Boam Bitch (1)

Hw 1 Due: Monday, Sep 12, 2011 8:00 am CST

Instructions: &platformYou might want to look at the following pages in the eBook as you work through the assignment.

• Populations and samples - pages 5 - 8

• Frequency distributions - pages 14 - 17

• Histograms - pages 17 - 20

• Mean, median, variance and standard deviation - pages 24 - 28

• Quartiles, interquartile range - pages 29 - 31

• Boxplots - pages 31 - 33

• (1 pts) 1 A sample refers to an entire group of interest. a subset of the entire group of interest. a number produced from a subset of the group of interest. a subset that includes all elements in the group of interest. a fixed unknown number that describes the entire group of interest.

• (1 pts) 1 If the shape of our data set is multimodal, we expect the mean to be larger than the median. the mean to be less than the median. the mean and the median to be approximately the same. none of the above.

• (1 pts) 1 The data below show the wins (W) and losses (L) for a basketball team over 25 consecutive games.

• W W W L W W W W W W L W W L W W W W W W W W L W WWhat is the relative frequency of wins among the 25 games? Give your answer to two decimal places. Response: Correct: 0.84 Instructor feedback: There are 21 wins among the 25 games, so the relative frequency is 21/25 = 0.84.

• The data below contain the number of defects observed on each of 250 lcd screens.

• (1 pts) 1 What proportion of the screens have at least 3 defects? Give your answer to three decimal places. Response: Correct: 0.364 Instructor feedback: Counting these by hand can take a long time, so construct a

frequency table for the defects column using the Stat > Tables > Frequency option. You will see that there are a total of 91 of the 250 screens which have at least 3 defects which leads to a proportion of 91/250 = 0.364.

• (1 pts) 1 What propotion of the screens have fewer than 3 defects? Give your answer to three decimal places. Response: Correct: 0.636 Instructor feedback: Counting these by hand can take a long time, so construct a frequency table for the defects column using the Stat > Tables > Frequency option. You will see that there are a total of 159 of the 250 screens which have fewer than 3 defects which leads to a proportion of 159/250 = 0.636.

• (1 pts) 1 What proportion of the screens have at most 3 defects? Give your answer to three decimal places. Response: Correct: 0.804 Instructor feedback: Counting these by hand can take a long time, so construct a frequency table for the defects column using the Stat > Tables > Frequency option. You will see that there are a total of 201 of the 250 screens which have at most 3 defects which leads to a proportion of 201/250 = 0.804.

• (1 pts) 1 What proportion of the screens have exactly 3 defects? Give your answer to three decimal places. Response: Correct: 0.168 Instructor feedback: Counting these by hand can take a long time, so construct a frequency table for the defects column using the Stat > Tables > Frequency option. You will see that there are a total of 42 of the 250 screens which have exactly 3 defects which leads to a proportion of 42/250 = 0.168.

• (1 pts) 1 What proportion of the screens have more than 3 defects? Give your answer to three decimal places. Response: Correct: 0.196 Instructor feedback: Counting these by hand can take a long time, so construct a frequency table for the defects column using the Stat > Tables > Frequency option. You will see that there are a total of 49 of the 250 screens which have more than 3 defects which leads to a proportion of 49/250 = 0.196.

• The data below indicate the time (in seconds) that it takes 25 separate employees to complete a certain task.

Recompute the time in minutes by choosing the Data > Transform data option. Specify a

Transformation of seconds/60 and then click Transform. Use StatCrunch to compute each of the following statistics.

• (1 pts) 1 What is the mean of the data in seconds? Give your answer to three decimal places. Response: Correct: 244.28 Instructor feedback: Choose the Stat > Summary Stats > Columns option. Select the seconds column and click Calculate.

• (1 pts) 1 What is the median of the data in seconds? Response: Correct: 248 Instructor feedback: Choose the Stat > Summary Stats > Columns option. Select the seconds column and click Calculate.

• (1 pts) 1 What is the variance of the data in seconds? Give your answer to three decimal places. Response: Correct: 1427.21 Instructor feedback: Choose the Stat > Summary Stats > Columns option. Select the seconds column and click Calculate.

• (1 pts) 1 What is the standard deviation of the data in seconds? Give your answer to three decimal places. Make sure you take your answer from StatCrunch and do not simply take the square root of your answer above. Response: Correct: 37.778 Instructor feedback: Choose the Stat > Summary Stats > Columns option. Select the seconds column and click Calculate.

• (1 pts) 1 What is the mean of the data in minutes? Give your answer to three decimal places. Response: Correct: 4.071 Instructor feedback: Choose the Stat > Summary Stats > Columns option. Select the seconds/60 column and click Calculate. Notice the answer here is the mean in seconds divided by 60.

• (1 pts) 1 What is the median of the data in minutes? Give your answer to three decimal places. Response: Correct: 4.133 Instructor feedback: Choose the Stat > Summary Stats > Columns option. Select

the seconds/60 column and click Calculate. Notice the answer here is the median in seconds divided by 60.

• (1 pts) 1 What is the variance of the data in minutes? Give your answer to three decimal places. Response: Correct: 0.396 Instructor feedback: Choose the Stat > Summary Stats > Columns option. Select the seconds/60 column and click Calculate. Notice the answer here is the variance in seconds divided by 60 squared (3600).

• (1 pts) 1 What is the standard deviation of the data in minutes? Give your answer to three decimal places. Make sure you take your answer from StatCrunch and do not simply take the square root of your answer above. Response: Correct: 0.63 Instructor feedback: Choose the Stat > Summary Stats > Columns option. Select the seconds/60 column and click Calculate. Notice the answer here is the standard deviation in seconds divided by 60.

• The data below indicate the contamination in parts per million in each of 50 samples of drinking water at a specific location.

• (1 pts) 1 What is the first quartile of the data? Response: Correct: 383

• (1 pts) 1 What is the third quartile of the data? Response: Correct: 398

• (1 pts) 1 What is the median of the data? Response: Correct: 390.5

• (1 pts) 1 What is the mean of the data? Give your answer to three decimal places. Response: Correct: 420.62

• (1 pts) 1 Values that are greater than Q3 + 1.5 IQR or less than Q1 - 1.5 IQR are typically considered outliers. What value is an outlier in this data? Response: Correct: 1942

• (1 pts) 1 Delete the outlier by clicking inside its cell and hitting the Delete key. What is the median after the outlier is deleted? Response: Correct: 390

• (1 pts) 1 What is the mean after the outlier is deleted? Give your answer to three decimal places. Response: Correct: 389.571

• The histogram shown below represents 40 student scores on a statistics exam.

• (1 pts) 1 Which of the following best describes the shape of the scores? symmetric left skewed right skewed bimodal

• (1 pts) 1 What proportion of the scores are below 80? Give your answer to three decimal places. Response: Correct: 0.65

• The boxplots below are for test scores from two sections (A and B) of the same course.

• (1 pts) 1 Which section has the highest median score? A B The medians are the same

• (1 pts) 1 Using the IQR as your measure, which section has the most spread in scores? A B The IQRs are the same

• (1 pts) 1 Roughly what percentage of section B has scores above 70? Do not include a % sign in your answer. Response: Correct: 75


Instructions: This assignment covers Chapter 3 and part of Chapter 15 from your text. You might want to look at the following pages in the eBook as you work through the assignment.

• Basic probability - pages 55 - 63

• Conditional probability and independence - pages 66 - 72

• Bayes rule - pages 72 - 75

• Basic reliability - pages 549 - 551

• A group of 200 students have been cross classified into the following two way table in terms of gender and pierced ears. A student is to be randomly selected from the group.

Pierced Not piercedFemale 79 37Male 19 65

•

• (1 pts) 1 What is the probability the student selected is male? Give your answer to three decimal places. Response: Correct: 0.42 Instructor feedback: P(male) = (19 + 65)/200.

• (1 pts) 1 What is the probability the student selected is a male without pierced ears? Give your answer to three decimal places. Response: Correct: 0.325 Instructor feedback: P(male AND not pierced) = 65/200. Note this is the probability of an intersection.

• (1 pts) 1 What is the probability the student selected is a female or has pierced ears? Give your answer to three decimal places. Response: Correct: 0.675 Instructor feedback: P(female OR pieced) = (79 + 37 + 19)/200. Note this is the probability of a union

• (1 pts) 1 What is the probability the student selected is neither female nor has pierced ears? Give your answer to three decimal places. Response: Correct: 0.325 Instructor feedback: P((female OR pierced)c) = 1 - P(female OR pierced). Note this uses the complements rule. The event not female or pierced is equivalent to an unpierced male from b part.

• (1 pts) 1 Given the student selected does not have pierced ears, what is the probability the student selected is male? Give your answer to three decimal places. Response: Correct: 0.637 Instructor feedback: P(male|not pierced) = P(male AND not pierced)/P(not pierced). P(male AND not pierced) was obtained in b part. P(not pierced) = (37 + 65)/200.

• (1 pts) 1 Is the following statement true or false? The event that the student selected is male is independent of the event that the student selected does not have pierced ears. TRUE FALSE Instructor feedback: The events can not be independent because P(male | not pierced) from e is not equal to P(male) from a.

• A college student is taking two courses. The probability she passes the first course is 0.75. The probability she passes the second course is 0.73. The probability she passes at least one of the courses is 0.9325.

• (1 pts) 1 What is the probability she passes both courses? Give your answer to four decimal places. Response: Correct: 0.5475 Instructor feedback: Let A be the event the she passes the first course and let B be the event that she passes the second. We want P(A B) = P(A) + P(B) - P(A B) = 0.75 + 0.73 - 0.9325 = 0.5475.

• (1 pts) 1 Is the event she passes one course independent of the event that she passes the other course? TRUE FALSE Instructor feedback: The events are independent if P(A B) = P(A)P(B). In this case, does 0.5475 equal (0.75)(0.73)?

• (1 pts) 1 What is the probability she does not pass either course? Give your answer to four decimal places. Response: Correct: 0.0675 Instructor feedback: In this case, we want P( (A B)c ) = 1 - P(A

B) = 1 - 0.9325 = 0.0675.

• (1 pts) 1 What is the probability she does not pass both courses? Give your answer to four decimal places. Response: Correct: 0.4525 Instructor feedback: In this case, we want P( (A B)c ) = 1 - P(A

B) = 1 - 0.5475 = 0.4525.

• (1 pts) 1 What is the probability she passes exactly one course? Give your answer to four decimal places. Response: Correct: 0.385 Instructor feedback: Consider the diagram below which displays the probability of being in A but not B (call it x), the probability of being in both A and B (call it y), and the probability of being in B but not A (call it z). The information given in the question tells us that x+y+z=0.9325, x+y=0.75 and y+z=0.73. Solve the system simultaneously to find x and z. In this case, our answer will be x + z = 0.385.

• (1 pts) 1 Given she passes the first course, what is the probability she passes the second? Give your answer to four decimal places. Response: Correct: 0.73 Instructor feedback: P(B | A) = P(A B)/P(A) = 0.5475/0.75 = 0.73.

