Question

You measure 20 dogs' weights, and find they have a mean weight of 41 ounces. Assume the population standard deviation is 11.9 ounces. Based on this, what is the maximal margin of error associated with a 99% confidence interval for the true population mean dog weight.

Give your answer as a decimal, to two places

Answer 1

(34.15, 47.85)

Explanation:

`n = ((z*stddev)/(ME) )^(2)`

stddev = 11.9

z = invNorm(

area = (1 + confidence interval)/2

mean = 0

stddev = 1

)

= invNorm(

0.995

0

1

)

≈ 2.576

n = 20

square root both sides

`√(20) = 2.576 * 11.9 / ME`

multiply both sides by ME, divide both sides by √20 to isolate ME
ME = 2.576 * 11.9/√20

≈ 6.85

interval = mean ± ME
= (41-6.85, 41+6.85)

= (34.15, 47.85)