FINAL EXAM REVIEW STA 2023 SP18 PDF
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This is a review for a statistics exam. The document contains questions related to observational studies, designed experiments, and correlation, plus some general statistics topics. It has multiple choice and free-response questions.
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Student: _____________________ Instructor: Olga Timofeeva Assignment: FINAL EXAM REVIEW - Date: _____________________ Course: STA 2023 SP18 05321 NOT GRADED (35 Quesions) 1. What is an observational study? What is a designed expe...
Student: _____________________ Instructor: Olga Timofeeva Assignment: FINAL EXAM REVIEW - Date: _____________________ Course: STA 2023 SP18 05321 NOT GRADED (35 Quesions) 1. What is an observational study? What is a designed experiment? Which allows the researcher to claim causation between an explanatory variable and a response variable? What is an observational study? A. An observational study is when a researcher assigns individuals to a certain group, intentionally changing the value of an explanatory variable, and then recording the value of the response variable for each group. B. An observational study is a list of all individuals in a population along with certain characteristics of each individual. C. An observational study measures the value of the response variable without attempting to influence the value of either the response or explanatory variables. What is a designed experiment? A. A designed experiment is when a researcher assigns individuals to a certain group, intentionally changing the value of an explanatory variable, and then recording the value of the response variable for each group. B. A designed experiment is a list of all individuals in a population along with certain characteristics of each individual. C. A designed experiment measures the value of the response variable without attempting to influence the value of either the response or explanatory variables. Which allows the researcher to claim causation between an explanatory variable and a response variable? (1) allows the researcher to claim causation between an explanatory variable and a response variable (1) A designed experiment An observational study 2. Determine whether the study depicts an observational study or an experiment. Sixty college students are divided into two groups. One group is exposed to rock music while studying. The other is not. After one week, both groups are questioned about their scores on final mathematical examinations. Is the study an observational study or an experiment? A. The study is an observational study because the study examines individuals in a sample, but does not try to influence the response variable. B. The study is an experiment because the study examines individuals in a sample, but does not try to influence the variable of interest. C. The study is an experiment because the researchers control one variable to determine the effect on the response variable. D. The study is an observational study because the researchers control one variable to determine the effect on the response variable. 3. Researchers initiated a long-term study of the population of American black bears. One aspect of the study was to develop a model that could be used to predict a bear's weight (since it is not practical to weigh bears in the field). One variable thought to be related to weight is the length of the bear. The accompanying data represent the lengths and weights of 12 American black bears. Complete parts (a) through (d) below. 1 Click the icon to view the data table. 2 Click the icon to view the critical values table. (a) Which variable is the explanatory variable based on the goals of the research? A. The number of bears B. The weight of the bear C. The length of the bear (b) Draw a scatter diagram of the data. Choose the correct graph below. A. B. C. D. Weight (kg) Length (cm) Weight (kg) Weight (kg) 180 180 180 180 40 40 40 40 100 200 100 200 100 200 100 200 Length (cm) Weight (kg) Length (cm) Length (cm) (c) Determine the linear correlation coefficient between weight and length. The linear correlation coefficient between weight and length is r =. (Round to three decimal places as needed.) (d) Does a linear relation exist between the weight of the bear and its length? Because the correlation coefficient is (1) and the absolute value of the correlation coefficient, , is (2) than the critical value for this data set, , (3) linear relation exists between the weight of the bear and its length. (Round to three decimal places as needed.) 1: Data table Total Length (cm) Weight (kg) 139.0 110 135.0 60 139.0 90 120.5 60 149.0 85 141.0 105 141.0 95 150.0 85 166.0 155 151.5 140 129.5 105 150.0 110 2: critical values for the correlation coefficient Critical Values for Correlation Coefficient n 3 0.997 4 0.950 5 0.878 6 0.811 7 0.754 8 0.707 9 0.666 10 0.632 11 0.602 12 0.576 13 0.553 14 0.532 15 0.514 16 0.497 17 0.482 18 0.468 19 0.456 20 0.444 21 0.433 22 0.423 23 0.413 24 0.404 25 0.396 26 0.388 27 0.381 28 0.374 29 0.367 30 0.361 (1) negative (2) not greater (3) a negative positive greater a positive no 4. A polling organization contacts 1562 teenagers who are 16 to 19 years of age and live in North America and asks whether or not they had driven a car on a highway recently. What is the population in the study? A. Teenagers who are 16 to 19 years of age and have driven a car on a highway. B. Teenagers who are 16 to 19 years of age. C. Teenagers who are 16 to 19 years of age and live in North America and have driven a car on a highway. D. Teenagers who are 16 to 19 years of age and live in North America. What is the sample in the study? A. Teenagers who are 16 to 19 years of age and live in North America. B. The 1562 teenagers who are 16 to 19 years of age and have driven a car on a highway. C. The 1562 teenagers who are 16 to 19 years of age and live in North America. D. Teenagers who are 16 to 19 years of age. 5. A(n) (1) is a numerical summary of a sample. A(n) (2) is a numerical summary of a population. (1) continuous variable (2) parameter statistic discrete variable parameter continuous variable discrete variable statistic 6. Determine whether the quantitative variable is discrete or continuous. Weight of gravel in a pile Is the variable discrete or continuous? A. The variable is discrete because it is countable. B. The variable is continuous because it is countable. C. The variable is continuous because it is not countable. D. The variable is discrete because it is not countable. 7. Determine whether the quantitative variable is discrete or continuous. Length of a snake Is the variable discrete or continuous? A. The variable is continuous because it is countable. B. The variable is discrete because it is not countable. C. The variable is discrete because it is countable. D. The variable is continuous because it is not countable. 8. Find the sample variance and standard deviation. 8, 50, 16, 48, 35, 27, 28, 28, 31, 32 Choose the correct answer below. Fill in the answer box to complete your choice. (Round to two decimal places as needed.) A. 2 σ = B. s2 = Choose the correct answer below. Fill in the answer box to complete your choice. (Round to one decimal place as needed.) A. σ = B. s = 9. An insurance company crashed four cars in succession at 5 miles per hour. The cost of repair for each of the four crashes was $426, $456, $402, $222. Compute the range, sample variance, and sample standard deviation cost of repair. The range is $. 