An insurance company has a portfolio of car insurance. For a given period of time, it is assumed that the drivers are classified in two categories: 'good drivers' and 'bad drivers'. The proportion of good drivers is 75% of the total drivers. We also know that the number of claims for a each driver is following Poisson with parameter A where A =1 for good drivers and A = 3 for bad drivers.
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Contingency Table
A contingency table can be defined as the visual representation of the relationship between two or more categorical variables that can be evaluated and registered. It is a categorical version of the scatterplot, which is used to investigate the linear relationship between two variables. A contingency table is indeed a type of frequency distribution table that displays two variables at the same time.
Binomial Distribution
Binomial is an algebraic expression of the sum or the difference of two terms. Before knowing about binomial distribution, we must know about the binomial theorem.
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- A negative correlation indicates: the values on the two variables being analyzed move in opposite directions. a higher score on one variable is related to a higher score on the other variable. the scores of each case in a sample simultaneously on two variables the value on the two variables being analyzed move in the same direction.You are a big soccer fan and you decide to gather information on the goal differential (goals scored - goals allowed) for your favorite team in the Premier League during the last season. You find that the average goal differential for home games (games that take place on the team's own stadium) was 1.37 and the average goal differential for away games (games that take place on the opponent team's stadium) was 0.32. With the data you have, if you regress goal differential (call it variable G) on a dummy variable (call it variable D) that takes the value 1 if the game is a home game and takes the value zero if the game is an away game, and obtain Ĝi = Bo + B₁ D₁, then, Bo will be equal to equal to and ₁ will beRia wants to know if there is a significant difference between the average monthly hair growth using Shampoo A vs Shampoo B. She recruited eight participants and compared their average monthly hair growth (in centimeters) using the two different brands. Her data is assumed to be normally distributed which produced the following values: Shampoo A 0.50 1.00 1.50 1.70 0.75 0.60 1.25 1.25 Shampoo B 1.00 1.25 2.00 2.00 1.50 1.00 1.50 1.75 Choose the appropriate test statistic and make a conclusion about the problem. Show your output below in a step-by-step manner: 21 - 22. State the null and alternative hypotheses. 23- 48. Make a decision about the null hypothesis: (23 - 26) Calculate the degree of freedom (27.) Assume that a 05 (two-tailed). Identify the critical values and state a decision rule.
- The amount of income spent on housing is an important component of the cost of living. The total costs of housing for homeowners might include mortgage payments, property taxes, and utility costs (water, heat, electricity). An economist selected a sample of 20 homeowners in New England and then calculated these total housing costs as a percent of monthly income, 5 years ago and now. The information is reported below. Is it reasonable to conclude the percent is less now than 5 years ago? Homeowner Five Years Ago Now Homeowner Five Years Ago Now 1 17 % 10 % 11 35 % 32 % 2 20 39 12 16 32 3 29 37 13 23 21 4 43 27 14 33 12 5 36 12 15 44 40 6 43 41 16 44 42 7 45 24 17 28 22 8 19 26 18 29 19 9 49 28 19 39 35 10 49…The following tableshows the typical depth (rounded to the nearest foot) for nonfailed wells in geological formations in Baltimore County (The Journal of Data Science, 2009, Vol. 7, pp. 111-127). Geological Formation Group Number of Nonfailed Wells Nonfailed Well Depth Gneiss 1,515 255 Granite 26 218 Loch Raven Schist 3,290 317 Mafic 349 231 Marble 280 267 Prettyboy Schist 1,343 255 Other schists 887 267 Serpentine 36 217 Total 7,726 2,027 Let the random variableX denote the depth (rounded to the nearest foot) for nonfailed wells. Detemine the cumulative distribution function for X. Round your answers to four decimal places (e.g. 98.7654). x< 217 217FOR EACH FXERCISE . A study was conducted to determine the relationship between a person's monthly income in dollars and the number of meals that person eats away from per month. The data from the sample are shown here. (The information in this exercise will be used for Exercises 14 and 36 in Section 10-4 and Exercise 16 in Section 10-5.) SKETCH SCAITERPLOT home - FIND SUMMARY STATISTICS FIND CORRELATION Income x 500 1200 1500 945 850 400 540 Meals y 8. 