WebAssign Math 3680 Homework 5 Devore Fall 2013 (Homework)

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1 WebAssign Math 3680 Homework 5 Devore Fall 2013 (Homework) Current Score : / 129 Due : Friday, October :59 PM CDT Mirka Martinez Applied Statistics, Math 3680-Fall 2013, section 2, Fall 2013 Instructor: John Quintanilla The due date for this assignment is past. Your work can be viewed below, but no changes can be made. Important! Before you view the answer key, decide whether or not you plan to request an extension. Your Instructor may not grant you an extension if you have viewed the answer key. Automatic extensions are not granted if you have viewed the answer key. Request Extension /18 points Previous Answers DevoreStat8 4.E.089. Construct a normal probability plot for the following sample of observations on coating thickness for low-viscosity paint Determine the z percentile associated with each sample observation. (Round your answers to two decimal places.) Sample observation z percentile Sample observation z percentile Choose the plot that best represents the normal probability plot for the sample of observations. Page 1 of 10

2 Would you feel comfortable estimating population mean thickness using a method that assumed a normal population distribution? Yes, the probability plot is reasonably straight. Yes, the probability plot is not straight. No, the probability plot is reasonably straight. No, the probability plot is not straight. Page 2 of 10

3 /11 points Previous Answers There are two traffic lights on a commuter's route to and from work. Let X 1 be the number of lights at which the commuter must stop on his way to work, and X 2 be the number of lights at which he must stop when returning from work. Suppose that these two variables are independent, each with the pmf given in the accompanying table (so X 1, X 2 is a random sample of size n = 2). x μ = 1, σ 2 = 0.8 p(x 1 ) (a) Determine the pmf of T o = X 1 + X 2. t o p(t o ) (b) Calculate μ To. μ To = 2 2 How does it relate to μ, the population mean? μ To = 2 2 μ (c) Calculate σ To 2. σ To 2 = How does it relate to σ 2, the population variance? σ To 2 = 2 2 σ 2 (d) Let X 3 and X 4 be the number of lights at which a stop is required when driving to and from work on a second day assumed independent of the first day. With T o = the sum of all four X i 's, what now are the values of E(T o ) and V(T o )? E(T o ) = 4 4 V(T o ) = /10 points Previous Answers DevoreStat8 5.AE.023. Example 5.23 Consider a simulation experiment in which the population distribution is quite skewed. The figure below shows the density curve for lifetimes of a certain type of electronic control [this is actually a lognormal distribution with E(ln(X)) = 3 and V(ln(X)) = ]. The statistic of interest is the sample mean X. The experiment utilized 500 replications and considered these sample sizes: n = 5, n = 10, n = 20, and n = 30. The resulting histograms along with a normal probability plot from MINITAB for the x values based on n = 30 are shown in the figures below. Page 3 of 10

4 Unlike the normal case, these histograms all differ in shape. In particular, they become progressively less less skewed as the sample size n increases. The average of the 500 x values for the four different sample sizes are all quite close to the mean value of the population distribution. If each histogram had been based on an unending sequence of x values rather than just 500, all four would have been centered at exactly Thus different values of n change the shape but not the center of the sampling distribution of X. Comparison of the four histograms in the figures below also shows that as n increases, the spread of the histograms decreases decreases. Increasing n results in a greater degree of concentration about the population mean value and makes the histogram look more like a normal curve. The histogram for n = 30 and the normal probability plot below provide convincing evidence that a sample size of n = 30 is sufficient to overcome the skewness of the population distribution and give an approximately normal X sampling distribution. Page 4 of 10

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6 /10 points Previous Answers DevoreStat8 5.AE.024. Example 5.24 In a notched tensile fatigue test on a titanium specimen, the expected number of cycles to first acoustic emission (used to indicate crack initiation) is μ = 28,000, and the standard deviation of the number of cycles is σ = Let X 1, X 2,..., X 25 be a random sample of size 25, where each X i is the number of cycles on a different randomly selected specimen. Then the expected value of the sample mean sample mean number of cycles until first emission is E(X) = μ = 28,000, and the expected total number of cycles for the specimens is E(T o ) = nμ = 25(28,000) = 700,000. The standard deviations of X and T o are σ X = σ/ n = = n σ To = n σ = 25 (2000) = ,000 If the sample size increases to n = 100, E(X) is unchanged but σ X = , half of its previous value (the sample size must be quadrupled quadrupled to halve the standard deviation of X.) /10 points Previous Answers DevoreStat8 5.E.046. The inside diameter of a randomly selected piston ring is a random variable with mean value 9 cm and standard deviation 0.04 cm. (a) If X is the sample mean diameter for a random sample of n = 16 rings, where is the sampling distribution of X centered and what is the standard deviation of the X distribution? (Enter your standard deviation to five decimal places.) center 9 9 cm standard deviation cm (b) Answer the questions posed in part (a) for a sample size of n = 64 rings. (Enter your standard deviation to five decimal places.) center 9 9 cm standard deviation cm (c) For which of the two random samples, the one of part (a) or the one of part (b), is 9 cm? Explain your reasoning. X more likely to be within 0.01 cm of X is more likely to be within 0.01 cm of 9 cm in sample (b) because of the decreased variability with a larger sample size. X is more likely to be within 0.01 cm of 9 cm in sample (b) because of the increased variability with a larger sample size. X is more likely to be within 0.01 cm of 9 cm in sample (a) because of the increased variability with a smaller sample size. X is more likely to be within 0.01 cm of 9 cm in sample (a) because of the decreased variability with a smaller sample size. Page 6 of 10