2 s = dollars2 (Round to the nearest whole number as needed.) s=$ (Round to two decimal places as needed.) 10. The following data represent the weights (in grams) of a simple random sample of a candy. 0.83 0.81 0.81 0.93 0.83 0.86 0.84 0.87 0.80 0.80 Determine the shape of the distribution of weights of the candies by drawing a frequency histogram and computing the mean and the median. Which measure of central tendency best describes the weight of the candy? Choose the correct frequency histogram below. A. B. C. 5 5 5 4 4 4 Frequency Frequency Frequency 3 3 3 2 2 2 1 1 1 0 0 0.79.85.91.79.85.91.79.85.91 Weight (grams) Weight (grams) Weight (grams) Is the histogram for the data set skewed right, skewed left, or symmetric? skewed right skewed left symmetric The mean weight of the candies is grams. (Type an integer or decimal rounded to the nearest thousandth as needed.) The median weight of the candies is grams. (Type an integer or decimal rounded to the nearest thousandth as needed.) Which measure of central tendency best describes the weight of the candy? mean median mode 11. A random sample of 15 college students were asked "How many hours per week typically do you 6 30 2 work outside the home?" Their responses are shown on the right. 6 7 14 Determine the shape of the distribution of hours worked by drawing a frequency histogram and 18 10 15 computing the mean and median. Which measure of central tendency better describes hours worked? 19 11 25 11 6 13 Choose the correct frequency histogram below. A. Hours Worked B. Hours Worked C. Hours Worked per Week per Week per Week Frequency Frequency Frequency 4 4 4 2 2 2 0 0 0 0 10 20 30 0 10 20 30 0 10 20 30 hours hours hours Is the histogram for the data set skewed right, skewed left, or symmetric? skewed left skewed right symmetric The mean number of hours worked by college students outside the home is approximately hours. (Type an integer or decimal rounded to three decimal places as needed.) The median number of hours worked by college students outside the home is hours. Which measure of central tendency better describes hours worked? mean median mode 12. The data in the accompanying table represent the ages of the presidents of a country on their first days in office. Complete parts (a) and (b). 3 Click the icon to view the data table. (a) Construct a stem-and-leaf plot. Choose the correct answer below. A. B. C. D. President Ages President Ages President Ages President Ages 4 23 4 23 4 23 4 667899 4 667899 4 667899 4 667899 4 23 5 1001112244444 5 0011112244444 5 0011112244444 5 0011112244444 5 555566677778 5 555566677778 5 555566677778 5 555566677778 6 0111244 6 0111244 6 01112445 6 0111244 6 579 6 579 6 79 6 579 Legend: 4 | 2 Legend: 4 | 2 Legend: 4 | 2 Legend: 4 | 6 represents 42 represents 42 represents 42 represents 46 years years years years (b) Describe the shape of the distribution. Choose the correct answer below. Skewed right Uniform Bell shaped Skewed left 3: President Ages 42 48 51 52 54 56 57 61 65 43 49 51 54 55 56 57 61 67 46 49 51 54 55 56 58 62 69 46 50 51 54 55 57 60 64 47 50 52 54 55 57 61 64 13. Suppose babies born after a gestation period of 32 to 35 weeks have a mean weight of 2900 grams and a standard deviation of 800 grams while babies born after a gestation period of 40 weeks have a mean weight of 3100 grams and a standard deviation of 455 grams. If a 34-week gestation period baby weighs 3275 grams and a 40-week gestation period baby weighs 3475 grams, find the corresponding z-scores. Which baby weighs more relative to the gestation period? Find the corresponding z-scores. Which baby weighs relatively more? Select the correct choice below and fill in the answer boxes to complete your choice. (Round to two decimal places as needed.) A. The baby born in week 40 weighs relatively more since its z-score, , is larger than the z-score of for the baby born in week 34. B. The baby born in week 40 weighs relatively more since its z-score, , is smaller than the z-score of for the baby born in week 34. C. The baby born in week 34 weighs relatively more since its z-score, , is larger than the z-score of for the baby born in week 40. D. The baby born in week 34 weighs relatively more since its z-score, , is smaller than the z-score of for the baby born in week 40. 14. The accompanying data represent the monthly rate of return of a certain company's common stock for the past few years. Complete parts (a) and (b) below. 4 Click the icon to view the data table. (a) Determine and interpret the quartiles. The first quartile is Q1 =. (Round to four decimal places as needed.) The second quartile is Q2 =. (Round to four decimal places as needed.) The third quartile is Q3 =. (Round to four decimal places as needed.) Interpret the quartiles. Choose the correct answer below. A. All monthly returns within one standard deviation of the mean are contained in the first quartile, all monthly returns within two standard deviations of the mean are contained in the second quartile, and all monthly returns within three standard deviations of the mean are contained in the third quartile. B. The first quartile is the lower bound of plausible monthly returns, and the third quartile is the upper bound of plausible monthly returns. Any monthly returns outside of these bounds are outliers. The second quartile represents the most common monthly return. C. Of the monthly returns, 25% are less than or equal to the first quartile, 50% are less than or equal to the second quartile, and 75% are less than or equal to the third quartile. D. The first quartile is one standard deviation below the mean (or average monthly return), the second quartile is the mean, and the third quartile is one standard deviation above the mean. (b) Check the data set for outliers. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The outlier(s) is/are. (Type an integer or a decimal. Do not round. Use a comma to separate answers as needed.) B. There are no outliers in the data set. 4: Rate of Return 0.26 0.26 0.03 0.06 0.06 − 0.04 − 0.04 0.22 0.48 0.06 − 0.14 0.19 0.05 0.18 0.09 0.02 − 0.05 − 0.02 0.08 0.02 − 0.02 0.13 − 0.08 − 0.02 0.06 − 0.01 0.07 − 0.05 0.01 − 0.10 0.02 0.03 0.01 0.11 − 0.11 0.09 0.10 0.25 − 0.02 0.03 15. According to a survey, the probability that a randomly selected worker primarily drives a van to work is 0.829. The probability that a randomly selected worker primarily takes public transportation to work is 0.056. Complete parts (a) through (d). (a) What is the probability that a randomly selected worker primarily drives a van or takes public transportation to work? P(worker drives a van or takes public transportation to work) = (Type an integer or decimal rounded to three decimal places as needed.) (b) What is the probability that a randomly selected worker primarily neither drives a van nor takes public transportation to work? P(worker neither drives a van nor takes public transportation to work) = (Type an integer or decimal rounded to three decimal places as needed.) (c) What is the probability that a randomly selected worker primarily does not drive a van to work? P(worker does not drive a van to work) = (Type an integer or decimal rounded to three decimal places as needed.) (d) Can the probability that a randomly selected worker primarily walks to work equal 0.20? Why or why not? A. Yes. If a worker did not primarily drive or take public transportation, the only other method to arrive at work would be to walk. B. Yes. The probability a worker primarily drives, walks, or takes public transportation would equal 1. C. No. The probability a worker primarily drives, walks, or takes public transportation would be greater than 1. D. No. The probability a worker primarily drives, walks, or takes public transportation would be less than 1. 