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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

Respuesta :

sarkhan2018
Using normal distribution [tex] \alpha = 1-0.95=0.05[/tex]
critical probability [tex] p^{*} = 1 - \frac{ \alpha }{2} [/tex]
critical value for the critical probability is CV = 1.96
standard error [tex]SE = \frac{3.2}{ \sqrt{41} } = 0.5[/tex]
margin of error = CV*SE = 1.96 * 0.5 = 0.98
95% confidence interval is between μ + 0.98 and μ - 0.98, which is (19.62, 21.58)

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