Question 1102542: Use the confidence level and sample data to find a confidence interval for estimating the population μ. Round your answer to the same number of decimal places as the sample mean.
39 packages are randomly selected from packages received by a parcel service. The sample has a mean weight of 29.9 pounds and a standard deviation of 3.0 pounds. What is the 95% confidence interval for the true mean weight, μ, of all packages received by the parcel service?
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! sample mean is 29.9, sample standard deviation is 3.0 pounds and sample size is 39
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since sample size is > 30, we can use a normal distribution
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alpha(a) = 1 - (95/100) = 0.05
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critical probability(p*) = 1 - (a/2) = 0.975
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critical value(CV) is the z-score associated with p*, the CV is 1.96
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standard error(SE) = 3/square root(39) = 0.4804
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margin of error(ME) = CV * SE = 1.96 * 0.4804 = 0.9416 approximately 0.9
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95% confidence interval = 29.9 + or - 0.9, which is (29.0, 30.8)
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