16. What does it mean to say that two variables are positively associated? Negatively associated? What does it mean to say that two variables are positively associated? A. There is a linear relationship between the variables, and whenever the value of one variable increases, the value of the other variable increases. B. There is a linear relationship between the variables. C. There is a relationship between the variables that is not linear. D. There is a linear relationship between the variables, and whenever the value of one variable increases, the value of the other variable decreases. What does it mean to say that two variables are negatively associated? A. There is a linear relationship between the variables, and whenever the value of one variable increases, the value of the other variable increases. B. There is a linear relationship between the variables. C. There is a linear relationship between the variables, and whenever the value of one variable increases, the value of the other variable decreases. D. There is a relationship between the variables that is not linear. 17. For the accompanying data set, (a) draw a scatter diagram of the data, (b) by hand, compute the correlation coefficient, and (c) determine whether there is a linear relation between x and y. Click here to view the data set.5 Click here to view the critical values table.6 (a) Draw a scatter diagram of the data. Choose the correct graph below. A. B. C. D. y y y y 20 20 20 20 x x x x 0 0 0 0 0 10 0 10 0 10 0 10 (b) By hand, compute the correlation coefficient. The correlation coefficient is r =. (Round to three decimal places as needed.) (c) Determine whether there is a linear relation between x and y. Because the correlation coefficient is (1) and the absolute value of the correlation coefficient, , is (2) than the critical value for this data set, , (3) linear relation exists between x and y. (Round to three decimal places as needed.) 5: Data set x 2 4 6 6 7 y 4 8 12 13 19 6: Critical values for the correlation coefficient Critical Values for Correlation Coefficient n 3 0.997 4 0.950 5 0.878 6 0.811 7 0.754 8 0.707 9 0.666 10 0.632 11 0.602 12 0.576 13 0.553 14 0.532 15 0.514 16 0.497 17 0.482 18 0.468 19 0.456 20 0.444 21 0.433 22 0.423 23 0.413 24 0.404 25 0.396 26 0.388 27 0.381 28 0.374 29 0.367 30 0.361 n (1) positive (2) not greater (3) a negative negative greater no a positive 18. A pediatrician wants to determine the relation that may exist Height (in.) Head Circumference (in.) between a child's height and head circumference. She 27.25 17.3 randomly selects 8 children from her practice, measures 25.75 17 their height and head circumference, and obtains the data 26.5 17.2 shown in the table. Complete parts (a) through (e) below. 25.5 17 7 Click here to see the Table of Critical Values for 27.5 17.6 Correlation Coefficient. 26.5 17.3 25.75 17.1 27.25 17.3 (a) If the pediatrician wants to use height to predict head circumference, determine which variable is the explanatory variable and which is the response variable. Choose the correct answer below. The explanatory variable is head circumference and the response variable is height. The explanatory variable is height and the response variable is head circumference. (b) Draw a scatter diagram. Choose the correct graph below. A. B. C. D. 17.6 28 17.6 28 Height (in.) Height (in.) Circ. (in.) Circ. (in.) 16.9 25 16.9 25 25 28 16.9 17.6 25 28 16.9 17.6 Circ. (in.) Height (in.) Height (in.) Circ. (in.) (c) Compute the linear correlation coefficient between the height and head circumference of a child. r= (Round to three decimal places as needed.) (d) Does a linear relation exist between height and head circumference? Select the correct choice below and fill in the answer box to complete your choice. (Round to three decimal places as needed.) A. No, there is no linear association since r is positive and is less than the critical value,. B. Yes, there appears to be a negative linear association because r is negative and is less than the negative of the critical value,. C. Yes, there appears to be a positive linear association because r is positive and is greater than the critical value,. D. Yes, there appears to be a positive linear association because r is positive and is less than the critical value,. (e) Convert the data to centimeters (1 inch = 2.54 cm), and recompute Head Circumference the linear correlation coefficient. What effect did the conversion have on Height (centimeters) (centimeters) the linear correlation coefficient? Convert the first four data values to centimeters. (Type integers or decimals. Do not round. List the terms in the same order as they appear in the original list.) Convert the last four data values to centimeters. Head Circumference Height (centimeters) (centimeters) (Type integers or decimals. Do not round. List the terms in the same order as they appear in the original list.) The new linear correlation coefficient is r =. The conversion to centimeters (1) (Round to three decimal places as needed.) 7: Data Table Critical Values for Correlation Coefficient n 3 0.997 4 0.950 5 0.878 6 0.811 7 0.754 8 0.707 9 0.666 10 0.632 11 0.602 12 0.576 13 0.553 14 0.532 15 0.514 16 0.497 17 0.482 18 0.468 19 0.456 20 0.444 21 0.433 22 0.423 23 0.413 24 0.404 25 0.396 26 0.388 27 0.381 28 0.374 29 0.367 30 0.361 n (1) made the value of r decrease. reversed the sign of r. had no effect on r. made the value of r increase. 19. The data below represent the number of days absent, x, and the final grade, y, for a sample of college students at a large university. Complete parts (a) through (e) below. No. of absences, x 0 1 2 3 4 5 6 7 8 9 Final grade, y 87.8 84.9 81.9 79.4 76.3 71.9 62.1 66.7 63.7 60.8 (a) Find the least-squares regression line treating the number of absences, x, as the explanatory variable and the final grade, y, as the response variable. y= x+ (Round to three decimal places as needed.) (b) Interpret the slope and y-intercept, if appropriate. Interpret the slope. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. (Round to three decimal places as needed.) A. For a final score of zero, the number of days absent is predicted to be days. B. For every unit change in the final grade, the number of days absent falls by days, on average. C. For every day absent, the final grade falls by , on average. D. For zero days absent, the final score is predicted to be. E. It is not appropriate to interpret the slope. Interpret the y-intercept. