Midterm Solutions David M. Rocke BIM 105, Fall 2018 David M. Rocke - PowerPoint PPT Presentation

Midterm Solutions David M. Rocke BIM 105, Fall 2018 David M. Rocke Midterm Solutions November 6, 2018 1 / 23

Midterm Scores Min 1Q Median 3Q Max 47.5 79.5 87.5 92.5 100 The course grade is 30% homework, 30% midterm exam, and 40% final exam. In general, A grades are 90–100, B grades are 80–90, etc. If you did not do as well as you would have liked... David M. Rocke Midterm Solutions November 6, 2018 2 / 23

Paths to Improvement Come to class Go to section Come to office hours Read Lecture 1, slides 16–18, “How to Succeed in this Class” Keep up with the work. You can’t catch up on weeks of class in a couple of days of cramming This is especially true of the second half of the course, where the methods are more complex. David M. Rocke Midterm Solutions November 6, 2018 3 / 23

Problem 1 An implantable cardiac pacemaker device is supposed to produce pulses with average voltages between 2.5 and 3.0 volts. A random sample of 5 from a large batch of pacemakers had measured voltages of 3.3, 2.4, 2.8, 2.9, and 3.1. (a) Find the sample mean, sample variance, and sample standard deviation. David M. Rocke Midterm Solutions November 6, 2018 4 / 23

x ) 2 x i − ¯ ( x i − ¯ x i x 3.3 0.4 0.16 2.4 -0.5 0.25 x = 14 . 5 / 5 = 2 . 9 ¯ 2.8 -0.1 0.01 s 2 = 0 . 46 / 4 = 0 . 115 x √ 2.9 0 0 s = 0 . 115 = 0 . 3391 3.1 0.2 0.04 14.5 0 0.46 David M. Rocke Midterm Solutions November 6, 2018 5 / 23

(b) Test the hypothesis that the population mean is 2.5 volts, two-sided, at the 5% level. Is the hypothesis rejected? x − µ ¯ t 4 = s / √ n 2 . 9 − 2 . 5 √ = = 2 . 6377 0 . 3391 / 5 t . 025 , 4 = 2 . 776 Don’t reject the hypothesis. David M. Rocke Midterm Solutions November 6, 2018 6 / 23

(c) Find a 95%confidence interval for the mean voltage of the population. Is this consistent with the results in part (b)? √ 2 . 90 ± (2 . 776)(0 . 3391) / 5 2 . 90 ± 0 . 4210 (2 . 479 , 3 . 321) Since 2.5 is in the interval, this is consistent with the hypothesis not being rejected. David M. Rocke Midterm Solutions November 6, 2018 7 / 23

Problem 2 An additional requirement on the pacemaker devices from Problem 1 is that the proportion of devices from each batch for which the measured voltage is between 2.5 and 3.0 volts should be at least 90%. A random sample of 150 pacemakers had 140 that had voltages in the required range. (a) Test the hypothesis that the true proportion of devices with voltages in the required range is 0.90. You may use either a one-side or two-sided test. What is the p-value of the test? Is the null hypothesis rejected at the 5% level? David M. Rocke Midterm Solutions November 6, 2018 8 / 23

We have n = 150, x = 140, and ˆ p = 0 . 9333. With p 0 = 0 . 90, we have that the standard deviation of ˆ p is � p 0 (1 − p 0 ) / n = 0 . 0245 and z = 0 . 93333 − 0 . 9 = 1 . 3608 0 . 0245 so for the one sided test, the p-value is 0.0869 from Table A2, and this is 0.1738 for the two-sided test. In neither case is the hypothesis rejected. David M. Rocke Midterm Solutions November 6, 2018 9 / 23

(b) Find a 95% confidence interval for the true proportion of devices with voltages in the required range. You may use either the “traditional” or the “modern” method. For the traditional method, the standard deviation of ˆ p is � p (1 − ˆ ˆ p ) / n = 0 . 0204 so the 95% confidence interval is 0 . 9333 ± (1 . 960)(0 . 0204) 0 . 9333 ± 0 . 0399 (0 . 8934 , 0 . 9733) David M. Rocke Midterm Solutions November 6, 2018 10 / 23

For the modern method, we have ˜ n = 104, p = 142 / 154 = 0 . 9221 and the standard deviation of ˜ ˜ p is � p (1 − ˜ ˜ p ) / ˜ n = 0 . 0216 so the 95% confidence interval is 0 . 9221 ± (1 . 960)(0 . 0216) 0 . 9221 ± 0 . 0423 (0 . 8797 , 0 . 9644) David M. Rocke Midterm Solutions November 6, 2018 11 / 23

Problem 3 Failures in a medical device fabrication system occur as a Poisson process with a mean rate of 2 failures per 100 hours. This means that the count of failures in a given interval has a Poisson distribution and the waiting time until the next failure has an exponential distribution. A week has 168 hours. (a) Find the mean, variance, and standard deviation of the number of failures in a week. David M. Rocke Midterm Solutions November 6, 2018 12 / 23