• (1 pts) 1 Given she passes the first course, what is the probability she does not pass the second? Give your answer to four decimal places. Response: Correct: 0.27 Instructor feedback: P(Bc | A) = P(A Bc)/P(A) = 0.2025/0.75 = 0.27 (using the notation from the solution to e). Notice this is one

minus the value from f.

• The Transportation Safety Authority (TSA) has developed a new test to detect large amounts of liquid in luggage bags. Based on many test runs, the TSA determines that if a bag does contain large amounts of liquid, there is a probability of 0.99 the test will detect it. If a bag does not contain large amounts of liquid, there is a 0.07 probability the test will conclude that it does (a false positive). Suppose that in reality only 1 in 100 bags actually contain large amounts of liquid.

• (1 pts) 1 What is the probability a randomly selected bag will have a positive test? Give your answer to four decimal places. Response: Correct: 0.0792 Instructor feedback: Let B be the event of a positive test and let A be the event that the bag has a large amount of liquid. From the law of total probability, P(B) = P(B|A)P(A) + P(B|Ac)P(Ac) = 0.99*0.01 + 0.07*0.99 = 0.0792.

• (1 pts) 1 Given a randomly selected bag has a positive test, what is the probability it actually contains a large amount of liquid? Give your answer to four decimal places. Response: Correct: 0.125 Instructor feedback: From Bayes rule, P(A|B) = P(B|A)P(A)/P(B) = P(B|A)P(A)/(P(B|A)P(A) + P(B|Ac)P(Ac)) = 0.99*0.01/[0.99*0.01 + 0.07*0.99)] = 0.125. This probability is really low so you would probably not want to use this test in practice.

• (1 pts) 1 Given a randomly selected bag has a positive test, what is the probability it does not contain a large amount of liquid? Give your answer to four decimal places. Response: Correct: 0.875 Instructor feedback: From Bayes rule, P(Ac|B) = P(B|Ac)P(Ac)/P(B) = P(B|Ac)P(Ac)/(P(B|A)P(A) + P(B|Ac)P(Ac)) = 0.07*0.99/[0.99*0.01 + 0.07*0.99)] = 0.125. This probability is really high indicating that a large proportion of bags with positive tests do not contain large amounts of liquid. You should see from this example that the complements rule also applies to conditional probability as well since P(Ac|B) = 1 - P(A|B).

• Two professors are applying for grants. Professor Jane has a probability of 0.66 of being funded. Professor Joe has probability 0.24 of being funded. Since the grants are submitted to two different federal agencies, assume the outcomes for each grant are independent.

• (1 pts) 1 What is the probability that both professors get their grants funded? Give your answer to four decimal places. Response: Correct: 0.1584 Instructor feedback: Let A be the event that Professor Jane is funded, and let B be the event that Professor Joe is funded. Since the grants are independent, P(A and B) = P(A)P(B) = 0.66*0.24 = 0.1584.

• (1 pts) 1 What is the probability that at least one of the professors will be funded? Give your answer to four decimal places. Response: Correct: 0.7416 Instructor feedback: P(A or B) = P(A) + P(B) - P(A or B) = 0.66 + 0.24 - 0.1584 = 0.7416.

• (1 pts) 1 What is the probability that Professor Jane is funded but Professor Joe is not? Give your answer to four decimal places. Response: Correct: 0.5016 Instructor feedback: Using the law of total probability, P(A and Bc) = P(A) - P(A and B) = 0.66 - 0.1584 = 0.5016.

• (1 pts) 1 Given at least one of the professors is funded, what is the probability that Professor Jane is funded but Professor Joe is not? Give your answer to four decimal places. Response: Correct: 0.6764 Instructor feedback: P(A and Bc | A or B) = P( (A and Bc) and (A or B) )/P(A or B) = P(A and Bc)/P(A or B) = 0.5016/0.7416 = 0.6764 since the set (A and Bc) is contained in the set (A or B).

• (1 pts) 1 Consider the system shown below where the reliability of component 1 is 0.9, the reliability of component 2 is 0.92, the reliability of component 3 is 0.93, the reliability of component 4 is 0.71 and the reliability of component 5 is 0.79.

What is the reliability of the entire system? Give your answer to 4 decimal places. Response: Correct: 0.986 Instructor feedback: The first subsystem is composed of components 1, 2 and 3 as a series, so its reliability is 0.9*0.92*0.93 = 0.77. The second

subsystem is composed of components 4 and 5 in parallel so its reliability is 1 - (1-0.71)(1-0.79) = 0.9391. The entire system is composed of the first and second subsystem in parallel, so the overall reliability is 1 - (1-0.77)(1-0.9391) = 0.986.


Instructions: This assignment covers Chapter 4 from your text. You might want to look at the following pages in the eBook as you work through the assignment.

• Basics of discrete distributions - pages 81 - 84

• Mean and variance of a discrete distribution - pages 93 - 95, 97

• Binomial distribution - pages 85 - 89, 95-96, 98

• Poisson distribution and process - pages 104, 106 - 108

• Each test of a new type of plane can end in success (S) or failure (F). Consider the outcomes of the sequential tests of three new planes. Let X represent the number of successes in the three tests. Let Y equal one if the first plane tested was successful and 0 otherwise. Consider the table of outcomes shown below.

Outcome FFF FFS FSF SFF FSS SFS A SSSX 0 1 1 1 B 2 2 3Y 0 C 0 1 0 1 1 1

•

• (1 pts) 1 What is the value of A in the table above? Response: Correct: SSF

• (1 pts) 1 What is the value of B in the table above? Response: Correct: 2

• (1 pts) 1 What is the value of C in the table above? Response: Correct: 0

• (1 pts) 1 Consider the probability model given in the table below:

• Outcome Probability• 0 0.022• 1 0.086• 2 0.021• 3 ?

What must the probability of a 3 be so that this is a valid probability model? Response: Correct: 0.871 Instructor feedback: The probabilities must sum to 1, so the probability of a 3 is 1 minus the sum of the other probabilities.

• A company has 4 service trucks. Let X represent the number of trucks that are not working at any point in time. Consider the probability model shown below for X.

X 0 1 2 3 4Probability 0.29 0.25 0.12 0.09 0.25

•

• (1 pts) 1 What is the probability that exactly 1 of the trucks are not working at any point in time? Give your answer to two decimal places. Response: Correct: 0.25 Instructor feedback: From the probability model, P(X=1) = 0.25.

• (1 pts) 1 What is the probability that fewer than 1 of the trucks are not working at any point in time? Give your answer to two decimal places. Response: Correct: 0.29 Instructor feedback: From the probability model, P(X < 1) = 0.29 which is the sum of the probabilities for outcomes less than 1.

• (1 pts) 1 What is the probability that more than 1 of the trucks are not working at any point in time? Give your answer to two decimal places. Response: Correct: 0.46 Instructor feedback: From the probability model, P(X > 1) = 0.46 which is the sum of the probabilities for outcomes greater than 1.

• (1 pts) 1 What is the expected number of trucks that are not working at any point in time? Give your answer to two decimal places. Response: Correct: 1.76 Instructor feedback: We take each outcome and multiply it by its associated probability. We then sum up these products to get 0*P(X=0) + 1*P(X=1) + 2*P(X=2) + 3*P(X=3) + 4*P(X=4) = 1.76.

• (1 pts) 1 What is the expected value of X2? Give your answer to two decimal places. Response: Correct: 5.54 Instructor feedback: We take each outcome squared and multiply by the associated probability. We then sum up these products to get 02*P(X=0) + 12*P(X=1) + 22*P(X=2) + 32*P(X=3) + 42*P(X=4) = 5.54.

• (1 pts) 1 What is the variance of X? Give your answer to four decimal places. Response: Correct: 2.4424 Instructor feedback: Var(X) = E(X2) - (E(X))2 = 5.54 - 1.762 = 2.4424.

• (1 pts) 1 What is the standard deviation of X? Give your answer to four decimal places. Response: Correct: 1.5628 Instructor feedback: The standard deviation of X is the square root of the variance of X which in this case is 2.44241/2 = 1.5628.

• A company that produces DVD drives has a 15% defective rate. Let X represent the number of defectives in a random sample of 58 of their drives.

• (1 pts) 1 What is the probability the sample will contain exactly 7 defective drives? Give your answer to four decimal places. Response: Correct: 0.1292 Instructor feedback: X has a Binomial distribution with parameters 58 and 0.15. Using the StatCrunch calculator, P(X=7) = 0.1292.

• (1 pts) 1 What is the probability the sample will contain more than 7 defective drives? Give your answer to four decimal places. Response: Correct: 0.6572 Instructor feedback: X has a Binomial distribution with parameters 58 and 0.15. Using the StatCrunch calculator, P(X > 7) = 0.6572.

• (1 pts) 1 What is the probability the sample will contain less than 7 defective drives? Give your answer to four decimal places. Response: Correct: 0.2136 Instructor feedback: X has a Binomial distribution with parameters 58 and 0.15. Using the StatCrunch calculator, P(X < 7) = 0.2136.

• (1 pts) 1 What is the expected number of defective drives in the sample? Give your answer to two decimal places. Response: Correct: 8.7

Instructor feedback: For a Binomial random variable, E(X) = np = 58*0.15 = 8.7.

• (1 pts) 1 What is the variance of the number of defective drives in the sample? Give your answer to four decimal places. Response: Correct: 7.395 Instructor feedback: For a Binomial random variable, Var(X) = np(1-p) = 58*0.15*(1 - 0.15) = 7.395.

• (1 pts) 1 What is the standard deviation of the number of defective drives in the sample? Give your answer to four decimal places. Response: Correct: 2.7194 Instructor feedback: The standard deviation of X is the square root of the variance of X which in this case is 7.3951/2 = 2.7194.

• (1 pts) 1 Each defective drive costs the company 12 dollars. What is the expected cost to the company for the defective drives in the sample? Give your answer to two decimal places. Response: Correct: 104.4 Instructor feedback: E(cX) = cE(X) where c is a constant. In this case, we are interested in the expected value of 12X which is the cost associated with the defectives in the sample. So, E(12X) = 12E(X) = 12*58*0.15 = 104.4.

• (1 pts) 1 Each defective drive costs the company 12 dollars. What is the standard deviation of the cost to the company for the defective drives in the sample? Give your answer to four decimal places. Response: Correct: 32.6325 Instructor feedback: Var(cX) = c2Var(X) where c is a constant. So, the standard deviation of cX is the square root of the variance which is c times the standard deviation of X. In this case, we are interested in the standard deviation of 12X which is the cost associated with the defectives in the sample. This standard deviation is then 32.6325.

• Support requests arrive at a software company at the rate of 1 every 30 minutes. Assume that the requests arrive as events in a Poisson process.

• (1 pts) 1 What is the expected number of requests in an hour? Give an exact answer. Response: Correct: 2 Instructor feedback: The expected number in an hour is 60/30 = 2.

• (1 pts) 1 What is the probability that the number of requests in an hour is between 3 and 5 inclusive? Give your answer to four decimal places. Response: Correct: 0.3068 Instructor feedback: Let X represent the number of requests in an hour. X has a Poisson distribution with a mean of 3. P(3 < X < 5) = P(X < 5) - P(X < 3) = 0.3068 using the StatCrunch calculator.

