The article “Stochastic Estimates of Exposure and Cancer Risk from Carbon Tetrachloride Released to the Air from the Rocky Flats Plant” (A. Rood, P. McGavran, et al., Risk Analysis, 2001:675–695) models the increase in the risk of cancer due to exposure to carbon tetrachloride as lognormal with μ = −15.65 and σ = 0.79. a) Find the mean risk. b) Find the median risk. c) Find the standard deviation of the risk. d) Find the 5th percentile. e) Find the 95th percentile.
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The article “Stochastic Estimates of Exposure and Cancer Risk from Carbon Tetrachloride Released to the Air from the Rocky Flats Plant” (A. Rood, P. McGavran, et al., Risk Analysis, 2001:675–695) models the increase in the risk of cancer due to exposure to carbon tetrachloride as lognormal with μ = −15.65 and σ = 0.79. a) Find the
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- The article "Characteristics and Trends of River Discharge into Hudson, James, and Ungava Bays, 1964-2000" (S. Dery, M. Stieglitz, et al., Journal of Climate, 2005:2540-2557) presents measurements of discharge rate x (in kmlyr) andpeakflow y (in m/s) for 42 rivers that drain into the Hudson, James, and Ungava Bays. The data are shown in the following table: Discharge Peak Flow 94.24 4110.3 66.57 4961.7 59.79 10275.5 48.52 6616.9 40.00 7459.5 32.30 2784.4 31.20 3266.7 30.69 4368.7 26.65 1328.5 22.75 4437.6 21.20 1983.0 20.57 1320.1 19.77 1735.7 18.62 1944.1 17.96 3420.2 17.84 2655.3 16.06 3470.3 1561.6 14.69 11.63 869.8 11.19 936.8 11.08 1315.7 10.92 1727.1 9.94 768.1 7.86 483.3An article in the journal Air and Waste (Update on Ozone Trends in California's South Coast Air Basin, Vol. 43, 1993) investigated the ozone levels in the South Coast Air Basin of California for the years 1976-1991. The author believes that the number of days the ozone levels exceeded 0.20 ppm (the response) depends on the seasonal meteorological index, which is the seasonal average 850-millibar Temperature (the predictor). The following table gives the data. Year Index 1976 1977 1978 1979 1980 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990 1991 Days 91 105 106 108 88 91 58 82 81 65 61 48 61 43 33 36 16.7 17.1 18.2 18.1 17.2 18.2 16.0 17.2 18.0 17.2 16.9 17.1 18.2 17.3 17.5 16.6 (a) Construct a scatter diagram of the data. (b) Estimate the prediction equation. (c) Test for significance of regression. (d) Calculate the 95% CI and PI on for a seasonal meteorological index value of 17. Interpret these quantities.The body mass index (BMI) of a person is defined to be the person’s body mass divided by the square of the person’s height. The article “Influences of Parameter Uncertainties within the ICRP 66 Respiratory Tract Model: Particle Deposition” (W. Bolch, E. Farfan, et al., Health Physics, 2001:378–394) states that body mass index (in kg/m²) in men aged 25–34 is lognormally distributed with parameters μ = 3.215 and σ = 0.157. a) Find the mean BMI for men aged 25–34. b) Find the standard deviation of BMI for men aged 25–34. c) Find the median BMI for men aged 25–34. d) What proportion of men aged 25–34 have a BMI less than 22? e) Find the 75th percentile of BMI for men aged 25–34.