12 16 10 3The following table shows the typical depth (rounded to the nearest foot) for nonfailed wells in geological formations in Baltimore County (The Journal of Data Science, 2009, Vol. 7, pp. 111-127). Geological Formation Group Number of Nonfailed Wells Nonfailed Well Depth Gneiss 1,515 255 Granite 26 218 Loch Raven Schist 3,290 317 Mafic 349 231 Marble 280 267 Prettyboy Schist 1,343 255 Other schists 887 267 Serpentine 36 217 Total 7,726 2,027 Let the random variable X denote the depth (rounded to the nearest foot) for nonfailed wells. Detemine the cumulative distribution function for X. Round your answers to four decimal places (e.g. 98.7654). x < 217 217Suppose that the students just completing freshman year at Galton University have Math SAT scores and first-year GPAS with the following summary statistics according to the registrar's office. Average Math SAT score = 540 SD = 100 Average first-year GPA = 2.6 r = 0.58 SD = 0.8 The scatterplot of their first-year GPAS on Math SAT scores is an oval-shaped cloud of points. Suppose a student from this freshman class at Galton University takes the SAT again during the summer and gets a score 75 points higher than his original score. By how much will his first-year GPA increase?The following table lists highway miles (in MPG) and weight (in pounds) for a variety of cars, foreign and domestic (same data as before). CAR HWY WEIGHT Chev. Cavalier 31 2795 Lincoln Cont. 24 3930 Mitsubishi Eclipse 33 3235 Olds. Aurora 26 3995 Pontiac Grand Am 30 3115 Chev. Corvette 28 3220 BMW 3-Series 27 3225 Ford Crown Victoria 24 3985 Hyundai Accent If we let x be the weight and y be the Highway MPG, and we used Excel to compute the equation of the Least- 37 2290 Square regression line (which comes out to be y = -0.00672'x + 51.1156) as well as the correlation coefficient (which comes out to be -0.89). Then you used that equation to predict that the Highway Miles for a car that weighs just 1000 pounds would be 44.4 mpg. I think this prediction is accurate because the person used Excel for the computation. O I think this prediction is not accurate since a car that weighs 1000 pounds is not part of the original data list. O Ithink this prediction is not accurate because the person…Consider a two-dimensional scatterplot representing the relationship between two continuous variables. If the correlation coefficient is -1, then: a. All points lie in a straight line with a slope of -1. b. All points lie in a straight line with an unknown negative slope. c. All points do not lie in a straight line but the best fitting regression line has a slope of -1 d. There is a strong positive relationship between the two variables.Below shows the linear relationship between first year college GPA and high school GPA For both variables the GPA ranges from 0 to 4 in this sample. However, for the high schools sampled in this region, the true range of GPA is 0 to 5, how might this affect the correlation between these two variables?The authors of a paper compared two different methods for measuring body fat percentage. One method uses ultrasound, and the other method uses X-ray technology. Body fat percentages using each of these methods for 16 athletes (a subset of the data given in a graph that appeared in the paper) are given in the accompanying table. You can assume that the 16 athletes who participated in this study are representative of the population of athletes. Athlete X-ray Ultrasound 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 5.00 8.00 9.25 12.00 17.25 29.50 5.50 6.00 8.00 13.50 9.25 11.00 12.00 14.00 17.00 18.00 4.25 4.75 9.00 11.75 17.00 27.50 6.50 6.75 8.75 14.50 9.50 12.00 12.25 15.50 18.00 18.25 Use these data to estimate the difference in mean body fat percentage measurement for the two methods. Use a confidence level of 95%. (Use μ = MX-ray-Multrasound. Round your answers to three decimal places.) × % Interpret the interval in context. O There is a 95% chance that the true mean body fat percentage…SEE MORE QUESTIONSRecommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons IncProbability and Statistics for Engineering and th…StatisticsISBN:9781305251809Author:Jay L. DevorePublisher:Cengage LearningStatistics for The Behavioral Sciences (MindTap C…StatisticsISBN:9781305504912Author:Frederick J Gravetter, Larry B. 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