• (1 pts) 1 What is the expected number of requests in a 10 hour work day? Give an exact answer. Response: Correct: 20 Instructor feedback: If we expect 2 in an hour, then we expect 10*2 = 20 over a 10 hour period.

• (1 pts) 1 What is the probability that the number of requests in a 10 hour work day is between 28 and 36 inclusive? Give your answer to four decimal places. Response: Correct: 0.0521 Instructor feedback: Let Y represent the number of requests in 10 hours. Y has a Poisson distribution with a mean of 20. P(28 < Y < 36) = P(Y < 36) - P(Y < 28) = 0.0521 using the StatCrunch calculator.

• (1 pts) 1 What is the standard deviation of the number of requests in a 10 hour work day? Give your answer to four decimal places. Response: Correct: 4.4721 Instructor feedback: Let Y represent the number of requests in 10 hours. Y has a Poisson distribution with a

mean of 20 and a variance of 20. The standard deviation is then the square root of the variance which is 4.4721.

•

Hw 4 Due: Monday, Oct 3, 2011 8:00 am CST

Instructions: This assignment covers Chapter 5 from your text. You might want to look at the following pages in the eBook as you work through the assignment.

• Basics of continuous distributions - pages 119 - 124

• Normal distribution - pages 125 - 131

• Gamma distribution - pages 138 - 141

• Trainees must complete a specific task in less than 2 minutes. Consider the probability density function below for the time it takes a trainee to complete the task.

f(x) = 0.82 - 0.32x 0 < x < 2

• (1 pts) 1 What is the probability a trainee will complete the task in less than 1.5 minutes? Give your answer to four decimal places. Response: Correct: 0.87 Instructor feedback: Let X represent the time in minutes that it takes for the trainee to complete the task. We want P(X < 1.5) = F(1.5) where F(x) represents the cdf of X. In this case, the cdf is the integral of f(t) over t from 0 to x so that F(x) = 0.82x - 0.32x2/2. The answer then is F(1.5) = 0.87.

• (1 pts) 1 What is the probability that a trainee will complete the task in more than 1.5 minutes? Give your answer to four decimal places. Response: Correct: 0.13 Instructor feedback: In this case, we are looking for P(X > 1.5) = 1 - P(X < 1.5) = 1 - 0.87 = 0.13.

• (1 pts) 1 What is the probability it will take a trainee between 0.32

minutes and 1.5 minutes to complete the task? Give your answer to four decimal places. Response: Correct: 0.624 Instructor feedback: We want P(0.32 < X < 1.5) = F(1.5) - F(0.32) = 0.87 - 0.246 = 0.624.

• (1 pts) 1 What is the expected time it will take a trainee to complete the task? Give your answer to four decimal places. Response: Correct: 0.7867 Instructor feedback: E(X) is the integral of x times f(x) from 0 to 2. The integral in this case works out to be 0.82*22/2 - 0.32*23/3 = 0.7867.

• (1 pts) 1 If X represents the time it takes to complete the task, what is E(X2)? Give your answer to four decimal places. Response: Correct: 0.9067 Instructor feedback: E(X2) is the integral of x2f(x) from 0 to 2. The integral in this case works out to be 0.82*23/3 - 0.32*24/4 = 0.9067.

• (1 pts) 1 If X represents the time it takes to complete the task, what is Var(X)? Give your answer to four decimal places.

Response: Correct: 0.2878 Instructor feedback: Var(X) = E(X2) - (E(X))2 = 0.9067 - 0.78672 = 0.2878.

• Let X represent the time it takes from when someone enters the line for a roller coaster until they exit on the other side. Consider the probability model defined by the cumulative distribution function given below.

0 x < 3F(x) = (x-3)/1.61 3 < x < 4.61

1 x > 4.61•

• (1 pts) 1 What is E(X)? Give your answer to three decimal places. Response: Correct: 3.805 Instructor feedback: To answer this question, you must

first find the pdf of X. The pdf is the derivative of the cdf, so f(x) = 1/1.61 for x between 3 and 4.61. E(X) is defined as the integral of xf(x) = x/1.61 from 3 to 4.61 in this case which works out to be 4.612/(2*1.61) - 32/(2*1.61) = 3.805. Alternatively, you could recognize that the distribution above is Uniform on 3 to 4.61, so the mean is (3 + 4.61)/2 = 3.805.

• (1 pts) 1 What is the value c such that P(X < c) = 0.29? Give your answer to four decimal places. Response: Correct: 3.4669 Instructor feedback: We need to solve F(c) = P( X < c) = (c-3)/1.61 = 0.29 for c. So, c = 3 + 1.61*0.29 = 3.4669.

• (1 pts) 1 What is the probability that X falls within 0.36 minutes of its mean? Give your answer to four decimal places. Response: Correct: 0.4472 Instructor feedback: P(3.805 - 0.36 < X < 3.805 + 0.36) = F(3.805 + 0.36) - F(3.805 - 0.36) = 2*0.36/1.61 = 0.4472.

• (1 pts) 1 Consider the function given below:

f(x) = kx3 0 < x < 2.14

What must be the value k so that f(x) is a valid probability density function? Give your answer to four decimal places. Response: Correct: 0.1907 Instructor feedback: f(x) must integrate to 1 on 0 to 2.14. The integral of f(x) from 0 to 2.14 is k*2.143+1/(3+1). Setting this integral equal to 1 and solving gives k equal to 0.1907.

• Suppose that the weights of airline passenger bags are normally distributed with a mean of 47.83 pounds and a standard deviation of 3.42 pounds.

• (1 pts) 1 What is the probability that the weight of a bag will be less than the maximum allowable weight of 50 pounds? Give your answer to four decimal places. Response: Correct: 0.737124 Instructor feedback: Let X represent the weight of a bag. Using the StatCrunch Normal calculator under the

Stat > Calculators menu with a mean of 47.83 and a standard deviation of 3.42, we find P(X less than 50) = 0.737124.

• (1 pts) 1 Let X represent the weight of a randomly selected bag. For what value of c is P(E(X) - c < X < E(X) + c)=0.85? Give your answer to four decimal places. Response: Correct: 4.9232 Instructor feedback: Let X represent the weight of a bag. Since the normal pdf is symmetric about E(X), we are looking for the c such that P(E(X) - c < X) = P(X > E(X) + c) = (1-0.85)/2 = 0.075. Using the StatCrunch Normal calculator under the Stat > Calculators menu with a mean of 47.83 and a standard deviation of 3.42, we find the area to the left of 42.906802 is equal to 0.075. We then have 42.906802 = 47.83 - c which solving for c yields c = 4.9232.

• (1 pts) 1 Assume the weights of individual bags are independent. What is the expected number of bags out of a sample of 19 that weigh less than 50 lbs? Give your answer to four decimal places. Response: Correct: 14.0053 Instructor feedback: Let Y represent the number of bags out of a sample of 19 that weigh less than 50 lbs. Y has a Binomial distribution with parameters n=19 and p=0.737124. E(Y) = 19*0.85 = 14.0053.

• (1 pts) 1 Assuming the weights of individual bags are independent, what is the probability that 15 or fewer bags weigh less than 50 pounds in a sample of size 19? Give your answer to four decimal places. Response: Correct: 0.7764 Instructor feedback: Let Y represent the number of bags out of a sample of 19 that weigh less than 50 lbs. Y has a Binomial distribution with parameters n=19 and p=0.737124. P(Y < 15 ) = 0.7764 using the Binomial calculator in StatCrunch.

• A manufacturer of industrial solvent guarantees its customers that each drum of solvent they ship out contains at least 100 lbs of solvent. Suppose the amount of solvent in each drum is normally distributed with a mean of 102.4 pounds and a standard deviation of 3.58 pounds.

• (1 pts) 1 What is the probability that a drum meets the guarantee? Give your answer to four decimal places. Response: Correct: 0.748696 Instructor feedback: Let X represent the amount of solvent in a drum. Using the StatCrunch Normal calculator under the Stat > Calculators menu with a mean of 102.4 and a standard deviation of 3.58, we find P(X > 100) = 0.748696.

• (1 pts) 1 What would the standard deviation need to be so that the probability a drum meets the guarantee is 0.98? Give your answer to three decimal places. Response: Correct: 1.1686 Instructor feedback: You need to use standardization to work this problem. We want to find the s such that P(X > 100) = 0.98. Note that P(X > 100) = P((X - 102.4)/s > (100 - 102.4)/s) = P(Z > (100 - 102.4)/s) where Z is a standard Normal random variable. Using the StatCrunch Normal calculator with a mean of 0 and a standard deviation of 1, we find P(Z > -2.053749) = 0.98. We then set -2.053749 = (100 - 102.4)/s and solve to find s = 1.1686.

• A company is preparing to launch a new web site to sell its products. From past experience, they know that purchases at the site will occur as a Poisson process with on average 25 purchases every 10 minutes.

• (1 pts) 1 What is the expected time in minutes before the first purchase after the site opens? Give your answer to four decimal places. Response: Correct: 0.4 Instructor feedback: Let X represent the time in minutes before the first purchase. X is a Gamma random variable with a=1 and b=0.4. E(X) = ab = 0.4.

• (1 pts) 1 What is the probability that more than 0.27 minutes pass before the first purchase? Give your answer to four decimal places. Response: Correct: 0.509156 Instructor feedback: Let X represent the time in minutes before the first purchase. X is a Gamma random variable with a=1 and b=0.4. Using the Gamma calculator in

StatCrunch, P(X > 0.27) = 0.509156.

• (1 pts) 1 What is the expected time in minutes until the sixth purchase? Give your answer to four decimal places. Response: Correct: 2.4 Instructor feedback: Let Y represent the time in minutes before the sixth purchase. Y is a Gamma random variable with a=6 and b=0.4. E(Y) = ab = 2.4.

• (1 pts) 1 What is the probability that the sixth purchase will occur between 2 minutes and 2.8 minutes after the site opens? Give your answer to four decimal places. Response: Correct: 0.315252 Instructor feedback: Let Y represent the time in minutes before the sixth purchase. Y is a Gamma random variable with a=6 and b=0.4. Using the Gamma calculator in StatCrunch, P(2 < Y < 2.8) = P(Y < 2.8) - P(Y < 2) = 0.315252.

•


Instructions: This assignment covers Section 5.10 and part of Section 6.2 from your text. You might want to look at the following pages in the eBook as you work through the assignment.

• Joint distributions - pages 145 - 156

• Central Limit Theorem - pages 183 - 185

• A restaurant needs a staff of 3 waiters and 2 chefs to be properly staffed. The joint probability model for the number of waiters (X) and chefs (Y) that show up on any given day is given below.

X

Y 0 1 2 3

0 k 0.01 0.01 0.021 0.02 0.03 0.05 0.062 0.01 0.01 0.06 0.7

•• (1 pts) 1

What must the value of k be for this to be a valid probability model? Response: Correct: 0.02 Instructor feedback: The probabilities must add up to 1, so k = 1 - (0.01 + 0.01 + 0.02 + 0.02 + 0.03 + 0.05 + 0.06 + 0.01 + 0.01 + 0.06 + 0.7) = 0.02.