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. (Round to three decimal places as needed.) A. For every day absent, the final grade falls by , on average. B. For every unit change in the final grade, the number of days absent falls by days, on average. C. For zero days absent, the final score is predicted to be. D. For a final score of zero, the number of days absent is predicted to be days. E. It is not appropriate to interpret the y-intercept. (c) Predict the final grade for a student who misses five class periods and compute the residual. Is the observed final grade above or below average for this number of absences? The predicted final grade is. This observation has a residual of , which indicates that the final grade is (1) average. (Round to one decimal place as needed.) (d) Draw the least-squares regression line on the scatter diagram of the data. Choose the correct graph below. A. B. C. D. y y y y 100 100 100 100 x x x x 50 50 50 50 0 10 0 10 0 10 0 10 (e) Would it be reasonable to use the least-squares regression line to predict the final grade for a student who has missed 15 class periods? Why or why not? A. Yes—15 missed class periods is possible and within the scope of the model B. No—15 missed class periods is outside the scope of the model. C. No—15 missed class periods is not possible and outside the scope of the model. D. No—15 missed class periods is not possible. E. More information regarding the student is necessary to be able to make a decision. (1) below above 20. Determine if the following probability experiment represents a binomial experiment. A random sample of 15 middle school students is obtained, and the individuals selected are asked to state their hair length. Choose the correct answer below. A. No, this probability experiment does not represent a binomial experiment because the trials are not independent. B. Yes, this probability experiment represents a binomial experiment because each trial has two mutually exclusive outcomes. C. No, this probability experiment does not represent a binomial experiment because the variable is continuous, and there are not two mutually exclusive outcomes. D. Yes, this probability experiment represents a binomial experiment because the probability of success is the same for each trial of the experiment. 21. Determine whether the following probability experiment represents a binomial experiment and explain the reason for your answer. An investor randomly purchases 19 stocks listed on a stock exchange. Historically, the probability that a stock listed on this exchange will increase in value over the course of a year is 43%. The number of stocks that increase in value is recorded. Does the probability experiment represent a biniomial experiment? A. No, because the trials of the experiment are not independent. B. No, because the probability of success differs from trial to trial. C. Yes, because the experiment satisfies all the criteria for a binomial experiment. D. No, because there are more than two mutually exclusive outcomes for each trial. 22. According to an airline, flights on a certain route are on time 75% of the time. Suppose 24 flights are randomly selected and the number of on-time flights is recorded. (a) Explain why this is a binomial experiment. (b) Find and interpret the probability that exactly 15 flights are on time. (c) Find and interpret the probability that fewer than 15 flights are on time. (d) Find and interpret the probability that at least 15 flights are on time. (e) Find and interpret the probability that between 13 and 15 flights, inclusive, are on time. (a) Identify the statements that explain why this is a binomial experiment. Select all that apply. A. There are three mutually exclusive possibly outcomes, arriving on-time, arriving early, and arriving late. B. There are two mutually exclusive outcomes, success or failure. C. The probability of success is the same for each trial of the experiment. D. The experiment is performed until a desired number of successes is reached. E. The trials are independent. F. Each trial depends on the previous trial. G. The experiment is performed a fixed number of times. (b) The probability that exactly 15 flights are on time is. (Round to four decimal places as needed.) Interpret the probability. In 100 trials of this experiment, it is expected about to result in exactly 15 flights being on time. (Round to the nearest whole number as needed.) (c) The probability that fewer than 15 flights are on time is. (Round to four decimal places as needed.) Interpret the probability. In 100 trials of this experiment, it is expected about to result in fewer than 15 flights being on time. (Round to the nearest whole number as needed.) (d) The probability that at least 15 flights are on time is. (Round to four decimal places as needed.) Interpret the probability. In 100 trials of this experiment, it is expected about to result in at least 15 flights being on time. (Round to the nearest whole number as needed.) (e) The probability that between 13 and 15 flights, inclusive, are on time is. (Round to four decimal places as needed.) Interpret the probability. In 100 trials of this experiment, it is expected about to result in between 13 and 15 flights, inclusive, being on time. (Round to the nearest whole number as needed.) 23. State the criteria for a binomial probability experiment. Choose the correct answer below. Select all that apply. A. The trials are independent. B. The probability of success, p, remains constant for each trial of the experiment. C. Each trial has two possible mutually exclusive outcomes: success and failure. D. The experiment consists of a fixed number, n, of trials. 24. Assume the random variable X is normally distributed with mean μ = 50 and standard deviation σ = 7. Compute the probability. Be sure to draw a normal curve with the area corresponding to the probability shaded. P(35 < X < 57) 8 Click the icon to view a table of areas under the normal curve. Which of the following normal curves corresponds to P(35 < X < 57)? A. B. C. 35 50 57 35 50 57 35 50 57 P(35 < X < 57) = (Round to four decimal places as needed.) 8: Tables of Areas under the Normal Curve 25. The mean incubation time for a type of fertilized egg kept at a certain temperature is 17 days. Suppose that the incubation times are approximately normally distributed with a standard deviation of 1 day. Complete parts (a) through (e) below. Click here to view the standard normal distribution table (page 1).9 Click here to view the standard normal distribution table (page 2).10 (a) Draw a normal model that describes egg incubation times of these fertilized eggs. Choose the correct graph below. Click here to view graph d.11 Click here to view graph c.12 Click here to view graph b.13 Click here to view graph a.14 (b) Find and interpret the probability that a randomly selected fertilized egg hatches in less than 15 days. The probability that a randomly selected fertilized egg hatches in less than 15 days is. (Round to four decimal places as needed.) Interpret this probability. Select the correct choice below and fill in the answer box to complete your choice. A. The average proportion of the way to hatching of all eggs fertilized in the past 15 days is. (Round to two decimal places as needed.) B. In every group of 100 fertilized eggs, eggs will hatch in less than 15 days. (Round to the nearest integer as needed.) C. If 100 fertilized eggs were randomly selected, of them would be expected to hatch in less than 15 days. (Round to the nearest integer as needed.) (c) Find and interpret the probability that a randomly selected fertilized egg takes over 19 days to hatch. The probability that a randomly selected fertilized egg takes over 19 days to hatch is. (Round to four decimal places as needed.) Interpret this probability. Select the correct choice below and fill in the answer box to complete your choice. A. In every group of 100 fertilized eggs, eggs will hatch in more than 19 days. (Round to the nearest integer as needed.) B. If 100 fertilized eggs were randomly selected, of them would be expected to take more than 19 days to hatch. (Round to the nearest integer as needed.) C. The average proportion of the way to hatching of all eggs fertilized more than 19 days ago is. (Round to two decimal places as needed.) (d) Find and interpret the probability that a randomly selected fertilized egg hatches between 16 and 17 days. The probability that a randomly selected fertilized egg hatches between 16 and 17 days is. (Round to four decimal places as needed.) Interpret this probability. Select the correct choice below and fill in the answer box to complete your choice. A. The average proportion of the way to hatching of all eggs fertilized between 16 and 17 days is. (Round to two decimal places as needed.) B. If 100 fertilized eggs were randomly selected, of them would be expected to hatch between 16 and 17 days. (Round to the nearest integer as needed.) C. In every group of 100 fertilized eggs, eggs will hatch between 16 and 17 days. (Round to the nearest integer as needed.) (e) Would it be unusual for an egg to hatch in less than 14 days? Why? The probability of an egg hatching in less than 14 days is , so it (1) be unusual, since the probability is (2) than 0.05. (Round to four decimal places as needed.) 