Respuesta :
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 = [tex]z\frac{s}{\sqrt{n} }[/tex]
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 = [tex]z\frac{s}{\sqrt{n} }[/tex]
[tex]0.6=1.96\frac{5}{\sqrt{n} }[/tex]
[tex]\sqrt{n}=\frac{1.96*5}{0.6}[/tex]
[tex]\sqrt{n}=16.33[/tex]
n = 267
For the margin of error with 95% confidence interval be 0.6, it is needed 267 mice.
The total number of mice must be weighed so that a 95% confidence interval will have a margin of error of 0.6 grams is 267 and this can be determined by using the formula of margin of error.
Given :
- 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 a standard deviation of 5 grams.
- 95% confidence interval.
The formula of margin of error can be used in order to determine the total number of mice. The formula of margin of error is given by:
[tex]\rm MOE = z\dfrac{s}{\sqrt{n} }[/tex]
where z is the z-score, s is the standard deviation, and n is the sample size.
Now, substitute the known values in the above formula.
[tex]\rm 0.6 = 1.96\times \dfrac{5}{\sqrt{n} }[/tex]
Simplify the above expression in order to determine the value of 'n'.
[tex]\rm n =\left(\dfrac{1.96\times 5}{0.6}\right)^2[/tex]
n = 267
Therefore, the total number of mice must be weighed so that a 95% confidence interval will have a margin of error of 0.6 grams is 267.
For more information, refer to the link given below:
https://brainly.com/question/20982963