• (1 pts) 1

What is the probability that at least one waiter and at least one chef show up on any given day? Response: Correct: 0.91 Instructor feedback: P(X > 1, Y > 1) = P(X=1,Y=1) + P(X=2,Y=1) + P(X=3,Y=1) + P(X=1,Y=2) + P(X=2,Y=2) + P(X=3,Y=2) = 0.03 + 0.05 + 0.06 + 0.01 + 0.06 + 0.7 = 0.91.

• (1 pts) 1 What is the probability that more chefs show up than waiters on any given day? Response: Correct: 0.04 Instructor feedback: P(Y > X) = P(X=0,Y=1) + P(X=0,Y=2) + P(X=1,Y=2) = 0.02 + 0.01 + 0.01 = 0.04.

• (1 pts) 1 What is the probability that more than three total staff (waiters and chefs) will show up on any given day? Response: Correct: 0.82 Instructor feedback: P(X+Y > 3) = P(X=3,Y=1) + P(X=2,Y=2) + P(X=3,Y=2) = 0.06 + 0.06 + 0.7 = 0.82.

• (1 pts) 1 What is the expected total number of staff (waiters and chefs) that will show up on any given day? Response: Correct: 4.35 Instructor feedback: E(X+Y) = (0+0)*0.02 + (1+0)*0.01 + (2+0)*0.01 + (3+0)*0.02 + (0+1)*0.02 + (1+1)*0.03 + (2+1)*0.05 + (3+1)*0.06 + (0+2)*0.01 + (1+2)*0.01 + (2+2)*0.06 + (3+2)*0.7 = 4.35.

• (1 pts) 1 What is the probability that three waiters will show up on any given day? Response: Correct: 0.78 Instructor feedback: P(X=3) = P(X=3,Y=0) + P(X=3,Y=1) + P(X=3,Y=2) = 0.02 + 0.06 + 0.7 = 0.78.

• (1 pts) 1 What is the probability that two chefs will show up on any given day? Response: Correct: 0.78 Instructor feedback: P(Y=2) = P(X=0,Y=2) + P(X=1,Y=2) + P(X=2,Y=2) + P(X=3,Y=2) = 0.01 + 0.01 + 0.06 = 0.7 = 0.78.

• (1 pts) 1 X and Y are independent. TRUE or FALSE

TRUE FALSE Instructor feedback: X and Y are dependent because P(X=3,Y=2) = 0.7 is not equal to P(X=3)*P(Y=2) = 0.78*0.78.

• A manufacturer has designed a process to produce pipes that are 10 feet long. The distribution of the pipe length, however, is actually Uniform on the interval 10 feet to 10.71 feet. Assume that the lengths of individual pipes produced by the process are independent. Let X and Y represent the lengths of two different pipes produced by the process.

• (1 pts) 1 What is the joint pdf for X and Y? f(x,y) = xy 10 < x < 10.71, 10 < y < 10.71 f(x,y) = 1/(0.71)2 10 < x < 11, 10 < y < 11 f(x,y) = 1/(0.71)2 10 < x < 10.71, 10 < y < 10.71 f(x,y) = 1 10 < x < 10.71, 10 < y < 10.71

• (1 pts) 0 What is the probability that a single pipe will be between 10.13 feet and 10.44 feet long? Give your answer to four decimal places. Response: Correct: 0.43662 Instructor feedback: P(10.13 < X < 10.44) = int_10.13^10.44 1/0.71 dx = (10.44- 10.13)/0.71 = 0.43662.

• (1 pts) 0 What is the probability that both pieces of pipe are between 10.13 feet and 10.44 feet long? Give your answer to four decimal places. Hint: Try to avoid doing calculus to solve this problem. Response: Correct: 0.1906 Instructor feedback: P(10.13 < X < 10.44,10.13 < Y < 10.44) = int_10.13^10.44 int_10.13^10.44 1/(0.71)2 dx dy = (10.44 - 10.13)*(10.44 - 10.13)/(0.71)2 = 0.1906. Graphically, we are finding the volume under the pdf over a region like the one shown below with w=0.71, L=10.13 and U=10.44. Remember to think of the pdf itself as coming out of your computer screen to a height of 1/(0.71)2. The area is clearly a square with area (10.44 - 10.13)*(10.44 - 10.13), so we multiply this area to by 1/(0.71)2 to get the volume. Note that since X and Y are independent, P(10.13 < X < 10.44,10.13 < Y < 10.44) = P(10.13 < X < 10.44)*P(10.13 < Y < 10.44) = [(10.44- 10.13)/0.71]*[(10.44- 10.13)/0.71].

• (1 pts) 0 What is the expected length of a single pipe? Give your answer to three decimal places. Response: Correct: 10.355 Instructor feedback: E(X) = (10 + 10.71)/2 since X is Uniform on 10 to 10.71.

• (1 pts) 0 What is the expected total length of the two pieces of pipe? Give your answer to three decimal places. Response: Correct: 20.71 Instructor feedback: E(X + Y) = ∫_10^10.71 ∫_10^10.71 (x+y)/(0.712) dx dy = 1/(0.71)2 ∫_10^10.71 (10.71)2/2 - (10)2/2 + 10.71y - 10y dy = (10.71 + 10)/2 + (10.71 + 10)/2 = E(X) + E(Y) = 20.71

• (1 pts) 0 What is the variance of the length of a single pipe? Give your answer to four decimal places. Response: Correct: 0.042008 Instructor feedback: The variance of a Uniform 10 to 10.71 is (10.71 - 10)2/12 = 0.712/12 = 0.042008.

• (1 pts) 0 What is the variance of the total length of both pipes? Give your answer to four decimal places. Response: Correct: 0.084017 Instructor feedback: Let Z = X + Y. Recall that Var(Z) = E(Z2) - (E(Z))2. We found E(Z) = E(X) + E(Y) above, so all we need to find now is E(Z2). E(Z2) = ∫_10^10.71 ∫_10^10.71 (x+y)2/(0.712) dx dy = 1/(0.71)2 ∫_10^10.71 ∫_10^10.71 x2 + 2xy + y2 dx dy = 1/(0.71)2 ∫_10^10.71 (10.71)3/3 - (10)3/3 + 2y[(10.71)2/2 - (10)2/2] + 10.71y2 - 10y2 dy = 1/(0.71)2[(10.71)3/3 - (10)3/3 + 2((10.71)2/2 - (10)2/2)((10.71)2/2 - (10)2/2) + (10.71)3/3 - (10)3/3] = [(10.71)3/3 - (10)3/3]/0.71 + 2((10.71+10)/2)((10.71+10)/2) + [(10.71)3/3 - (10)3/3]/0.71 = [(10.71)3/3 - (10)3/3]/0.71 + 2E(X)E(Y) + [(10.71)3/3 - (10)3/3]/0.71 So, Var(Z) = [(10.71)3/3 - (10)3/3]/0.71 + 2E(X)E(Y) + [(10.71)3/3 - (10)3/3]/0.71 - [E(X)]2 - 2E(X)E(Y) - [E(Y)]2 = [(10.71)3/3 - (10)3/3]/0.71 - [E(X)]2 + [(10.71)3/3 - (10)3/3]/0.71 - [E(Y)]2 = E(X2) - [E(X)]2 + E(Y2) - [E(Y)]2 = Var(X) + Var(Y) = 0.042008 + 0.042008 = 0.084017. Note that E(X2) = E(Y2) = [(10.71)3/3 - (10)3/3]/0.71. Since X and Y are independent, the variance of the sum is the sum of the variances.

• (1 pts) 0 What is the probability that the second pipe (with length Y) is more than 0.34 feet

longer than the first pipe (with length X)? Give your answer to four decimal places. Hint: Do not use calculus to get your answer. Response: Correct: 0.1358 Instructor feedback: We want to find P(Y > X + 0.34). The key here is to draw the proper region. Reference the graph below with w=0.71 and a=0.34. In this case, our region of integration is the shaded triangle. Since the joint pdf is flat at the value of 1/(0.712) over the entire square. the probability we are trying to find is just the area of the rectangle (0.71-0.34)2/2 times the height of the function 1/(0.712) which turns out to be 0.1358.

• Calls to a customer service center last on average 2.4 minutes with a standard deviation of 2 minutes. An operator in the call center is required to answer 100 calls each day. Assume the call times are independent.

• (1 pts) 1 What is the expected total amount of time in minutes the operator will spend on the calls each day? Give an exact answer. Response: Correct: 240 Instructor feedback: The expected value of the sum of the 100 independent times is 100*2.4 = 240.

• (1 pts) 1 What is the standard deviation of the total amount of time in minutes the operator will spend on the calls each day? Give your answer to four decimal places. Response: Correct: 20 Instructor feedback: The standard deviation of the total time of the {n] independent calls is sqrt(100)*2 = 20.

• (1 pts) 0 What is the approximate probability that the total time spent on the calls will be less than 228 minutes? Give your answer to four decimal places. Use the standard deviation as you entered it above to answer this question. Response: Correct: 0.274253 Instructor feedback: We have a large sample so we use the normal approximation from the CLT. Using the StatCrunch normal calculator with the mean and standard deviation given above, P(total time less than 228) = 0.274253.

• (1 pts) 0 What is the value c such that the approximate probability that the total time spent on the calls each day is less than c is 0.9? Give your answer to four decimal places. Use the standard deviation as you entered it above to answer this question.

Response: Correct: 265.631

Instructor feedback: We have a large sample so we use the normal approximation from the CLT. Using the StatCrunch normal calculator with the mean and standard deviation given above, P(total time < 265.631) = 0.9.

• A particular variety of watermelon weighs on average 26.1 pounds with a standard deviation of 1.02 pounds. Consider the sample mean weight of 84 watermelons of this variety. Assume the individual watermelon weights are independent.

• (1 pts) 0 What is the expected value of the sample mean weight? Give an exact answer. Response: Correct: 26.1 Instructor feedback: The expected value of the sample mean is the same as the population mean which is 26.1.

• (1 pts) 0 What is the standard deviation of the sample mean weight? Give your answer to four decimal places. Response: Correct: 0.1113 Instructor feedback: The standard deviation of the sample mean is 1.02/sqrt(84) = 0.1113.

• (1 pts) 0 What is the approximate probability the sample mean weight will be less than 25.95? Give your answer to four decimal places. Use the standard deviation as you entered it above to answer this question. Response: Correct: 0.088876 Instructor feedback: We have a large sample so we use the normal approximation from the CLT. Using the StatCrunch normal calculator with the mean and standard deviation given above, P(sample mean less than 25.95) = 0.088876.

• (1 pts) 0 What is the value c such that the approximate probability the sample mean will be less than c is 0.91? Give your answer to four decimal places. Use the standard deviation as you entered it above to answer this question. Response: Correct: 26.2492 Instructor feedback: We have a large sample so we use the normal approximation from the CLT. Using the StatCrunch normal calculator with the mean and standard deviation given above, P(total time less than 26.2492) = 0.91.

•


Instructions: This assignment covers confidence intervals for a population mean, population variance and a population proportion. These topics are covered in portions of Chapters 6, 7 and 9. You might want to look at the following pages in the eBook as you work through the assignment.