9: Standard Normal Distribution Table (page 1) 10: Standard Normal Distribution Table (page 2) 11: Graph ~Ld2 X 11 13 15 17 19 21 23 12: Graph ~Ld1 X 14 15 16 17 18 19 20 13: Graph ~Ld4 X 11 12 13 14 15 16 17 14: Graph ~Ld3 X 17 18 19 20 21 22 23 (1) would not (2) greater would less 26. Steel rods are manufactured with a mean length of 30 centimeter (cm). Because of variability in the manufacturing process, the lengths of the rods are approximately normally distributed with a standard deviation of 0.08 cm. 15 Click the icon to view a table of areas under the normal curve. (a) What proportion of rods has a length less than 29.9 cm? (Round to four decimal places as needed.) (b) Any rods that are shorter than 29.82 cm or longer than 30.18 cm are discarded. What proportion of rods will be discarded? (Round to four decimal places as needed.) (c) Using the results of part (b), if 5000 rods are manufactured in a day, how many should the plant manager expect to discard? (Use the answer from part b to find this answer. Round to the nearest integer as needed.) (d) If an order comes in for 10,000 steel rods, how many rods should the plant manager expect to manufacture if the order states that all rods must be between 29.9 cm and 30.1 cm? (Round up to the nearest integer.) 15: Tables of Areas under the Normal Curve 27. Suppose a simple random sample of size n = 81 is obtained from a population with μ = 88 and σ = 27. (a) Describe the sampling distribution of x. (b) What is P x > 93.25 ? (c) What is P x ≤ 80.8 ? (d) What is P 84.25 < x < 93.7 ? (a) Choose the correct description of the shape of the sampling distribution of x. A. The distribution is skewed left. B. The distribution is skewed right. C. The distribution is uniform. D. The distribution is approximately normal. E. The shape of the distribution is unknown. Find the mean and standard deviation of the sampling distribution of x. x = μ x = σ (b) P x > 93.25 = (Round to four decimal places as needed.) (c) P x ≤ 80.8 = (Round to four decimal places as needed.) (d) P 84.25 < x < 93.7 = (Round to four decimal places as needed.) 28. Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean μ = 214 days and standard deviation σ = 11 days. Complete parts (a) through (f) below. (a) What is the probability that a randomly selected pregnancy lasts less than 210 days? The probability that a randomly selected pregnancy lasts less than 210 days is approximately. (Round to four decimal places as needed.) Interpret this probability. Select the correct choice below and fill in the answer box within your choice. (Round to the nearest integer as needed.) A. If 100 pregnant individuals were selected independently from this population, we would expect pregnancies to last less than 210 days. B. If 100 pregnant individuals were selected independently from this population, we would expect pregnancies to last more than 210 days. C. If 100 pregnant individuals were selected independently from this population, we would expect pregnancies to last exactly 210 days. (b) Suppose a random sample of 18 pregnancies is obtained. Describe the sampling distribution of the sample mean length of pregnancies. The sampling distribution of x is (1) with μx = and σx =. (Round to four decimal places as needed.) (c) What is the probability that a random sample of 18 pregnancies has a mean gestation period of 210 days or less? The probability that the mean of a random sample of 18 pregnancies is less than 210 days is approximately. (Round to four decimal places as needed.) Interpret this probability. Select the correct choice below and fill in the answer box within your choice. (Round to the nearest integer as needed.) A. If 100 independent random samples of size n = 18 pregnancies were obtained from this population, we would expect sample(s) to have a sample mean of 210 days or more. B. If 100 independent random samples of size n = 18 pregnancies were obtained from this population, we would expect sample(s) to have a sample mean of exactly 210 days. C. If 100 independent random samples of size n = 18 pregnancies were obtained from this population, we would expect sample(s) to have a sample mean of 210 days or less. (d) What is the probability that a random sample of 33 pregnancies has a mean gestation period of 210 days or less? The probability that the mean of a random sample of 33 pregnancies is less than 210 days is approximately. (Round to four decimal places as needed.) Interpret this probability. Select the correct choice below and fill in the answer box within your choice. (Round to the nearest integer as needed.) A. If 100 independent random samples of size n = 33 pregnancies were obtained from this population, we would expect sample(s) to have a sample mean of 210 days or more. B. If 100 independent random samples of size n = 33 pregnancies were obtained from this population, we would expect sample(s) to have a sample mean of exactly 210 days. C. If 100 independent random samples of size n = 33 pregnancies were obtained from this population, we would expect sample(s) to have a sample mean of 210 days or less. (e) What might you conclude if a random sample of 33 pregnancies resulted in a mean gestation period of 210 days or less? This result would be (2) so the sample likely came from a population whose mean gestation period is (3) 214 days. (f) What is the probability a random sample of size 17 will have a mean gestation period within 9 days of the mean? The probability that a random sample of size 17 will have a mean gestation period within 9 days of the mean is. (Round to four decimal places as needed.) (1) skewed right (2) unusual, (3) less than normal expected, equal to skewed left greater than 29. The reading speed of second grade students in a large city is approximately normal, with a mean of 88 words per minute (wpm) and a standard deviation of 10 wpm. Complete parts (a) through (f). (a) What is the probability a randomly selected student in the city will read more than 94 words per minute? The probability is. (Round to four decimal places as needed.) Interpret this probability. Select the correct choice below and fill in the answer box within your choice. A. If 100 different students were chosen from this population, we would expect to read exactly 94 words per minute. B. If 100 different students were chosen from this population, we would expect to read less than 94 words per minute. C. If 100 different students were chosen from this population, we