Based on this, what is the maximal margin of error associated with a 99% confidence interval for the true population mean dog weight.

Question

You measure 35 dogs' weights, and find they have a mean weight of 78 ounces. Assume the population standard deviation is 2.8 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

Because the population standard deviation is known, the 99% confidence interval is given by

[x̄ - z_0.005*(σ / √n), x̄ + z_0.005*(σ / √n)], where x̄ = 78 is the sample mean, z_0.005=2.576 is the z-score associated with the 99% confidence level, σ = 2.8 is the population standard deviation and n = 35 is the sample size.

The maximal margin of error is z_0.005*(σ / √n) = 2.576*(2.8/√35) ≈ 1.22.