• Sampling distribution of the sample mean - pages 187 - 189

• Sampling distribution of the sample variance - pages 189 - 190

• Confidence interval for a population mean - pages 209 - 212

• Confidence interval for a population variance - pages 268 - 270

• Confidence interval for a population proportion - pages 278 - 281

• The data below contain the number of defects observed on each

of 50 lcd screens made by a certain manufacturer.

Construct a 95% confidence interval for the mean number of defects per screen for all screens produced by this manufacturer.

• (1 pts) 1 What is the lower limit on the interval? Give your answer to three decimal places. Response: Correct: 1.692 Instructor feedback: Since the sample size is large (> 30), we use the Stat > Z statistics > One Sample option. Select Defects then click Next and choose the confidence interval option.

• (1 pts) 1 What is the upper limit on the interval? Give your answer to three decimal places. Response: Correct: 2.588 Instructor feedback: Since the sample size is large (> 30), we use the Stat > Z statistics > One Sample option. Select Defects then click Next and choose the confidence interval option.

• (1 pts) 1 Based on the interval above, would you believe that the mean number of defects per screen for all screens from this manufacturer is 1.77? No because 1.77 is not inside the interval. No because 1.77 is inside the interval. Yes because 1.77 is inside the interval. Yes because 1.77 is not inside the interval.

• (1 pts) 1 If we are very certain that the true standard deviation of the number of defects per screen is below 2, what sample size would be required so that the width of the 95% confidence interval for the mean number of defects per screen is at most 0.26? Make sure you enter a whole number below. Response: Correct: 910 Instructor feedback: Recall that z0.025=1.96. The required sample size is (2*1.96*2/0.26)2 = 910 when rounded up.

• A sanitation department is interested in estimating the mean amount of garbage per bin for all bins in the city. In a random

sample of 40 bins, the sample mean amount was 47.1 pounds and the sample standard deviation was 3.9 pounds. Construct 95% and 99% confidence intervals for the mean amount of garbage per bin for all bins in the city.

• (1 pts) 1 What is the lower limit of the 95% interval? Give your answer to three decimal places. Response: Correct: 45.891 Instructor feedback: Recall that z0.025=1.96. So the lower limit is 47.1 - 1.96*3.9/sqrt(40) = 45.891.

• (1 pts) 1 What is the upper limit of the 95% interval? Give your answer to three decimal places. Response: Correct: 48.309 Instructor feedback: Recall that z0.025=1.96. So the upper limit is 47.1 + 1.96*3.9/sqrt(40) = 48.309.

• (1 pts) 1 What is the lower limit of the 99% interval? Give your answer to three decimal places. Response: Correct: 45.512 Instructor feedback: Recall that z0.005=2.5758. So the lower limit is 47.1 - 2.5758*3.9/sqrt(40) = 45.512.

• (1 pts) 1 What is the upper limit of the 99% interval? Give your answer to three decimal places. Response: Correct: 48.688 Instructor feedback: Recall that z0.005=2.5758. So the upper limit is 47.1 + 2.5758*3.9/sqrt(40) = 48.688.

• (1 pts) 1 Consider the claim that the mean amount of garbage per bin is 48.4985 pounds. Is the following statement true or false? The decision about the claim would depend on whether we use a 95% or 99% confidence interval. TRUE FALSE

• Company A is trying to sell its website to Company B. As part of the sale, Company A claims that the average user of their site stays on the site for 10 minutes. To test this claim Company B collects the times (in minutes) below for a sample of 15 users.

Construct a 93% confidence interval for the true mean time spent on the web site.

• (1 pts) 1 What is the lower limit of the 93% interval? Give your answer to three decimal places. Enter 0 if your lower limit is less than 0. Response: Correct: 3.092 Instructor feedback: Since the sample size is small (< 30), we use the Stat > T statistics > One Sample option. Select Time then click Next and choose the confidence interval option with appropriate confidence level.

• (1 pts) 1 What is the upper limit of the 93% interval? Give your answer to three decimal places. Response: Correct: 9.642 Instructor feedback: Since the sample size is small (< 30), we use the Stat > T statistics > One Sample option. Select Time then click Next and choose the confidence interval option with appropriate confidence level.

• (1 pts) 1 Based on this data, do you believe the claim made by Company A? No because 10 is not inside the interval. No because 10 is inside the interval. Yes because 10 is inside the interval. Yes because 10 is not inside the interval.

• (1 pts) 1 Which of the following assumptions should be checked before constructing the above confidence interval? the data need to have small variance the data need to follow a t distribution the data need to follow a normal distribution the data need to be skewed Instructor feedback: In this case, an exponential distribution was used to simulate the data so the t confidence interval is really not appropriate.

• A university with a high water bill is interested in estimating the mean amount of time that students spend in the shower each day. In a sample of 19 students, the average time was 5.07 minutes and the standard deviation was 1.31 minutes. Using this sample information, construct a 99% confidence interval for the mean

amount of time that students spend in the shower each day.

• (1 pts) 1 What is the lower limit of the 99% interval? Give your answer to three decimal places. Response: Correct: 4.205 Instructor feedback: Since the sample size is small, we use a t interval. Using the StatCrunch t calculator, we find t0.005,18=2.8784. So the lower limit is 5.07 - 2.8784*1.31/sqrt(19) = 4.205.

• (1 pts) 1 What is the upper limit of the 99% interval? Give your answer to three decimal places. Response: Correct: 5.935 Instructor feedback: Since the sample size is small, we use a t interval. Using the StatCrunch t calculator, we find t0.005,18=2.8784. So the upper limit is 5.07 + 2.8784*1.31/sqrt(19) = 5.935.

• A company produces a women's bowling ball that is supposed to weigh exactly 14 pounds. Unfortunately, the company has a problem with the variability of the weight. In a sample of 12 of the bowling balls the sample standard deviation was found to be 0.68 pounds. Construct a 95% confidence interval for the variance of the bowling ball weight.

• (1 pts) 1 What is the lower limit of the 95% interval? Give your answer to three decimal places. Response: Correct: 0.232 Instructor feedback: Using the StatCrunch chi-square calculator, we find c2

0.975,11=21.92. So the lower limit is (12-1)*0.682/21.92 = 0.232.

• (1 pts) 1 What is the upper limit of the 95% interval? Give your answer to three decimal places. Response: Correct: 1.333 Instructor feedback: Using the StatCrunch chi-square calculator, we find c2

0.025,11=3.8157. So the upper limit is (12-1)*0.682/3.8157 = 1.333.

• (1 pts) 1 Which of the following assumptions should be checked before constructing the above confidence interval?

the data need to have small variance the data need to follow a chi-square distribution the data need to follow a normal distribution the data need to be right skewed

• A web based software company is interested in estimating the proportion of individuals who use the Firefox browser. In a sample of 200 of individuals, 18 users stated that they used Firefox. Using this data, construct a 99% confidence interval for the proportion of all individuals that use Firefox.

• (1 pts) 1 What is the lower limit on the 99% confidence interval? Give your answer to three decimal places. Response: Correct: 0.038 Instructor feedback: Recall that z0.005=2.5758. The sample proportion is 18/200 = 0.09, so the lower limit is 0.09 - 2.5758*sqrt(0.09*(1-0.09)/200) = 0.038.

• (1 pts) 1 What is the upper limit on the 99% confidence interval? Give your answer to three decimal places. Response: Correct: 0.142 Instructor feedback: Recall that z0.005=2.5758. The sample proportion is 18/200 = 0.09, so the upper limit is 0.09 + 2.5758*sqrt(0.09*(1-0.09)/200) = 0.142.

• (1 pts) 1 Google states that the proportion of all individuals that use Firefox is 0.2. Based on the interval above, does the Google claim seem reasonable? No because 0.2 is not inside the interval. No because 0.2 is inside the interval. Yes because 0.2 is inside the interval. Yes because 0.2 is not inside the interval.

• (1 pts) 1 What sample size would be required so that the width of the 99% confidence interval would be at most 0.04 units wide? Be as conservative as possible with your answer! Response: Correct: 4147 Instructor feedback: The required sample size is (2.5758/0.04)2 = 4147 when rounded up.

•


Instructions: This assignment covers hypothesis testing for a population mean and a population proportion. These topics are covered in portions of Chapters 6, 7 and 9. You might want to look at the following pages in the eBook as you work through the assignment.

• Sampling distribution of the sample mean - pages 187 - 189

• Basics of hypothesis testing - pages 222 - 227

• Hypothesis test for a population mean - pages 229 - 234

• Hypothesis test for a population proportion - pages 283 - 284

• A quality control engineer at a particular lcd screen manufacturer is studying the mean number of defects per screen. Based on historical evidence, the mean number of defects per screen was thought to be 2.63. There have recently been changes to the manufacturing process, and the engineer now feels that the mean number of defects per screen may be significantly smaller than 2.63. Using the number of defects on each of 50 sample screens shown below, conduct the appropriate hypothesis test using a 0.05 level of significance.

• (1 pts) 1 What are the appropriate null and alternative hypotheses?

H0: = 2.63 versus Ha: ≠ 2.63 H0: = 2.63 versus Ha: < 2.63 H0: = 2.63 versus Ha: > 2.63 H0: = 2.63 versus Ha: > 2.63

• (1 pts) 1 What is the test statistic? Give your answer to four decimal places. Response: Correct: -0.5276 Instructor feedback: Since n is large, we use the one sample Z test in StatCrunch (Stat > Z statistics > One sample). Select the defects column and then click Next. Specify 2.63 for the null mean and "<" for the alternative hypothesis. Click Calculate to find the Z statistic is -0.5276.

• (1 pts) 1 What is the P-value for the test? Give your answer to four decimal places.

Response: Correct: 0.2989 Instructor feedback: Since n is large, we use the one sample Z test in StatCrunch (Stat > Z statistics > One sample). Select the defects column and then click Next. Specify 2.63 for the null mean and "<" for the alternative hypothesis. Click Calculate to find the P-value is 0.2989. You can verify this answer using the Normal Calculator to find P(Z < -0.5276) = 0.2989.

• (1 pts) 1 What is the appropriate conclusion? Fail to reject the claim that the mean number of defects per screen is 2.63 because the P-value is smaller than 0.05. Fail to reject the claim that the mean number of defects per screen is 2.63 because the P-value is larger than 0.05. Reject the claim that the mean number of defects per screen is 2.63 because the P-value is larger than 0.05. Reject the claim that the mean number of defects per screen is 2.63 because the P-value is smaller than 0.05.

• A sanitation supervisor is interested in testing to see if the mean amount of garbage per bin is different from 50. In a random sample of 36 bins, the sample mean amount was 48.1 pounds and the sample standard deviation was 3.1 pounds. Conduct the appropriate hypothesis test using a 0.1 level of significance.


H0: = 50 versus Ha: ≠ 50 H0: = 50 versus Ha: < 50 H0: = 50 versus Ha: > 50 H0: = 50 versus Ha: ≠ 50

• (1 pts) 1 What is the test statistic? Give your answer to four decimal places. Response: Correct: -3.6774 Instructor feedback: We use the Z statistic since n is large. The Z statistic is (48.1 - 50)/(3.1/sqrt(36)) = -3.6774.

• (1 pts) 1 What is the P-value for the test? Give your answer to four decimal places. Response: Correct: 0.0002 Instructor feedback: For this two-sided test, the P-value is 2*P(Z > |-3.6774|) = 0.0002.

