Statistics 110 – Summer II 2006 Solutions

Statistics 110 – Summer II 2006 July 20, 2006 Name:____SOLUTIONS__________________ Test # 1 Please complete the following problems. Be sure to ask me if you have any questions or anything is unclear. Partial credit will be given, so please be sure to show all of your work. 1. (6 pts) The UCONN office of institutional research provides statistics on the racial composition of Undergraduates at UCONN. A pie chart of the undergraduate racial composition as of October 15, 2005 is given below: UCONN Undergraduate Racial Composition American Indian or Alask a Nativ e 0.4% Nonresident Aliens 1.0% Hispanic 4.5% Black , non-Hispanic 5.2% Asian or Pacific Islander 7.0% White, non-Hispanic 72.1% Unknown 9.8% Source: UCONN Office of Institutional Research (October 15, 2005) Interpret this pie chart, and describe the racial composition of UCONN undergraduates. Most of the students are non-Hispanic White (72.1%). The next largest group is the Asian and Pacific Islanders, at 7%. The race group with the smallest percentage of students is American Indian or Alaska Natives, with 0.4%. The percentages of Hispanics and Black, nonHispanic are similar (4.5% and 5.2%). Only 1% of undergraduates are non-resident aliens. Also, it is interesting that the race is unknown for a sizeable percentage of the undergraduate student body (9.8%). 2. (6 pts) The UCONN office of institutional research also tabulates statistics about student enrollment. The bar chart below shows the enrollment by school of students in the spring of 2006. Spring 2006 Enrollment at UCONN 9000 8000 Enrollment 7000 6000 5000 4000 3000 2000 1000 0 s y n g g ts re th ss es es es ee ck ac al tio rin di Ar di sin ltu ne nc gr Hi e i a e r u u m u e e c t e t H s i d u n ic ne S S Nu ar fe Bu Sc Fi gi ngr ed lif Ed ly al Ph i i l n o c r A l & t E N m A s ne Ra rt Fa Ge A al er b Li School Source: UCONN office of Institutional Research What does this bar chart tell you about the pattern of enrollment at UCONN? Most of the students are enrolled in the School of Liberal Arts and Sciences (~8500). The lowest numbers are enrolled in the Ratcliffe-Hicks and General Studies schools (~100 each). After liberal arts, the school with the next highest enrollment is business (~1750) and then engineering (~1600). 3. (10 pts) The following data represent the LSAT scores of some students in a Connecticut college in 2005. Use these data to answer the following questions: 157 139 151 141 145 151 132 136 126 142 138 122 153 127 137 130 142 122 131 137 141 141 138 133 a. Find the mean and standard deviation of these data. (1 pt) Mean = 138 (1 pt) Standard Deviation = 9.3017 b. Find the five-number summary of these data. (1 (1 (1 (1 (1 pt) pt) pt) pt) pt) Min Q1 Med Q3 Max = = = = = 122 131.5 138 142 157 c. Are there any outliers? (Use the 1.5 IQR rule here) The IQR = Q3 – Q1 = 142 – 131.5 = 10.5 1.5 * IQR = 1.5 * 10.5 = 15.75 Q1 – 1.5 IQR = 131.5 – 15.75 = 115.75 Q3 + 1.5 IQR = 142 + 15.75 = 157.75 (3 pts) There are no outliers. No data points are below 115.75 or above 157.75. Weight (lbs.) 4. (6 pts) The weights of two populations of mule deer have been measured1, one in the Northwest, and the other in the Northeast. Boxplots of the weights for the two groups are given below: 300 200 100 Northwest Northeast Use the boxplots to compare the distributions of weights for the two groups of deer. The weight of the Northwest deer population tends to be less than that of the Northeast deer population (IQR from 175 – 250 versus one from 175 – 325). The Northeast deer population shows more variability. Also, the Northwest deer population is skewed to the left while that of the Northeast is symmetric. 