maxium margin error

Question

You measure 26 dogs' weights, and find they have a mean weight of 36 ounces. Assume the population standard deviation is 10 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  ounces

Answer 1

margin of error = t_{a/2*s/sqrt(n) = 2.79*10/sqrt(26) = 5.47. ta/2 is the critical t value of an area tail of (1-.99)/2 = .005 with 25 degree of freedom.}