would expect to read more than 94 words per minute. (b) What is the probability that a random sample of 10 second grade students from the city results in a mean reading rate of more than 94 words per minute? The probability is. (Round to four decimal places as needed.) Interpret this probability. Select the correct choice below and fill in the answer box within your choice. A. If 100 independent samples of n = 10 students were chosen from this population, we would expect sample(s) to have a sample mean reading rate of more than 94 words per minute. B. If 100 independent samples of n = 10 students were chosen from this population, we would expect sample(s) to have a sample mean reading rate of less than 94 words per minute. C. If 100 independent samples of n = 10 students were chosen from this population, we would expect sample(s) to have a sample mean reading rate of exactly 94 words per minute. (c) What is the probability that a random sample of 20 second grade students from the city results in a mean reading rate of more than 94 words per minute? The probability is. (Round to four decimal places as needed.) Interpret this probability. Select the correct choice below and fill in the answer box within your choice. A. If 100 independent samples of n = 20 students were chosen from this population, we would expect sample(s) to have a sample mean reading rate of more than 94 words per minute. B. If 100 independent samples of n = 20 students were chosen from this population, we would expect sample(s) to have a sample mean reading rate of less than 94 words per minute. C. If 100 independent samples of n = 20 students were chosen from this population, we would expect sample(s) to have a sample mean reading rate of exactly 94 words per minute. (d) What effect does increasing the sample size have on the probability? Provide an explanation for this result. A. Increasing the sample size increases the probability because σx increases as n increases. B. Increasing the sample size decreases the probability because σx decreases as n increases. C. Increasing the sample size decreases the probability because σx increases as n increases. D. Increasing the sample size increases the probability because σx decreases as n increases. (e) A teacher instituted a new reading program at school. After 10 weeks in the program, it was found that the mean reading speed of a random sample of 21 second grade students was 90.1 wpm. What might you conclude based on this result? Select the correct choice below and fill in the answer boxes within your choice. (Type integers or decimals rounded to four decimal places as needed.) A. A mean reading rate of 90.1 wpm is unusual since the probability of obtaining a result of 90.1 wpm or more is. This means that we would expect a mean reading rate of 90.1 or higher from a population whose mean reading rate is 88 in of every 100 random samples of size n = 21 students. The new program is abundantly more effective than the old program. B. A mean reading rate of 90.1 wpm is not unusual since the probability of obtaining a result of 90.1 wpm or more is. This means that we would expect a mean reading rate of 90.1 or higher from a population whose mean reading rate is 88 in of every 100 random samples of size n = 21 students. The new program is not abundantly more effective than the old program. (f) There is a 5% chance that the mean reading speed of a random sample of 21 second grade students will exceed what value? There is a 5% chance that the mean reading speed of a random sample of 21 second grade students will exceed wpm. (Round to two decimal places as needed.) 30. A survey was conducted that asked 1006 people how many books they had read in the past year. Results indicated that x = 13.2 books and s = 16.6 books. Construct a 90% confidence interval for the mean number of books people read. Interpret the interval. 16 Click the icon to view the table of critical t-values. Construct a 90% confidence interval for the mean number of books people read and interpret the result. Select the correct choice below and fill in the answer boxes to complete your choice. (Use ascending order. Round to two decimal places as needed.) A. If repeated samples are taken, 90% of them will have a sample mean between and. B. There is 90% confidence that the population mean number of books read is between and. C. There is a 90% chance that the true mean number of books read is between and. 16: Table of Critical t-Values 31. The data shown below represent the repair cost for a low-impact collision in a simple random sample of mini- and micro-vehicles. Complete parts (a) through (d) below. $3148 $1033 $743 $663 $773 $1758 $3345 $2086 $2665 $1377 Click here to view the table of critical t-values.17 Click here to view page 1 of the standard normal distribution table.18 Click here to view page 2 of the standard normal distribution table.19 (a) Draw a normal probability plot to determine if it is reasonable to conclude the data come from a population that is normally distributed. Choose the correct answer below. A. B. C. D. 2 4 4 2 Expected z-score Expected z-score Expected z-score Expected z-score 1 2 2 1 0 0 0 0 -1 -2 -2 -1 -2 -4 -4 -2 0 2000 4000 0 2000 4000 0 2000 4000 0 2000 4000 Repair Cost ($) Repair Cost ($) Repair Cost ($) Repair Cost ($) Is it reasonable to conclude that the data come from a population that is normally distributed? A. No, because the plotted values are not linear. B. No, because there are not enough values to make a determination. C. Yes, because the plotted values are not linear. D. Yes, because the plotted values are approximately linear. (b) Draw a boxplot to check for outliers. Choose the correct answer below. A. B. C. D. 0 2000 4000 0 2000 4000 0 2000 4000 0 2000 4000 Does the boxplot suggest that there are outliers? A. No, there are no points that are outside of the 1.5(IQR) boundary. B. No, there are no points that are greater than the third quartile or less than the first quartile. C. Yes, there is at least one point that is greater than the third quartile or less than the first quartile. D. Yes, there is at least one point that is outside of the 1.5(IQR) boundary. (c) Construct and interpret a 95% confidence interval for population mean cost of repair. Select the correct choice and fill in the answer boxes to complete your choice. (Round to one decimal place as needed.) A. The lower bound is $ and the upper bound is $. We are 95% confident that the mean cost of repair is within the confidence interval. B. The lower bound is $ and the upper bound is $. We are 95% fid t th t th t f i i t id f th fid i t l (d) Suppose you obtain a simple random sample of size n = 10 of a specific type of mini-vehicle that was in a low-impact collision and determine the cost of repair. Do you think a 95% confidence interval would be wider or narrower? Explain. A. Wider, because taking a second random sample will always lead to a wider confidence interval. B. Narrower, because there is less variability in the data because any variability caused by the different types of vehicles has been removed. C. Wider, because there is more variability in the data because variability in the repair cost of the car has been added. D. Narrower, because taking a second random sample will always lead to a narrower confidence interval. 