1 This is fictional data. Don’t plan your deer hunting trip based on this. 5. (6 pts) Every fall, the US Census Bureau reports on health insurance coverage in the United States. A histogram of the percentage of people without health insurance among all 50 states and the District of Columbia is given below: State-Level Uninsured Rates 14 12 Frequency 10 8 6 4 2 0 8 12 16 20 Percent Uninsured 24 Source: 2005 CPS, US Census Bureau Based on the histogram above, describe the distribution of these data. (Consider such things as the center, skewness, spread, and presence of outliers.) The distribution is slightly skewed to the right. The median is around 14%, and the mean is slightly higher, around 16%. The rates range from 8% to 26%. There is a possible outlier. One state has an uninsured rate of 26%. 6. (2 pts) Give an example of a variable that might be skewed to the right. Salaries of baseball players on a team might be skewed to the right. The superstar may be making a lot more than the rest of the team. 7. (3 pts) You are studying meerkats in Africa, and record numerous variables for each of the members of a particular “gang” of meerkats. Three of these variables are: -- Gender, recorded as M or F -- Height, recorded in inches -- Status, recorded as 0 for juveniles and 1 for adults State whether each of these three variables is a categorical random variable or a quantitative random variable. (1 pt each) Gender and Status are categorical variables. Height is a quantitative variable. 8. (8 pts) Suppose that the number of traffic stops per day on Eagleville Road is normally distributed with a mean of 12 and a standard deviation of 2. a. Label the normal curve below with the appropriate numbers on the horizontal axis. (2 pts) (2 pts) b. Shade the area under the curve that corresponds to the probability that more than 15 traffic stops were made. c. Calculate the probability that more than 15 traffic stops were made. (4 pts) [calculator] normalcdf(15, 1 E 99, 12, 2) = 0.0668 or 6.68% [by hand] Standardize = z = x - µ = 15 – 12 = 1.5 σ 2 From Table A, P(Z < 1.5) = 0.9332 So, P(Z > 1.5) = 1 - 0.9332 = 0.0668 or 6.68% 9. (9 pts) Suppose that the life of a given brand of auto batteries is normally distributed, with a mean of 44 months and a standard deviation of 6 months. a. Find the percentage of batteries that will have a span of 41 to 52 months. [calculator] normalcdf(41, 52, 44, 6) = 0.6003 or 60.03% This differs from the result we get by hand because of roundoff errors. [by hand] Standardize x = 41: z = x - µ = 41 – 44 = -0.5 σ 6 From Table A, P(Z < -0.5) = 0.3085 Standardize x = 52: z = x - µ = 52 – 44 = 1.33 σ 6 From Table A, P(Z < 1.33) = 0.9082 So, P(41 < X < 52) = P(-0.5 < Z < 1.33) = 0.9082 – 0.3085 = 0.5997 59.97% of batteries will have a lifespan of 41 to 52 months. b. Find the percentage of batteries that will have a span of more than 58 months. [calculator] normalcdf(58, 1 E 99, 44, 6) = 0.0098 or 0.98% [by hand] Standardize: z = x - µ = 58 – 44 = 2.33 σ 6 From Table A, P(Z < 2.33) = 0.9901 So, P(X > 58) = P(Z > 2.33) = 1 - 0.9901 = 0.0099 or 0.99% c. Find the percentage of batteries that will have a span of less than 35 months. [calculator] normalcdf(-1 E 99, 35, 44, 6) = 0.0668 or 6.68% [by hand] Standardize: z = x - µ = 35 – 44 = -1.5 σ 6 From Table A, P(X < 35) = P(Z < -1.5) = 0.0668 or 6.68% 10. (6 pts) Every July, the famed Tour de France bike race circles France. For a certain stage that is 170 km (106 miles!), the tour organizers suppose that the finishing times (in hours) are normally distributed with a mean of 5.2 hours and a standard deviation of 0.2 hours. A team manager would like to get an idea of how fast his rider must finish in order to be in the fastest 10 percent of the riders. What time would put a rider into the fastest 10%? We want the X value that gives us the fastest, or lowest, 10% of times. This is an example of the “backwards” type of problem. [calculator] invNorm(0.10, 5.2, 0.2) = 4.9437. A rider with a time of 4.94 hours or less would be in the top 10%. [by hand] From Table A, find the Z value that gives us an area of 10% to the left of it. This is Z = -1.28. Then we “unstandardize”: X − X − 5.2 Z=  −1.28 =  −0.256 = X − 5.2  X = 4.944  0.2 11. (9 pts) Three scatterplots are given below. Below each one, give what you believe would be a plausible value for the correlation between X and Y. Explain why you chose those numbers. Scatterplot A r = 0.05 . 