41 packages are randomly selected from packages received by a parcel service. the sample has a mean weight of 20.6 pounds and a
standard deviation of 3.2 pounds. what is the best pint estimate for a confidence interval estimating the true mean weight, μ, of all packages received by the parcel service? answer: _____pounds
Using normal distribution critical probability critical value for the critical probability is CV = 1.96 standard error margin of error = CV*SE = 1.96 * 0.5 = 0.98 95% confidence interval is between μ + 0.98 and <span>μ - 0.98, which is (19.62, 21.58)</span>