The rate of failure is 1 per 50 hours, or 0.02 per hour. For a 168 hour week, that would be a mean rate of (168)(0 . 02) = 3 . 36 failures per week. The number of failures then is Poisson with λ = 3 . 36. The mean and variance are therefore both 3.36 and the standard √ deviation is 3 . 36 = 1 . 833. David M. Rocke Midterm Solutions November 6, 2018 13 / 23

(b) What is the chance of no failures in a week? What is the chance of exactly 2 failures in a week? The Poisson pdf is f ( x ) = λ x e − λ / x !. The chance of no failures is λ 0 e − 3 . 36 / 0! = e − 3 . 36 = 0 . 0347. The chance of two failures is λ 2 e − 3 . 36 / 2! = (3 . 36) 2 e − 3 . 36 / 2 = 0 . 1961 David M. Rocke Midterm Solutions November 6, 2018 14 / 23

(c) Find the mean, variance and standard deviation of the time from start of the week to the first failure. This is exponential with parameter λ = 1 / 50, so that the mean is 50 and so is the standard deviation. The variance is 2500. David M. Rocke Midterm Solutions November 6, 2018 15 / 23

(d) What is the chance that the first failure is at greater than 50 hours? What is the chance that it is at greater than 168 hours? The cdf of the exponential is 1 − e − λ t . The chance of being greater than t is therefore e − λ t = e − t / 50 . For t = 50, this is e − 1 = 0 . 3679. For t = 168, this is e − 168 / 50 = 0 . 0347. This probability is the same as the first one in part (b), and this is because the event no failures in the week is the same as the event that the first failure time is more than 168 hours. David M. Rocke Midterm Solutions November 6, 2018 16 / 23

Problem 4 The number of defective items per day in a medical device production facility averages 3 on the day shift (8am–4pm), 7 on the swing shift (4pm–12pm), and 5 on the night shift (12:00pm–8:00am). Assume that each is a Poisson random variable, statistically independent across shifts and days. In a 30 day month, then, there are 90 shifts. (a) Compute the mean, variance, and standard deviation of the total number of defectives for the month. David M. Rocke Midterm Solutions November 6, 2018 17 / 23

The total number of defectives is the sum of 30 Poisson random variables with mean and variance 3 plus the sum of 30 with mean and variance 7 plus 30 with mean and variance 5. Thus both the mean and the variance are (30)(3) + (30)(7) + (30)(5) = 450. The standard √ deviation is 450 = 21 . 21. David M. Rocke Midterm Solutions November 6, 2018 18 / 23

(b) What is the probability that the number of defectives is greater than 500. (Hint: The sum or the average of many random variables has approximately what distribution?) z = (500 − 450) / 21 . 21 = 2 . 357. From Table A.2, under z = 2 . 36, the probability is 1 − 0 . 9909 = 0 . 0091. We use the normal distribution because of the central limit theorem. We don’t double the probability because this is not a hypothesis test. David M. Rocke Midterm Solutions November 6, 2018 19 / 23

(c) If the average number of defectives did not vary by shift, but instead was 5 for every shift, what then would be the mean, variance, and standard deviation of defectives in 30 days? The sum of 90 Poisson random variables with mean 5 and variance 5 has mean 90 ∗ 5 = 450 and variance 90 ∗ 5 = 450, the same as in part (a). It turns out that the sum of independent Poisson random variables is Poisson, so the variables in part (a) and part (c) have the same distribution, because the distribution depends only on the mean. David M. Rocke Midterm Solutions November 6, 2018 20 / 23

Problem 5 A quality manager is keeping track of days on which the production line had a stoppage. She found that the chance that any given day had a production stoppage was 0.10. When there was a stoppage, 90% of such cases had no stoppages on the previous day. Is the event “stoppage today” statistically independent of the event “stoppage yesterday”? Justify your conclusion. David M. Rocke Midterm Solutions November 6, 2018 21 / 23

They are statistically independent and there are lots of ways to justify this. One argument is that the probability of a stoppage today given a stoppage yesterday is 0.10, the same as the unconditional probability of a stoppage today. Formally, let Y and T stand for the events that stoppages happened yesterday and today, respectively. We are given P ( Y ) = P ( T ) = 0 . 10 and P ( ¯ Y | T ) = 0 . 90, which means P ( Y | T ) = 0 . 10. Since P ( Y | T ) = P ( Y ) = 0 . 10, the events are independent. David M. Rocke Midterm Solutions November 6, 2018 22 / 23

One can also construct the two-by-two table and observe that the product of the probabilities of yesterday having a stoppage and today having one is the probability of that joint event. The green 0.09 is 90% of the probability of failure today as specified. The blue 0.01 is the product of the marginals. Failure Yesterday No Yes Total Failure No 0.81 0.09 0.90 Today Yes 0.09 0.01 0.10 Total 0.90 1.00 0.10 David M. Rocke Midterm Solutions November 6, 2018 23 / 23