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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)

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