Answered

Welcome to Westonci.ca, the place where your questions find answers from a community of knowledgeable experts. Connect with a community of experts ready to help you find accurate solutions to your questions quickly and efficiently. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

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

Sagot :

coolstick

Answer:

(34.15, 47.85)

Step-by-step explanation:

[tex]n = (\frac{z*stddev}{ME} )^{2}[/tex]

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

[tex]\sqrt{20} = 2.576 * 11.9 / ME[/tex]

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)

We hope this was helpful. Please come back whenever you need more information or answers to your queries. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.