample Size ( n ) = 238.9

Sample Size ( n ) = 239

For 99 % 239 mice must be weighed.

Note - Rounding z value to 2 decimal place instead of 3 decimal.Sample Size will also change.