17: Table of Critical t-Values 18: Standard Normal Distribution (Page 1) 19: Standard Normal Distribution (Page 2) 32. A random sample of 1029 adults in a certain large country was asked "Do you pretty much think televisions are a necessity or a luxury you could do without?" Of the 1029 adults surveyed, 512 indicated that televisions are a luxury they could do without. Complete parts (a) through (e) below. Click here to view the standard normal distribution table (page 1).20 Click here to view the standard normal distribution table (page 2).21 (a) Obtain a point estimate for the population proportion of adults in the country who believe that televisions are a luxury they could do without. p= (Round to three decimal places as needed.) (b) Verify that the requirements for constructing a confidence interval about p are satisfied. The sample (1) a simple random sample, the value of (2) is , which is (3) 10, and the (4) (5) less than or equal to 5% of the (6) (Round to three decimal places as needed.) (c) Construct and interpret a 95% confidence interval for the population proportion of adults in the country who believe that televisions are a luxury they could do without. Select the correct choice below and fill in any answer boxes within your choice. (Type integers or decimals rounded to three decimal places as needed. Use ascending order.) A. There is a % chance the proportion of adults in the country who believe that televisions are a luxury they could do without is between and. B. We are % confident the proportion of adults in the country who believe that televisions are a luxury they could do without is between and. (d) Is it possible that a supermajority (more than 60%) of adults in the country believe that television is a luxury they could do without? Is it likely? It is (7) that a supermajority of adults in the country believe that television is a luxury they could do without because the 95% confidence interval (8). (Type an integer or a decimal. Do not round.) (e) Use the results of part (c) to construct a 95% confidence interval for the population proportion of adults in the country who believe that televisions are a necessity. The 95% confidence interval is ( , ). (Round to three decimal places as needed.) 20: Standard Normal Distribution Table (page 1) 21: Standard Normal Distribution Table (page 2) (1) can be assumed to be (2) np p 1−p (3) less than is stated to be np 1 − p greater than or equal to cannot be assumed to be n is stated to not be p (4) sample proportion (5) cannot be assumed to be (6) population proportion. sample size can be assumed to be sample proportion. population size is stated to not be population size. population proportion is stated to be sample size. (7) possible, but not likely (8) does not contain not possible contains likely 33. Several years ago, 42% of parents who had children in grades K-12 were satisfied with the quality of education the students receive. A recent poll asked 1,025 parents who have children in grades K-12 if they were satisfied with the quality of education the students receive. Of the 1,025 surveyed, 492 indicated that they were satisfied. Construct a 95% confidence interval to assess whether this represents evidence that parents' attitudes toward the quality of education have changed. What are the null and alternative hypotheses? H0 : p (1) versus H1 : p (2) (Round to two decimal places as needed.) Use technology to find the 95% confidence interval. The lower bound is. The upper bound is. (Round to two decimal places as needed.) What is the correct conclusion? A. Since the interval does not contain the proportion stated in the null hypothesis, there is insufficient evidence that parents' attitudes toward the quality of education have changed. B. Since the interval contains the proportion stated in the null hypothesis, there is sufficient evidence that parents' attitudes toward the quality of education have changed. C. Since the interval contains the proportion stated in the null hypothesis, there is insufficient evidence that parents' attitudes toward the quality of education have changed. D. Since the interval does not contain the proportion stated in the null hypothesis, there is sufficient evidence that parents' attitudes toward the quality of education have changed. (1) > (2) > = = < < ≠ ≠ 34. In a study, researchers wanted to measure the effect of alcohol on the hippocampal region, the portion of the brain responsible for long-term memory storage, in adolescents. The researchers randomly selected 14 adolescents with alcohol use disorders to determine whether the hippocampal volumes in the alcoholic adolescents were less than the normal 3 volume of 9.02 cm. An analysis of the sample data revealed that the hippocampal volume is approximately normal with 3 3 x = 8.14 cm and s = 0.8 cm. Conduct the appropriate test at the α = 0.01 level of significance. State the null and alternative hypotheses. H0 : μ (1) H1 : μ (2) (Type integers or decimals. Do not round.) Identify the t-statistic. t0 = (Round to two decimal places as needed.) Identify the P-value. P-value = (Round to three decimal places as needed.) Make a conclusion regarding the hypothesis. (3) the null hypothesis. There (4) sufficient evidence to claim that the mean 3 hippocampal volume is (5) cm. (1) = (2) < (3) Fail to reject (4) is (5) less than > ≠ Reject is not greater than < > equal to ≠ = 35. The mean waiting time at the drive-through of a fast-food restaurant from the time an order is 107.9 79.1 placed to the time the order is received is 87.9 seconds. A manager devises a new drive-through 68.1 96.1 system that he believes will decrease wait time. As a test, he initiates the new system at 56.0 87.3 his restaurant and measures the wait time for 10 randomly selected orders. The wait times are 76.9 71.9 provided in the table to the right. Complete parts (a) and (b) below. 67.9 88.8 22 Click the icon to view the table of correlation coefficient critical values. (a) Because the sample size is small, the manager must verify that the wait time is normally distributed and the sample does not contain any outliers. The normal probability plot is shown below and the sample correlation coefficient is known to be r = 0.989. Are the conditions for testing the hypothesis satisfied? Expected z-score (1) the conditions (2) satisfied. The normal probability plot 2 (3) linear enough, since the correlation coefficient is (4) than 1 the critical value. 0 60 75 90 105 -1 -2 Time (sec) (b) Is the new system effective? Conduct a hypothesis test using the P-value approach and a level of significance of α = 0.1. First determine the appropriate hypotheses. H0 : (5) (6) 87.9 H1 : (7) (8) 87.9 Find the test statistic. t0 = (Round to two decimal places as needed.) Find the P-value. The P-value is. (Round to three decimal places as needed.) Use the α = 0.1 level of significance. What can be concluded from the hypothesis test? A. The P-value is greater than the level of significance so there is not sufficient evidence to conclude the new system is effective. B. The P-value is less than the level of significance so there is not sufficient evidence to conclude the new system is effective. C. The P-value is greater than the level of significance so there is sufficient evidence to conclude the new system is effective. D. The P-value is less than the level of significance so there is sufficient evidence to conclude the new system is effective. 