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

Midterm Solutions

David M. Rocke BIM 105, Fall 2018

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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...

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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.

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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.

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xi xi − ¯ x (xi − ¯ x)2 3.3 0.4 0.16 2.4

• 0.5

0.25 2.8

• 0.1

0.01 2.9 3.1 0.2 0.04 14.5 0.46 ¯ x = 14.5/5 = 2.9 s2

x

= 0.46/4 = 0.115 s = √ 0.115 = 0.3391

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(b) Test the hypothesis that the population mean is 2.5 volts, two-sided, at the 5% level. Is the hypothesis rejected? t4 = ¯ x − µ s/√n = 2.9 − 2.5 0.3391/ √ 5 = 2.6377 t.025,4 = 2.776 Don’t reject the hypothesis.

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(c) Find a 95%confidence interval for the mean voltage

• f 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.

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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?

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We have n = 150, x = 140, and ˆ p = 0.9333. With p0 = 0.90, we have that the standard deviation of ˆ p is

• p0(1 − p0)/n = 0.0245 and

z = 0.93333 − 0.9 0.0245 = 1.3608 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.

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(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

• ˆ

p(1 − 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)

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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)

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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.

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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.

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(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) = λxe−λ/x!. The chance of no failures is λ0e−3.36/0! = e−3.36 = 0.0347. The chance of two failures is λ2e−3.36/2! = (3.36)2e−3.36/2 = 0.1961

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(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.

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(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−λt. The chance of being greater than t is therefore e−λt = 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.

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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

• f the total number of defectives for the month.

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The total number of defectives is the sum of 30 Poisson random variables with mean and variance 3 plus the sum

• f 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.

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(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.

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(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

• nly on the mean.

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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.

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They are statistically independent and there are lots of ways to justify this. One argument is that the probability

• f 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.

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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 0.10 1.00

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