A sample of 25 items is selected from a very large shipment. It is found to have a mean weight of 310 gm and standard deviation equal to 9 gm. State and compute the 95% confidence limits for the population mean weight.
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Expert's answer
2017-02-20T11:39:12-0500
1 - alpha = 0.95 -> alpha = 0.05 1 - alpha/2 = 1 - 0.05/2 = 0.975 0.975 quantile z of normal distribution is equal to 1.96 n = 25, mean = 310, sd = 9 so the 95% confidence limits are [310 - 1.96*9 / sqrt(25), 310 + 1.96*9 / sqrt(25)] = [306.472, 313.528]
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