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Unformatted text preview: Solutions for STAT 101, Homework 8 Professor Rakhlin Solutions prepared by Julie, Tung, and Wei December 3, 2011 Total points = 68 points Question. Chapter 14, #26 A bottler carefully weighs bottles coming off its production line to check that the system is filling the bottles with the correct amount of beverage. By design, the system is set to slightly overfill the bottles to allow for random variation in the packaging. The content weight is designed to be 1,020 g wit standard deviation 8 g. Assume that the weights are normally distributed. (Use tables or software for the normal distribution.) (a) A proposed system shuts down the facility if the contents of a bottle weigh less than 1,000 g. What is the chance of a Type I error with this system? (b) Suppose the mean content level of bottles were to fall to 1,000 g.What is the chance that the proposed system will miss the drop in average level and commit a type II error? Answer (a)[2 points] From the design, the weight of a bottle W ∼ N ( μ = 1020 ,σ = 8). Hence the probability of incorrectly stopping the production is P ( W < 1000) = P ( Z < (1000 1020) / 8 = 2 . 5) = 0 . 0062. (b)[2 points] In this question, the new μ is 1,000 g. Therefore, the probability of making a type II error is P ( W > 1 , 000) = 1 / 2 since only half of the production would lie below the threshold at 1,000 g. Question. Chap 14, # 30 (2 points) Where should the control limits in an Xbar chart be placed if the design of the process sets α = 0 . 01 with the following parameters (assume that the sample size condition for control charts has been verified)? (a) (1 point) μ = 100, σ = 20, and n = 25 cases per batch Using the formula in the book, the control limits are μ z α/ 2 σ √ n and μ + z α/ 2 σ √ n (0.5 point). For α = 0 . 01, z . 01 / 2 = 2 . 5758294, our control limits are 88.84054 and 111.5195 (0.5 point, any reasonable rounding is okay). 1 (b) (1 point) μ = 2000, σ = 2000, and n = 100 cases per batch Using the formula in the book, the control limits are μ z α/ 2 σ √ n and μ + z α/ 2 σ √ n (0.5 point). For α = 0 . 01, z . 01 / 2 = 2 . 5758294, our control limits are 848.0541 and 3151.946 (0.5 point, any reasonable rounding is okay). Note to graders: If students explicitly stated the formula for at least one of the questions, give them full credit for the formula for both parts. Question. Chap 15, # 26 (2 points) Which is more likely to contain μ , the zinterval ¯ X ± 1 . 96 σ/ √ n or the tinterval ¯ X ± t . 025 ,n 1 S/ √ n Both are 95% confidence intervals for μ (1 point), but with different distributions; the first one is using the normal distribution and the second one is using the tdistribution....
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 Fall '08
 Heller
 Statistics, Standard Deviation, Variance, σ

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