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Statistics homework week 2

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selects simple random samples of 30 finished products and computes the sample mean product weight. If test results over a long period of time show that 5% of the values are over 2.1 pounds and 5% ar... e under 1.9 pounds, what are the mean and the standard deviation for the population of products produced with this process? Assuming a "Normal" distribution Mean = (1/2)(2.1+1.9) = 2.0 With 5% in each tail, we have 90% in the middle of the distribution. This translates into a z-value of P(Z < z*) = 0.05 has a z* = 1.645 2.0 + (1.645)(s) = 2.1 s = 0.1 / 1.645 = 0.06079 5. In an effort to estimate the mean amount spent per customer for dinner at a major Atlanta restaurant, data were collected for a sample of 49 customers. Assume a population standard deviation of $5. a. At 95% confidence, what is the margin of error? The margin of error is found using the following formula: In order to find the margin of error we need to find first. Since we want a confidence level of 95% this implies that = 5%, which means that = 0.025. =-1.959963986 Margin of error is equal to (-1.9599)(5/(49) = -1.399928571 b. If the sample mean is $24.80, what is the 95% confidence interval for the population mean? σ = 5 x = 24.8 ̅ n= 49 Confidence level = .95 Z Interval (23.4, 26.2) x =24.8 ̅ This study source was downloaded by 100000829219024 from CourseHero.com on 08-1 [Show More]

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