• (1 pts) 1 What is the appropriate conclusion? Fail to reject the claim that the mean amount per bin is 50 pounds because the P-value is smaller than 0.1. Fail to reject the claim that the mean amount per bin is 50 pounds because the P-value is larger than 0.1. Reject the claim that the mean amount per bin is 50 pounds because the P-value is larger than 0.1. Reject the claim that the mean amount per bin is 50 pounds because the P-value is smaller than 0.1.

• Company A is trying to sell its website to Company B. As part of the sale, Company A claims that the average user of their site stays on the site for 10 minutes. Company B is concerned that the mean time is significantly less than 10 minutes. Company B collects the times (in minutes) below for a sample of 11 users.

Conduct the appropriate hypothesis test for Company B using a 0.08 level of significance.



• (1 pts) 1 What is the test statistic? Give your answer to four decimal places. Response: Correct: -5.9524 Instructor feedback: We use a T statistic because the sample size is small. Using Stat > T statistics > One Sample, we select the Time column and then click Next. Specify "<" for the alternative hypothesis and 10 for the null mean. Click Calculate to find the T statistic of -5.9524.

• (1 pts) 1 What is the critical value for the test? Give your answer to four decimal places. Response: Correct: -1.5179 Instructor feedback: Use stat > Calculators > T and enter 0.08 for the probability and "<" for the direction, to find P(T < -1.5179) = 0.08.

• (1 pts) 1 What is the appropriate conclusion? Fail to reject the claim that the mean time is 10 minutes because the test statistic is smaller than the critical point. Fail to reject the claim that the mean time is 10 minutes because the test statistic is larger than the critical point. Reject the claim that the mean time is 10 minutes because the test statistic is larger than the critical point. Reject the claim that the mean time is 10 minutes because the test statistic is smaller than the critical point.

• A facilities manager at a university reads in a research report that the mean amount of time spent in the shower by an adult is 5 minutes. He decides to collect data to see if the mean amount of time that college students spend in the shower is significantly different from 5 minutes. In a sample of 13 students, he found the average time was 5.39 minutes and the standard deviation was 0.75 minutes. Using this sample information, conduct the appropriate hypothesis test at the 0.1 level of significance.



• (1 pts) 1 What is the test statistic? Give your answer to four decimal places. Response: Correct: 1.8749 Instructor feedback: We use the T statistic since n is small. The T statistic is (5.39 - 5)/(0.75/sqrt(13)) = 1.8749.

• (1 pts) 1 What is the P-value for the test? Give your answer to four decimal places. Response: Correct: 0.0853 Instructor feedback: For this two-sided test, the P-value is 2*P(T > |1.8749|) = 0.0853 where T has 13 - 1 degrees of freedom.

• (1 pts) 1 What is the appropriate conclusion? Fail to reject the claim that the mean time is 5 minutes because the P-value is smaller than 0.1. Fail to reject the claim that the mean time is 5 minutes because the P-value is larger than 0.1.

Reject the claim that the mean time is 5 minutes because the P-value is larger than 0.1. Reject the claim that the mean time is 5 minutes because the P-value is smaller than 0.1.

• A research report claims that 20% of all individuals use Firefox to browse the web. A software company is trying to determine if the proportion of their users who use Firefox is significantly different from 0.2. In a sample of 200 of their users, 30 users stated that they used Firefox. Using this data, conduct the appropriate hypothesis test using a 0.01 level of significance.

• (1 pts) 1 What are the appropriate hypotheses? H0: p = 0.2 versus Ha: p ≠ 0.2 H0: p = 0.2 versus Ha: p < 0.2 H0: p = 0.2 versus Ha: p > 0.2 H0: = 0.2 versus Ha: > 0.2

• (1 pts) 1 What is the test statistic? Give your answer to four decimal places. Response: Correct: -1.7678 Instructor feedback: The Z statistic is (0.15 - 0.2)/sqrt(0.2*(1-0.2)/200) = -1.7678.

• (1 pts) 1 What is the P-value for the test? Give your answer to four decimal places. Response: Correct: 0.0771 Instructor feedback: The P-value for this two-sided test is 2*P(Z > |-1.7678|) = 0.0771.

• (1 pts) 1 What is the appropriate conclusion? Fail to reject the claim that the Firefox proportion is 0.2 because the P-value is smaller than 0.01. Conclude that the Firefox proportion is 0.2 because the P-value is larger than 0.01. Conclude that the Firefox proportion is not 0.2 because the P-value is larger than 0.01. Conclude that the Firefox proportion is not 0.2 because the P-value is smaller than 0.01.

• Company A makes a large shipment to Company B. Company B can reject the shipment if they can conclude that the proportion of defective items in the shipment is larger than 0.1. In a sample of 400 items from the shipment, Company B finds that 59 are defective. Conduct the appropriate hypothesis test for Company

B using a 0.01 level of significance.

• (1 pts) 1 What are the appropriate hypotheses? H0: p = 0.1 versus Ha: p ≠ 0.1 H0: p = 0.1 versus Ha: p < 0.1 H0: p = 0.1 versus Ha: p > 0.1 H0: = 0.1 versus Ha: > 0.1

• (1 pts) 1 What is the test statistic? Give your answer to four decimal places. Response: Correct: 3.1667 Instructor feedback: The Z statistic in this case is (0.1475 - 0.1)/sqrt(0.1*(1-0.1)/400) = 3.1667.

• (1 pts) 0 What is the critical point for the test? Give your answer to four decimal places. Response: Correct: 2.3263 Instructor feedback: The critical point is z0.01=2.3263 which can be found using the Normal calculator.

• (1 pts) 1 What is the appropriate conclusion? Fail to reject the claim that the defective proportion in the shipment is 0.1 because the test statistic is larger than the critical point. Fail to reject the claim that the defective proportion in the shipment is 0.1 because the test statistic is smaller than the critical point. Conclude that the defective proportion in the shipment is greater than 0.1 because the test statistic is smaller than the critical point. Conclude that the defective proportion in the shipment is greater than 0.1 because the test statistic is larger than the critical point.

•

Hw 8 Due: Monday, Nov 14, 2011 8:00 am CST

Instructions: This assignment covers two sample confidence intervals and hypothesis tests for comparing two population means, two population variances and two population proportions. You might want to look at the following pages in the eBook as you work through the assignment.

• Confidence intervals/hypothesis tests for two means - pages 246 - 261

• Hypothesis test for two variances - pages 272 - 275

• An engineer in charge of water rationing for the U.S. Army wants to determine if the average male soldier spends less time in the shower than the average female soldier. Let mm represent the average time in the shower of male soldiers and mf represent the average time in the shower of female soldiers.

• (1 pts) 1 What are the appropriate hypotheses for the engineer? H0: mm = mf versus Ha: mm ≠ mf H0: mm = mf versus Ha: mm < mf H0: mm = mf versus Ha: mm > mf H0: sm = sf versus Ha: sm > sf

• (1 pts) 1 Among a sample of 56 male soldiers the average shower time was found to be 2.48 minutes and the standard deviation was found to be 0.66 minutes. Among a sample of 59 female soldiers the average shower time was found to be 2.55 minutes and the standard deviation was found to be 0.74 minutes. What is the test statistic? Give your answer to three decimal places. Response: Correct: -0.5359 Instructor feedback: Z = (2.48 - 2.55)/sqrt(0.662/56 + 0.742/59) = -0.5359.

• (1 pts) 1 What is the P-value for the test? Give your answer to four decimal places. Response: Correct: 0.296 Instructor feedback: P-value = P(Z < -0.5359) = 0.296.

• (1 pts) 1 Using a 0.05 level of significance, what is the appropriate conclusion? Reject the claim that the average shower times are different for male and female soldiers because the P-value is greater than 0.05. Fail to reject the claim that the average shower times are the same for male and female soldiers because the P-value is greater than 0.05. Conclude that the average shower time for males is equal to the average shower time for females because the P-value is less than 0.05. Conclude that the average shower time for males is less

than the average shower time for females because the P-value is less than 0.05.

• Reserve Capacity (RC) is the number of minutes a fully charged battery at 80 degrees will discharge 25 amps until the battery drops below 10.5 volts. An automobile manufacturer wants to compare the mean RC for batteries from two different suppliers. RC data (in minutes) for samples of 50 batteries from each supplier are shown below. Construct a 95% confidence interval for the difference between the mean RC for supplier 1 and the mean RC for supplier 2.

• (1 pts) 1 What is the lower limit of the 95% confidence interval? Give your answer to three decimal places. Response: Correct: 0.313 Instructor feedback: Since the sample sizes are both large, use the Stat > Z statistics > Two sample option in StatCrunch. Select Supplier 1 for Sample 1 and Supplier 2 for Sample 2. Click Next and enter the value of 95 divided by 100 for the confidence level. Then click Calculate to get the interval.

• (1 pts) 1 What is the upper limit of the 95% confidence interval? Give your answer to three decimal places. Response: Correct: 7.847 Instructor feedback: Since the sample sizes are both large, use the Stat > Z statistics > Two sample option in StatCrunch. Select Supplier 1 for Sample 1 and Supplier 2 for Sample 2. Click Next and enter the value of 95 divided by 100 as a decimal for the confidence level. Then click Calculate to get the interval.

• (1 pts) 1 Based on this interval, does the claim that mean RC rating is the same for both suppliers seem reasonable? No because 0 is not inside the interval. No because 0 is inside the interval. Yes because 0 is inside the interval. Yes because 0 is not inside the interval.

• A hospital wishes to justify the benefits of nutrition programs for pregnant women using birth weight data from newborns. The hospital hopes to show that the mean birth weight for newborns from mothers who complete the program is higher than the birth

weight for newborns from mothers who do not complete the program. A group of 16 pregnant women were randomly divided into two groups; the first group received the nutrition program and the second group did not receive the program. The resulting weights (in grams) of the newborn babies from each group are shown below.

• (1 pts) 1 Let m1 represent the mean associated with the nutrition program, and let m2 represent the mean associated with no nutrition program. What are the proper hypotheses? H0: m1 = m2 versus Ha: m1 ≠ m2 H0: m1 = m2 versus Ha: m1 < m2 H0: m1 = m2 versus Ha: m1 > m2 H0: m1 > m2 versus Ha: m1 = m2

• (1 pts) 1 Before performing the primary test of interest, what is the P-value for the test where the null is that the variability in birth weight is the same regardless of whether the mother completes the program and the alternative is two sided? Give your answer to four decimal places. Response: Correct: 0.6868 Instructor feedback: Choose the Stat > Variance > Two sample option from StatCrunch. Select Group 1 for Sample 1 and Group 2 for Sample 2. Click Compute. By default a two sided test is computed.

• (1 pts) 1 For the primary test of interest specified in Part a, use the test that assumes equal variances since the P-value in Part b is not significant. What is the test statistic? Give your answer to four decimal places. Response: Correct: 1.5094 Instructor feedback: Since the sample sizes are both small, choose the Stat > T Statistics > Two sample option from StatCrunch. Select Group 1 for Sample 1 and Group 2 for Sample 2. Click Next and select the greater than alternative. Click Compute. By default the variances are pooled. By the way, a QQ plot shows that both groups are roughly normally distributed.

