Two hundred fish caught in Cayuga Lake had a mean length of 14.5 inches. The population...
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Two hundred fish caught in Cayuga Lake had a mean length of 14.5 inches. The population standard deviation is 2.4 inches. (Give your answer correct to two decimal places.) (a) Find the 90% confidence interval for the population mean length. (b) Find the 98% confidence interval for the population mean length. 2. How large a sample should be taken if the population mean is to be estimated with 99% confidence to within $83? The population has a standard deviation of $903. (Round your answer up to the next whole number.) 3. The owner of a local chain of grocery stores is always trying to minimize the time it takes her customers to check out. In the past, she has conducted many studies of the checkout times, and they have displayed a normal distribution with a mean time of 12.3 minutes. and a standard deviation of 2.5 minutes. She has implemented a new schedule for cashiers in hopes of reducing the mean checkout time. A random sample of 32 customers visiting her store this week resulted in a mean of 10.9 minutes. Does she have sufficient evidence to claim the mean checkout time this week was less than 12.3 minutes? Use a = 0.02. (a) Find z. (Round your answer to two decimal places.) (ii) Find the p-value. (Round your answer to four decimal places.) Two hundred fish caught in Cayuga Lake had a mean length of 14.5 inches. The population standard deviation is 2.4 inches. (Give your answer correct to two decimal places.) (a) Find the 90% confidence interval for the population mean length. (b) Find the 98% confidence interval for the population mean length. 2. How large a sample should be taken if the population mean is to be estimated with 99% confidence to within $83? The population has a standard deviation of $903. (Round your answer up to the next whole number.) 3. The owner of a local chain of grocery stores is always trying to minimize the time it takes her customers to check out. In the past, she has conducted many studies of the checkout times, and they have displayed a normal distribution with a mean time of 12.3 minutes. and a standard deviation of 2.5 minutes. She has implemented a new schedule for cashiers in hopes of reducing the mean checkout time. A random sample of 32 customers visiting her store this week resulted in a mean of 10.9 minutes. Does she have sufficient evidence to claim the mean checkout time this week was less than 12.3 minutes? Use a = 0.02. (a) Find z. (Round your answer to two decimal places.) (ii) Find the p-value. (Round your answer to four decimal places.)
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Lets solve each part of the problem step by step 1 For the confidence intervals a For a 90 confidence interval Given Sample mean x 145 inches Population standard deviation 24 inches Sample size n 200 ... View the full answer
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