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LiRa [457]
3 years ago
12

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
Mathematics
1 answer:
lina2011 [118]3 years ago
6 0
Using normal distribution \alpha = 1-0.95=0.05
critical probability p^{*} = 1 -  \frac{ \alpha }{2}
critical value for the critical probability is CV = 1.96
standard error SE =  \frac{3.2}{ \sqrt{41} } = 0.5
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>
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