7 /10 points Previous Answers DevoreStat8 5.E.047. The inside diameter of a randomly selected piston ring is a random variable with mean value 9 cm and standard deviation 0.08 cm. Suppose the distribution of the diameter is normal. (Round your answers to four decimal places.) (a) Calculate P(8.99 X 9.01) when n = 16. P(8.99 X 9.01) = (b) How likely is it that the sample mean diameter exceeds 9.01 when n = 25? P(X 9.01) = /10 points Previous Answers DevoreStat8 5.E.049. There are 44 students in an elementary statistics class. On the basis of years of experience, the instructor knows that the time needed to grade a randomly chosen first examination paper is a random variable with an expected value of 5 min and a standard deviation of 4 min. (Round your answers to four decimal places.) (a) If grading times are independent and the instructor begins grading at 6:50 P.M. and grades continuously, what is the (approximate) probability that he is through grading before the 11:00 P.M. TV news begins? (b) If the sports report begins at 11:10, what is the probability that he misses part of the report if he waits until grading is done before turning on the TV? Page 7 of 10

8 /10 points Previous Answers DevoreStat8 5.E.050. The breaking strength of a rivet has a mean value of 10,000 psi and a standard deviation of 500 psi. (a) What is the probability that the sample mean breaking strength for a random sample of 40 rivets is between 9,900 and 10,200? (Round your answer to four decimal places.) (b) If the sample size had been 15 rather than 40, could the probability requested in part (a) be calculated from the given information? Explain your reasoning. Yes, the probability in part (a) can still be calculated from the given information. No, n should be greater than 30 in order to apply the Central Limit Theorem. No, n should be greater than 20 in order to apply the Central Limit Theorem. No, n should be greater than 50 in order to apply the Central Limit Theorem /10 points Previous Answers DevoreStat8 5.E.051. Time taken by a randomly selected applicant for a mortgage to fill out a certain form has a normal distribution with mean value 9 min and standard deviation 3 min. If five individuals fill out a form on one day and six on another, what is the probability that the sample average amount of time taken on each day is at most 11 min? (Round your answer to four decimal places.) Page 8 of 10

9 /10 points Previous Answers DevoreStat8 5.E.054. Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.67 and standard deviation (a) If a random sample of 25 specimens is selected, what is the probability that the sample average sediment density is at most 3.00? Between 2.67 and 3.00? (Round your answers to four decimal places.) at most between 2.67 and (b) How large a sample size would be required to ensure that the probability in part (a) is at least 0.99? (Round your answer up to the nearest whole number.) specimens /10 points Previous Answers DevoreStat8 7.E.001. Consider a normal population distribution with the value of σ known. (a) What is the confidence level for the interval % x ± 2.88σ/? n (Round your answer to one decimal place.) (b) What is the confidence level for the interval % x ± 1.44σ/? n (Round your answer to one decimal place.) (c) What value of zα/2 in the CI formula below results in a confidence level of 99.7%? (Round your answer to two decimal places.) σ x zα/2, x + zα/2 n zα/2 = σ n (d) Answer the question posed in part (c) for a confidence level of 78%. (Round your answer to two decimal places.) zα/2 = Page 9 of 10

10 /10 points Previous Answers DevoreStat8 7.E.002. Each of the following is a confidence interval for μ = true average (i.e., population mean) resonance frequency (Hz) for all tennis rackets of a certain type: (111.4, 112.6) (111.1, 112.9) (a) What is the value of the sample mean resonance frequency? Hz (b) Both intervals were calculated from the same sample data. The confidence level for one of these intervals is 90% and for the other is 99%. Which of the intervals has the 90% confidence level, and why? The first interval has the 90% confidence level because it is a wider interval. The second interval has the 90% confidence level because it is a wider interval. The second interval has the 90% confidence level because it is a narrower interval. The first interval has the 90% confidence level because it is a narrower interval. Page 10 of 10