• (1 pts) 1 What is the P-value associated with the test statistic? Give your answer to four decimal places. Response:

Correct: 0.0767 Instructor feedback: Since the sample sizes are both small, choose the Stat > T Statistics > Two sample option from StatCrunch. Select Group 1 for Sample 1 and Group 2 for Sample 2. Click Next and select the greater than alternative. Click Compute. By default the variances are pooled. By the way, a QQ plot shows that both groups are roughly normally distributed.

• (1 pts) 1 What is the appropriate conclusion for the hospital using a 0.1 level of significance? Reject the claim that the mean birth weight with the program is higher than the mean birth weight without the program because the P-value is less than 0.1. Fail to reject the claim that the mean birth weight with the program is equal to the mean birth weight without the program because the P-value is greater than 0.1. Fail to reject the claim that the mean birth weight with the program is equal to the mean birth weight without the program because the P-value is less than 0.1. Conclude that the mean birth weight with the program is higher than the mean birth weight without the program because the P-value is less than 0.1.

• A professor is interested in comparing the average amount spent on textbooks for freshmen and sophomores. A random sample of 10 freshmen yielded a sample mean amount of $1062 and a sample standard deviation of $54. A random sample of 10 sophomores yielded a sample mean amount of $1243 and a sample standard deviation of $310. Construct a 95% confidence interval for the difference between the mean amount spent on textbooks by freshmen and the mean amount spent by sophomores (ie. do freshmen minus sophomores). Since the sample standard deviations are wildly different, use a method which does not assume the populations have the same variance.

• (1 pts) 1 What is the appropriate degrees of freedom in this case? Give your answer to four decimal places. Response: Correct: 9.5457 Instructor feedback: v = (542/10 + 3102/10)2/((542/10)2/(10-1) + (3102/10)2/(10-1)) = 9.5457.

• (1 pts) 1 What is the lower confidence limit on the interval? Give your answer to two decimal places. Response: Correct: -404.154 Instructor feedback: The critical point is t0.05/2,9.5457 = 2.2426, so the lower endpoint on the interval is 1062 -

1243 - 2.2426*sqrt(542/10 + 3102/10) = -404.154.

• (1 pts) 1 What is the upper confidence limit on the interval? Give your answer to two decimal places. Response: Correct: 42.154 Instructor feedback: The critical point is t0.05/2,9.5457 = 2.2426, so the upper endpoint on the interval is 1062 - 1243 + 2.2426*sqrt(542/10 + 3102/10) = 42.154.

• (1 pts) 1 Using this interval, is it reasonable to conclude that the average sophomore spends more than the average freshman? Yes because the upper endpoint on the interval is below 0. Yes because the upper endpoint on the interval is above 0. No because the interval contains 0. Yes because the interval contains 0.

• A consumer advocacy group feels that Walmart provides a less safe shopping environment than Target. To try to prove their point, they randomly selected 24 Walmart stores and found the annual number of crime reports filed for these stores. Each of these Walmart stores was then paired with a Target store within a 10 mile radius, and the annual crime reports for the associated Target stores were also recorded. The data is shown below. Test to to see if the mean number of crime incidents at Walmart stores is significantly larger than the mean number of crime incidents at Target stores. Please note the data shown below is not real, but if you would like to see the real example, check out walmartcrimereport.com. Unfortunately, the real data was collected in a way that was not fair to Walmart.

• (1 pts) 1 Let mW represent the mean incidents associated with Walmart, and let mT represent the mean incidents associated with Target. What are the proper hypotheses? H0: mW = mT versus Ha: mW ≠ mT H0: mW = mT versus Ha: mW < mT H0: mW = mT versus Ha: mW > mT H0: mW > mT versus Ha: mW = mT

• (1 pts) 1 What is the test statistic? Give your answer to four decimal places.

Response: Correct: 1.8763 Instructor feedback: The data is clearly paired in this case, so we need to do a paired t test. To perform this paired t test in StatCrunch, select Stat > T Statistics > Paired option. Specify Walmart for Sample 1 and Target for Sample 2. Click Next and choose the > option for the Alternative. Click Calculate to get the output. The test statistic is shown to be 1.8763.

• (1 pts) 1 Using a 0.05 level of significance, what is the critical point for this test? Give your answer to four decimal places. Response: Correct: 1.7139 Instructor feedback: Use the Stat > Calculators > T option. Specify a degrees of freedom to 24-1=23, ≥ for the inequality and 0.05 for the value of the probability. Click Compute to get the critical value of 1.7139.

• (1 pts) 1 What is the appropriate conclusion? Reject the claim that mW is greater than mT because the test statistic is larger than the critical point. Fail to reject the claim that mW is equal to mT because the test statistic is less than the critical point. Fail to reject the claim that mW is equal to mT because the test statistic is larger than the critical point. Conclude that mW is greater than mT because the test statistic is larger than the critical point.

• Shortly after September 11th 2001, a researcher wanted to determine if the proportion of females that favored war with Iraq was significantly different from the proportion of males that favored war with Iraq. In a sample of 65 females, 26 favored war with Iraq. In a sample of 67 males, 40 favored war with Iraq.

• (1 pts) 1 Let pF represent the proportion of females that favor the war, pM represent the proportion of males that favor the war. What are the proper hypotheses? H0: pF = pM versus Ha: pF ≠ pM H0: pF = pM versus Ha: pF < pM H0: pF = pM versus Ha: pF > pM H0: pF < pM versus Ha: pF = pM

• (1 pts) 1 What is the test statistic? Compute the statistic using male statistics subtracted from female statistics. Give your answer to four decimal places. Response:

Correct: -2.2633 Instructor feedback: First note that the sample proportion for females is px = 26/65 = 0.4 and the sample proportion for males is py = 40/67 = 0.597. The combined sample proportion is pbar = (26 + 40)/(65 + 67) = 0.5. The test statistic is Z = (0.4 - 0.597)/sqrt(0.5*(1-0.5)*(1/65 + 1/67)) = -2.2633.

• (1 pts) 1 What is the P-value for the test? Give your answer to four decimal places. Response: Correct: 0.0236 Instructor feedback: The two-sided P-Value is 2*P(Z > |-2.2633|) which equals 0.0236 using the StatCrunch Normal calculator. You can also find this P-value using the Stat > Proportions > Two sample > with summary option. Specify the values for both samples and click Compute to get the test results.

• (1 pts) 1 Using a 0.01 level of significance, what conclusion should be reached? The proportion of females that favor the war and the proportion of males that favor the war are not significantly different because the P-value is less than 0.01. The proportion of females that favor the war and the proportion of males that favor the war are not significantly different because the P-value is greater than 0.01. The proportion of females that favor the war and the proportion of males that favor the war are significantly different because the P-value is greater than 0.01. The proportion of females that favor the war and the proportion of males that favor the war are significantly different because the P-value is less than 0.01.

• (1 pts) 1 What is the lower endpoint of a 99% confidence interval for the difference between the proportion of females that favor the war and the proportion of males that favor the war? Give your answer to four decimal places. Response: Correct: -0.417 Instructor feedback: The critical point for the interval is z0.01/2 = 2.5758. The lower limit is then given by 0.4 - 0.597 - 2.5758*sqrt(0.4*(1-0.4)/65 + 0.597*(1-0.597)/67) = -0.417.

• (1 pts) 1

What is the upper endpoint of a 99% confidence interval for the difference between the proportion of females that favor the war and the proportion of males that favor the war? Give your answer to four decimal places. Response: Correct: 0.023 Instructor feedback: The critical point for the interval is z0.01/2 = 2.5758. The upper limit is then given by 0.4 - 0.597 + 2.5758*sqrt(0.4*(1-0.4)/65 + 0.597*(1-0.597)/67) = 0.023.

•

Hw 9 Due: Monday, Nov 21, 2011 8:00 am CST

Instructions: This assignment covers the ANOVA procedure for comparing several population means. These topics are covered in Sections 12.1 and 12.2 of your eBook. You might want to look at the following pages in the eBook as you work through the assignment.

• Background - pages 354 - 357

• ANOVA - pages 357 - 367

• A trucking company conducted an ANOVA analysis comparing the mean mileage associated with several different brands of tires. The ANOVA table is shown below.

Source DF SS MS FTreatment 2 3.483 1.7415 2.498924Error 21 14.635 CTotal 23 B

•

• (1 pts) 1 How many brands of tires did the company compare? Response: Correct: 3 Instructor feedback: The number of treatments (in this case brands compared) is DFTRT + 1 = 2 + 1 = 3.

• (1 pts) 1 What is the value of B in the ANOVA table? Give your answer to three decimal places. Response: Correct: 18.118 Instructor feedback: SStot = SStrt + SSerr = 3.483 + 14.635 = 18.118.

• (1 pts) 1 What is the value of C in the ANOVA table? Give your answer to four decimal places. Response: Correct: 0.6969 Instructor feedback: MSerr = SSerr/DFerr = 14.635/21 = 0.6969.

• (1 pts) 1 Using a 0.1 level of significance, what is the critical point compared to the F statistic in order to make a conclusion? Give your answer to four decimal places. Response: Correct: 2.574569 Instructor feedback: Using the StatCrunch F calculator, F0.1,2,21 = 2.574569.

• (1 pts) 1 Using a 0.05 level of significance, what conclusion should the company reach? Fail to reject the hypothesis that one brand has a different mean mileage than the other brands. Fail to reject the hypothesis that all brands have the same mean mileage. Reject the hypothesis that one brand has a different mean mileage than the other brands. Reject the hypothesis that all brands have the same mean mileage.

• A researcher wants to determine the impact of soil type on the growth of a certain type of plant. She grows three plants in each of four different types of soil and measures the growth in inches for each plant after one month resulting in the data below.

• (1 pts) 1 What null hypothesis is the researcher testing if she runs an ANOVA with this data?

One type of soil has a higher mean growth for the plant than the others. The mean growth of the plant in each type of soil is the same. Soil 3 provides a lower mean growth for the plant than the other types of soil. The mean growth of the plant is different in each type of soil. The variability in growth of the plant in each type of soil is the same.

• (1 pts) 1 What is the SStrt for the ANOVA? Give your answer to at least three decimal places. Response: Correct: 10.2066667 Instructor feedback: From the StatCrunch ANOVA,

Source DF SS MS F P-valueTreatment 3 10.2066667 3.4022222 2.907882 0.1011Error 8 9.36 1.17Total 11 19.5666667

•

• (1 pts) 1 What is DFerr for the ANOVA? Response: Correct: 8 Instructor feedback: From the ANOVA output in b, DFerr = 8.

• (1 pts) 1 What is the value of the F statistic for the ANOVA? Give your answer to at least three decimal places. Response: Correct: 2.907882 Instructor feedback: For the ANOVA output in b, F = 2.907882 = 3.4022222/1.17.

• (1 pts) 1 Using a 0.1 level of significance, what conclusion should the researcher reach? Soil 1 has a higher mean growth for the plant than the other types of soil. There is not enough evidence to reject the claim that the mean growth of the plant is the same in each type of soil. Soil 3 has a lower mean growth for the plant than the other types of soil. The mean growth of the plant is not the same for all soil types.

