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Problem 1 In a sample of 100 steel wires the average breaking strength is 50 kN, with a standard deviation of 2 kN. n: Find

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Problem 1 In a sample of 100 steel wires the average breaking strength is 50 kN, with a standard deviation of 2 kN. n: Find a 95% confidence interval for the mean breaking strength of this type of wire. An engineer claims that the mean breaking strength is between 49.7 kN and 50.3 kN. With what condence can this statement be made? How many wires must be sampled so that a 95% confidence interval specifies the mean breaking point to within i0.5 kN? Problem 2 An article in Journai of the American Statisticai Association (1990, Vol. 85, pp. 972-985) measured weight of 30 rats under experiment controls. Suppose that there are 12 underweight rats. a. Calculate a 90% two-sided confidence interval on the true proportion of rats that would show underweight from the experiment. Round your answers to 3 decimal places. b. Using the point estimate of ,0 obtained from the preliminary sample, what sample size is needed to be 90% confident that the error in estimating the true value ofp is no more than 0.02? c. How large must the sample be if we wish to be at least 90% confident that the error in estimating p is less than 0.02, regardless of the true value ofp? Problem 3 The United States Golf Association tests golf balls to ensure that they conform to the rules of golf. Balls are tested for weight, diameter, roundness, and overall distance. The overall distance test is conducted by hitting balls with a driver swung by a mechanical device nicknamed "Iron Byron\" after the legendary great Byron Nelson, whose swing the machine is said to emulate. Following are 100 distances (in yards) achieved by a particular brand of golf ball in the overall distance test. 261.3 258.1 254.2 257.7 237.9 255.8 241.4 259.4 265.7 270.6 274.2 261.4 254.5 283.7 270.5 255.1 268.9 267.4 253.6 234.3 263.2 270.7 233.7 263.5 244.5 251.8 259.5 257.5 272.6 253.7 262.2 252.0 280.3 274.9 233.7 274.0 264.5 244.8 264.0 268.3 272.1 260.2 260.7 245.5 279.6 237.8 278.5 273.3 263.7 260.6 280.3 272.7 261.0 260.0 279.3 252.1 244.3 272.2 248.3 278.7 236.0 271.2 279.8 245.6 241.2 251.1 267.0 273.4 247.7 254.8 272.8 270.5 254.4 232.1 271.5 242.9 273.6 256.1 251.6 256.8 273.0 240.8 276.6 264.5 264.5 226.8 255.3 266.6 250.2 255.8 285.3 255.4 240.5 255.0 273.2 251.4 276.1 277.8 266.8 268.5 a. Can you support a claim that mean distance achieved by this particular golf ball exceeds 280 yards? Test the hypothesis H0 2 ,u = 280 versus H141 > 280. Use a = 0.05. Is it possible to reject H0 hypothesis at the 0.05 level of significance? Find the P-value. b. Check the normality assumption i.e. check if the data satisfy the normality assumption. Problem 4 Data from the National Association of Realtors indicates that mean price of a home in Denver, Colorado, in the first three months of 2018 was $255,300. We can write this as 255.3 with units of thousands of dollars for our calculations. A random sample of 60 homes sold in 2020 had a mean price of 235 (thousand dollars). Assume the standard deviation 5 = 140 (thousand dollars). Can we conclude that the mean price in 2020 differs from the mean price in the first 3 months of 2018? Use the a = 0.05 level of significance. Parameter Null hypothesis Alternative hypothesis This test is: One-sided Two-sided Test statistic p-value Circle a decision: Reject H0 Do not reject H0 Conclusion: At the 0.05 level of significance, we have sufficient evidence to say that The following is a boxplot of the sample data. Should we assume the population of prices is normally distributed? Yes No {:7 x x l l I l l n 200 400 (:00 300 Is the inference still valid? Yes No

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