22: Critical values Sample Size, n Critical Value Sample Size, n Critical Value 5 0.880 16 0.941 6 0.888 17 0.944 7 0.898 18 0.946 8 0.906 19 0.949 9 0.912 20 0.951 10 0.918 21 0.952 11 0.923 22 0.954 12 0.928 23 0.956 13 0.932 24 0.957 14 0.935 25 0.959 15 0.939 30 0.960 (1) No, (2) are not (3) is not (4) greater (5) p (6) < Yes, are is less σ ≠ μ = > (7) μ (8) > p = σ ≠ < 1. C. An observational study measures the value of the response variable without attempting to influence the value of either the response or explanatory variables. A. A designed experiment is when a researcher assigns individuals to a certain group, intentionally changing the value of an explanatory variable, and then recording the value of the response variable for each group. (1) A designed experiment 2. C. The study is an experiment because the researchers control one variable to determine the effect on the response variable. 3. C. The length of the bear Weight (kg) 180 40 100 200 D. Length (cm) 0.725 (1) positive 0.725 (2) greater 0.576 (3) a positive 4. D. Teenagers who are 16 to 19 years of age and live in North America. C. The 1562 teenagers who are 16 to 19 years of age and live in North America. 5. (1) statistic (2) parameter 6. C. The variable is continuous because it is not countable. 7. D. The variable is continuous because it is not countable. 2 8. B. s = 161.12 B. s = 12.7 9. 234 11,097 105.34 10. 5 4 Frequency 3 2 1 0.79.85.91 C. Weight (grams) skewed right 0.838 0.83 median 11. Hours Worked per Week Frequency 4 2 0 0 10 20 30 A. hours skewed right 12.867 11 median 12. President Ages 4 23 4 667899 5 0011112244444 5 555566677778 6 0111244 6 579 Legend: 4 | 2 represents 42 B. years Bell shaped 13. A. The baby born in week 40 weighs relatively more since its z-score, 0.82 , is larger than the z-score of 0.47 for the baby born in week 34. 14. − 0.0200 0.0300 0.0950 C. Of the monthly returns, 25% are less than or equal to the first quartile, 50% are less than or equal to the second quartile, and 75% are less than or equal to the third quartile. A. The outlier(s) is/are 0.48. (Type an integer or a decimal. Do not round. Use a comma to separate answers as needed.) 15. 0.885 0.115 0.171 C. No. The probability a worker primarily drives, walks, or takes public transportation would be greater than 1. 16. A. There is a linear relationship between the variables, and whenever the value of one variable increases, the value of the other variable increases. C. There is a linear relationship between the variables, and whenever the value of one variable increases, the value of the other variable decreases. 17. y 20 x 0 B. 0 10 0.955 (1) positive 0.955 (2) greater 0.878 (3) a positive 18. The explanatory variable is height and the response variable is head circumference. Circ. (in.) 17.6 16.9 25 28 C. Height (in.) 0.902 C. Yes, there appears to be a positive linear association because r is positive and is greater than the critical value, 0.707. 69.215 43.942 65.405 43.18 67.31 43.688 64.77 43.18 69.85 44.704 67.31 43.942 65.405 43.434 69.215 43.942 0.902 (1) had no effect on r. 19. − 3.174 87.833 C. For every day absent, the final grade falls by 3.174 , on average. C. For zero days absent, the final score is predicted to be 87.833. 72.0 − 0.1 (1) below y 100 x 50 C. 0 10 B. No—15 missed class periods is outside the scope of the model. 20. C. No, this probability experiment does not represent a binomial experiment because the variable is continuous, and there are not two mutually exclusive outcomes. 21. C. Yes, because the experiment satisfies all the criteria for a binomial experiment. 22. B. There are two mutually exclusive outcomes, success or failure., C. The probability of success is the same for each trial of the experiment., E. The trials are independent., G. The experiment is performed a fixed number of times. 0.0667 7 0.0547 5 0.9453 95 0.1141 11 23. A. The trials are independent., B. The probability of success, p, remains constant for each trial of the experiment., C. Each trial has two possible mutually exclusive outcomes: success and failure., D. The experiment consists of a fixed number, n, of trials. 24. B. 35 50 57 0.8253 25. Click here to view graph c. 0.0228 C. If 100 fertilized eggs were randomly selected, 2 of them would be expected to hatch in less than 15 days. (Round to the nearest integer as needed.) 0.0228 B. If 100 fertilized eggs were randomly selected, 2 of them would be expected to take more than 19 days to hatch. (Round to the nearest integer as needed.) 0.3413 B. If 100 fertilized eggs were randomly selected, 34 of them would be expected to hatch between 16 and 17 days. (Round to the nearest integer as needed.) 0.0013 (1) would (2) less 26. 0.1056 0.0244 122 12,680 27. D. The distribution is approximately normal. 88 3 0.0401 0.0082 0.8657 28. 0.3581 A. If 100 pregnant individuals were selected independently from this population, we would expect 36 pregnancies to last less than 210 days. (1) normal 214 2.5927 0.0614 C. If 100 independent random samples of size n = 18 pregnancies were obtained from this population, we would expect 6 sample(s) to have a sample mean of 210 days or less. 0.0184 C. If 100 independent random samples of size n = 33 pregnancies were obtained from this population, we would expect 2 sample(s) to have a sample mean of 210 days or less. (2) unusual, (3) less than 0.9993 29. 0.2743 C. If 100 different students were chosen from this population, we would expect 27 to read more than 94 words per minute. 0.0289 A. If 100 independent samples of n = 10 students were chosen from this population, we would expect 3 sample(s) to have a sample mean reading rate of more than 94 words per minute. 0.0036 A. If 100 independent samples of n = 20 students were chosen from this population, we would expect 0 sample(s) to have a sample mean reading rate of more than 94 words per minute. B. Increasing the sample size decreases the probability because σx decreases as n increases. B. A mean reading rate of 90.1 wpm is not unusual since the probability of obtaining a result of 90.1 wpm or more is 0.1679. This means that we would expect a mean reading rate of 90.1 or higher from a population whose mean reading rate is 88 in 17 of every 100 random samples of size n = 21 students. The new program is not abundantly more effective than the old program. 91.59 30. B. There is 90% confidence that the population mean number of books read is between 12.34 and 14.06. 2 Expected z-score 1 0 -1 -2 0 2000 4000 31. D. Repair Cost ($) D. Yes, because the plotted values are approximately linear. B. 0 2000 4000 A. No, there are no points that are outside of the 1.5(IQR) boundary. A. The lower bound is $ 1033.9 and the upper bound is $ 2484.3. We are 95% confident that the mean cost of repair is within the confidence interval. B. Narrower, because there is less variability in the data because any variability caused by the different types of vehicles has been removed. 32. 0.498 (1) is stated to be (2) np 1 − p 257.244 (3) greater than or equal to (4) sample size (5) can be assumed to be (6) population size. B. We are 95 % confident the proportion of adults in the country who believe that televisions are a luxury they could do without is between 0.467 and 0.528. (7) possible, but not likely (8) does not contain 0.6 0.472 0.533 33. (1) = 0.42 (2) ≠ 0.42 0.45 0.51 D. Since the interval does not contain the proportion stated in the null hypothesis, there is sufficient evidence that parents' attitudes toward the quality of education have changed. 34. (1) = 9.02 (2) < 9.02 − 4.12 0.001 (3) Reject (4) is (5) less than 9.02 35. (1) Yes, (2) are (3) is (4) greater (5) μ (6) = (7) μ (8) < − 1.63 0.069 D. The P-value is less than the level of significance so there is sufficient evidence to conclude the new system is effective.