• The NIH wants to compare the mean weight loss for five different diets. They measure the weight loss (in pounds) of 5 men assigned to each of the diets for one month. The resulting data set is below.

• (1 pts) 1 Which of the following is NOT an assumption that should be checked before performing an ANOVA? The average weight loss should be roughly the same for each diet. The variability in weight loss should be roughly the same for each diet. Weight loss should be roughly normally distributed for each diet.

• (1 pts) 1 What is the F statistic for the ANOVA? Give your answer to at least three decimal places. Response: Correct: 3.656907 Instructor feedback: From the StatCrunch ANOVA,

Source DF SS MS F P-valueTreatment 4 11.6904 2.9226 3.656907 0.0216Error 20 15.984 0.7992Total 24 27.6744

•

• (1 pts) 1 Using a 0.01 level of significance, what is the critical point that one would compare to the F statistic in order to make a conclusion? Give your answer to three decimal places. Response: Correct: 4.43069 Instructor feedback: From the StatCrunch F calculator, F0.01,4,20 = 4.43069.

• (1 pts) 1 What is the P-value from the ANOVA? Give your answer to four decimal places. Response: Correct: 0.0216 Instructor feedback: See the ANOVA output from b part.

• (1 pts) 1 What is the proper conclusion for NIH in this case? Diet 4 has the largest mean weight loss because the F statistic is larger than the critical point.

Fail to reject the claim that all diets are equivalent in terms of mean weight loss because the F statistic is smaller than the critical point. Diet 5 has the largest mean weight loss because the F statistic is larger than the critical point. Reject the claim that all diets are equivalent in terms of mean weight loss because the F statistic is larger than the critical point. Fail to reject the claim that all diets are equivalent in terms of mean weight loss because the F statistic is larger than the critical point.

• A pizza lover wants to compare the average delivery times for four local pizza restaurants. Over the course of a few weeks, he orders a number of pizzas from each restaurant, and he records the time it takes for each pizza to be delivered.

• (1 pts) 1 When performing an ANOVA with this data, what is the alternative hypothesis? All of the restaurants have different mean delivery times Two of the restaurants have different mean delivery times

At least two of the restaurants have different mean delivery times One of the restaurants has a different mean delivery time than the others

• (1 pts) 1 A partial ANOVA table for his data is shown below. What is the value of B?

Source DF SS MS F P-valueTreatment B 18.216 D F GError C 14.522 ETotal 18 32.738

•

Response: Correct: 3 Instructor feedback: B is DFtrt = 4 - 1 = 3 since there are four restaurants being compared.

• (1 pts) 1 What is the value of C in the ANOVA table? Response: Correct: 15 Instructor feedback: C is DFerr = DFtot - DFtrt = 15.

• (1 pts) 1 What is the value of D in the ANOVA table? Give your answer to three decimal places.

Response: Correct: 6.072 Instructor feedback: D is MStrt = SStrt/DFtrt = 6.072.

• (1 pts) 1 What is the value of E in the ANOVA table? Give your answer to three decimal places. Response: Correct: 0.9681 Instructor feedback: D is MSerr = SSerr/DFerr = 0.9681.

• (1 pts) 1 What is the value of F in the ANOVA table? Give your answer to two decimal places. Response: Correct: 6.272079 Instructor feedback: F = MStrt/MSerr = 6.272079.

• (1 pts) 1 What is the value of G in the ANOVA table? Give your answer to four decimal places. Response: Correct: 0.0057 Instructor feedback: G is the P-value = P(F > 6.272079) = 0.0057 using the StatCrunch calculator with numerator degrees of freedom 3 and denominator degrees of freedom 15.

• (1 pts) 1 Using a 0.05 level of significance, what should his conclusion be in this case? He should conclude that at least two of the restaurants have different mean delivery times because the P-value is greater than 0.05. He should fail to reject the claim that at all of the restaurants have the same mean delivery times because the P-value is greater than 0.05. He should conclude that at all of the restaurants have the same mean delivery times because the P-value is less than 0.05. He should conclude that at least two of the restaurants have different mean delivery times because the P-value is less than 0.05.

•

Hw 10 Due: Monday, Dec 5, 2011 8:00 am CST

Instructions: This assignment covers the simple linear regression procedure. This topic is covered in Chapter 11. You might want to look at the following pages in the eBook as you work through the assignment.

• Least squares principle - pages 301 - 309

• Hypothesis tests/confidence intervals/prediction intervals for regression - pages 309 - 315

• A random sample of 20 individuals who graduated from college five years ago were asked to report the total amount of debt (in $) they had when they graduated from college and the total value of their current investments (in $) resulting in the data set below.

• (1 pts) 1 Which statement best describes the relationship between these two variables? College debt is not associated with current investment. As college debt increases current investment increases. As college debt increases current investment decreases. As college debt decreases current investment decreases. Instructor feedback: Construct a scatter plot in StatCrunch to answer this question. You will see a negative association (downward trend) in the scatter plot.

• (1 pts) 1 Develop a regression equation for predicting current investment based on college debt. What is the expected change in current investment for each additional dollar of

college debt? Give your answer to four decimal places. Response: Correct: -1.880283 Instructor feedback: Use StatCrunch to regress Invested on Debt with the Stat > Regression > Simple Linear option. The Y-Variable is Invested and the X-Variable is Debt. The value asked for in this question is the slope estimate from the regression.

• (1 pts) 1 When testing for a significant linear relationship in your regression analysis, what is the proper conclusion at the 0.1 level of significance? We fail to reject the claim of no linear relationship between college debt and current investment because the P-value is less than 0.1. We fail to reject the claim of no linear relationship between college debt and current investment because the P-value is greater than 0.1. There is a significant linear relationship between college debt and current investment because the P-value is greater than 0.1. There is a significant linear relationship between college debt and current investment because the P-value is less than 0.1. Instructor feedback: The P-value for the test that the true slope is zero versus the two sided alternative is approximately 0. If the P-value is less than 0.1 we determine there is a significant linear relationship. Otherwise, we fail to reject the claim of no linear relationship.

• (1 pts) 1 What is the predicted current investment for an individual who had a college debt of $5000? Give your answer to two decimal places. Response: Correct: 59534.082848 Instructor feedback: You can use StatCrunch to get this answer by plugging in a predicted value of 5000 or you can simply plug this value into your estimated regression equation to find a predicted value of 68935.499973 + -1.880283*5000 = 59534.082848. Note that the values shown here may be slightly different than those shown in StatCrunch due to differences in rounding.

• (1 pts) 1 What proportion of the variation in current investment is explained by college debt? Give your answer to four decimal places.

Response: Correct: 0.9851 Instructor feedback: This is the value of R-sq = 0.9851 from the StatCrunch regression output.

• A small Internet company wants to determine how the money they spend on Google Adwords impacts their monthly revenue. Over 6 consecutive months, they vary the amount they spend on their Adwords campaign (in $) and record the associated revenue (in $) for each month. The data is shown below.

• (1 pts) 1 Develop a regression equation for predicting monthly revenue based on the amount spent with Adwords. What is the y-intercept? Give your answer to two decimal places. Response: Correct: 522.142857 Instructor feedback: From the StatCrunch regression output, the y-intercept is 522.142857.

• (1 pts) 1 What is the proper interpretation of the y-intercept in the regression equations? The y-intercept describes the expected increase in revenue for each additional dollar spent on Adwords. The y-intercept describes the expected decrease in revenue for each additional dollar spent on Adwords. The y-intercept describes the expected revenue if the company spends $25 in a given month on Adwords. The y-intercept describes the expected revenue if the company does not spend any money in a given month on Adwords.

• (1 pts) 1 What is the sample correlation between these two variables? Give your answer to two decimal places. Response: Correct: -0.187115 Instructor feedback: The sample correlation coefficient can be found in the regression output to be -0.187115.

• (1 pts) 1 What is the slope of your regression equation? Give your answer to two decimal places. Response: Correct: -0.325714

Instructor feedback: From the StatCrunch regression output, the slope is -0.325714.

• (1 pts) 1 Using a 0.1 level of significance, does this regression equation appear to have any value for predicting revenue based on Adwords expenditures? No because there is a significant linear relationship between the two quantities. No because there is not a significant linear relationship between the two quantities. Yes because there is not a significant linear relationship between the two quantities. Yes because there is a significant linear relationship between the two quantities. Instructor feedback: The P-value for the test of significance is 0.7226 in this case. If the P-value is less than 0.1, then we conclude there is a significant linear relationship.

• An exercise science major wants to try to use body weight to predict how much someone can bench press. He collects the data shown below on 30 male students. Both quantities are measured in pounds.

• (1 pts) 1 What type of association does there appear to be between these two variables? no association positive association negative association can not be determined Instructor feedback: Use a scatter plot to see the upward trend relating these two variables.

• (1 pts) 1 Compute a 95% confidence interval for the average bench press of 150 pound males. What is the lower limit? Give your answer to two decimal places. Response: Correct: 135.271135 Instructor feedback: When doing the regression, click Next after selecting the appropriate columns, select the "Predict Y for X=" box and enter 150 as the value.

• (1 pts) 1

Compute a 95% confidence interval for the average bench press of 150 pound males. What is the upper limit? Give your answer to two decimal places. Response: Correct: 143.28625 Instructor feedback: When doing the regression, click Next after selecting the appropriate columns, select the "Predict Y for X=" box and enter 150 as the value.

• (1 pts) 1 Compute a 95% prediction interval for the bench press of a 150 pound male. What is the lower limit? Give your answer to two decimal places. Response: Correct: 119.764545 Instructor feedback: When doing the regression, click Next after selecting the appropriate columns, select the "Predict Y for X=" box and enter 150 as the value.

• (1 pts) 1 Compute a 95% prediction interval for the bench press of a 150 pound male. What is the upper limit? Give your answer to two decimal places. Response: Correct: 158.792841 Instructor feedback: When doing the regression, click Next after selecting the appropriate columns, select the "Predict Y for X=" box and enter 150 as the value.

• A used car dealer wants to develop a regression equation that determines mileage as a function of the age of a car in years. He collects the data shown below for the 12 cars he has on his lot.

• (1 pts) 1 What is the slope of the regression equation? Give your answer to two decimal places. Response: Correct: 8874.2625 Instructor feedback: See the StatCrunch regression output.

• (1 pts) 1 What is the value of the correlation coefficient? Give your answer to two decimal places. Response: Correct: 0.977722

Instructor feedback: See the StatCrunch regression output.

• (1 pts) 1 A 4 year old car is delivered to his lot with 160000 miles. Manually enter these values in the data table above and rerun the regression analysis. What is the value of the slope? Give your answer to two decimal places. Response: Correct: 2820.242629 Instructor feedback: See the StatCrunch regression output.

• (1 pts) 1 Including the additional car, what is the value of the correlation coefficient? Give your answer to two decimal places. Response: Correct: 0.325442 Instructor feedback: See the StatCrunch regression output.

• (1 pts) 1 Did the additional car strengthen or weaken the linear relationship between age and mileage? It strengthened the linear relationship. It weakened the linear relationship. Can not be determined. Instructor feedback: The correlation coefficient dropped a great deal.

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