70 60 The scatterplot shows a random pattern. The correlation should be close to 0. 50 Y 40 30 20 10 0 0 5 10 15 20 25 30 35 X Scatterplot B r = 0.91 . 80 The scatterplot shows a strong, positive association. The correlation should be close to 1. 70 60 50 40 30 20 10 0 0 10 20 30 40 X Scatterplot C r = -0.62 . 70 The correlation shows a moderately strong, negative association. So, the correlation should be somewhere in the middle of 0 and -1. 60 50 Y 40 30 20 10 0 0 5 10 15 20 X 25 30 35 12. (6 pts) You are studying the relationship between acreage and the number of trees. A scatterplot of the data is given below: Scatterplot of Tree Count vs. Acreage 300 Tree Count 250 200 influential 150 100 outlier 50 0 40 50 60 70 80 90 100 110 Acreage a. Are there any outliers or influential data values in this scatterplot? Circle and label them. (2 pts) Yes, there is one of each. See the graph above. b. Describe the scatterplot. Pay particular attention to form, direction, and strength. (4 pts) There seems to be a positive association between acreage and tree count. The association in moderately strong, with a correlation maybe somewhere around 0.6. As noted above, there is a possible outlier and an influential point. 13. (10 pts) Suppose you are studying the relationship between rainfall (measured in inches) and wheat yield (measured in bushels/acre). You do a regression line with rainfall as the X variable and wheat yield as the Y variable, and get the following regression equation: yˆ = 15.25 + 2.14x a. List the y-intercept and slope for this line. (2 pts) y-intercept = a = 15.25 slope = b = 2.14 b. Interpret the slope of this line. That is, how does a change in rainfall (X) affect the wheat yield (Y)? (2 pts) One additional inch of rainfall results in 2.14 additional bushels of wheat per acre. Recall that b = slope = change in y 2.14 bushels = change in x 1 inch c. What yield would you predict for Pocatello, Idaho’s average July rainfall total of 0.7 inches? (2 pts) Pocatello: Ŷ = 15.25 + 2.14 (0.7) = 16.75 bushels/acre What yield would you predict for Storrs, Connecticut’s average July rainfall total of 4.1 inches? (2 pts) Storrs: Ŷ = 15.25 + 2.14 (4.1) = 24.02 bushels/acre d. If the R2 value for this regression is R2 = 0.796, what does this tell you about your regression line? (2 pts) This is close to 1, which tells me that the regression line is working pretty well. 14. (13 pts) Suppose you wish to use the following data about expenses covered by an insurance carrier and the number of days of confinement for a random sample of 10 patients. Number of Days of Confinement 1 3 6 7 2 4 12 15 5 9 Expense (in $) 50 175 180 200 60 140 420 540 170 300 Expense ($) a. Draw a scatterplot of these data on the axes given below: (3 pts) 600 500 400 300 200 100 0 0 5 10 15 20 Days of Confinement b. Calculate the correlation between the days of confinement (X) and the expense (Y). (2 pts) r = 0.9803 . c. Describe the relationship between these variables. Pay particular attention to form, direction, strength, and the presence of outliers. (3 pts) There is a strong, positive, and linear association between days of confinement and expense. I don’t think there are any outliers in this dataset. d. Write down the equation of the regression line for predicting the expense (Y) from the number of days of confinement (X). (3 pts) Let y = expense, and x = days of confinement. Ŷ = 6.347 + 33.930 x e. What would be the predicted expense for a patient in the hospital for 8 days? (2 pts) Ŷ = 6.347 + 33.930 (8) = $277.79 15. Bonus Question (4 pts): The equation for a regression line between two variables X and Y is: yˆ = 23.26 − 1.29x We are also given the following summary statistics: x = 8.62 s x = 1.24 y = 12.14 s y = 2.36 What is the R2 value for this regression line? We can use the equation for finding b to find r, and then find R2. b = -1.29 from the regression equation given in the problem. b=r Sy Sx 2.36 1.24 − 1.29 = 1.9032 r r = −0.6778 − 1.29 = r r 2 = (− 0.6778) = 0.4594 2 So, R2 for this line is 0.4594, a moderately good fit.