- A study was performed looking at the risk of fractures in three rural Iowa communities according to whether their drinking water was “higher calcium,” “higher fluorides,” or “control” as determined by water samples. Table 11.10 presents data comparing the rate of fractures (over 5 years) between the higher-calcium vs the control communities for women ages 20–35 and 55–80, respectively . Table 11.10 Relationship of calcium content of drinking water to the rate of fractures in rural Iowa Ages 20-35 Number of women with fractures Total number of women Ages 55-80 Number of woemn with fractures Total number of women Control 3 37 Control 11 121 High calcium 1 33 High calcium 21 148 13.1 What test can be used to compare the fracturerates in these two communities while controlling for age? 13.2 Implement the test in Problem 13.1, report a p-value, and make a conclusion on relationship between drinking water calcium concentration and rate of fracture based on the p-value.A study was performed looking at the risk of fractures in three rural Iowa communities according to whether their drinking water was “higher calcium,” “higher fluorides,” or “control” as determined by water samples. Table 11.10 presents data comparing the rate of fractures (over 5 years) between the higher-calcium vs the control communities for women ages 20–35 and 55–80, respectively . Table 11.10 Relationship of calcium content of drinking water to the rate of fractures in rural Iowa Ages 20-35 Number of women with fractures Total number of women Ages 55-80 Number of woemn with fractures Total number of women Control 3 37 Control 11 121 High calcium 1 33 High calcium 21 148 13.1 What test can be used to compare the fracturerates in these two communities while controlling for age? 13.2 Implement the test in Problem 13.1, report a p-value, and make a conclusion on relationship between drinking water calcium concentration and rate of fracture based on the p-value. 13.3…The article "Lead Dissolution from Lead Smelter Slags Using Magnesium Chloride Solutions" (A. Xenidis, T. Lillis, and I. Hallikia, The AusIMM Proceedings, 1999:37-14) discusses an investigation of leaching rates of lead in solutions of magnesium chloride. The data in the following table (read from a graph) present the percentage of lead that has been extracted at various times (in minutes). Time (t) 4 8 16 30 60 120 Percent extracted (v) |1.2 1.6 2.3 2.8 3.6 4.4 a. The article suggests fitting a quadratic model y = Bo + B,t + Bz² + ɛ to these data. Fit this model, and compute the standard deviations of the coefficients. b. The reaction rate at time t is given by the derivative dy/dt = B, + 2B,t. Estimate the time at which the reaction rate will be equal to 0.05. c. The reaction rate at t = Oisequal to B1. Find a 95% confidence interval for the reaction rate at t = 0. d. Can you conclude that the reaction rate is decreasing with time? Explain.
- Ocean currents are important in studies of climate change, as well as ecology studies of dispersal of plankton. Drift bottles are used to study ocean currents in the Pacific near Hawaii, the Solomon Islands, New Guinea, and other islands. Let x represent the number of days to recovery of a drift bottle after release and y represent the distance from point of release to point of recovery in km/100. The following data are representative of one study using drift bottles to study ocean currents. Σx = 476, Σy = 87.1, Σx2 = 62,290, Σy2 = 2046.87, Σxy = 11121.3,and r ≈ 0.94367. x days 72 76 32 91 205y km/100 14.7 19.6 5.3 11.7 35.8 a) Use a 1% level of significance to test the claim ? > 0.(Use 2 decimal places.)t =critical t= b) Find the predicted distance (km/100) when a drift bottle has been floating for 60 days. (Use 2 decimal places.)________ km/100 c) Find a 90% confidence interval for your prediction of part (d). (Use 1 decimal place.)lower limit = _____…A paper gives data on x = change in Body Mass Index (BMI, in kilograms/meter2) and y = change in a measure of depression for patients suffering from depression who participated in a pulmonary rehabilitation program. The table below contains a subset of the data given in the paper and are approximate values read from a scatterplot in the paper. BMI Change (kg/m²) 0.5 -0.5 0 0.1 0.7 0.8 1 1.5 1.2 1 0.4 0.4 Depression Score Change -1 9 4 4 5 8 13 14 17 18 12 14 The accompanying computer output is from Minitab. Fitted Line Plot Depression score change = 6.512 + 5.472 BMI change 20 S 5.26270 R-Sq 27.16% R-Sq (adj) 19.88% 15- : 10- -0.5 0.0 1.5 Ⓡ S 5.26270 Coefficients Term Coef VIF SE Coef 2.26 T-Value 2.88 P-Value 0.0164 Constant 6.512 BMI change 5.472 2.83 1.93 0.0823 1.00 Regression Equation Depression score change = 6.512 + 5.472 BMI change (a) What percentage of observed variation in depression score change can be explained by the simple linear regression model? (Round your answer to…Cell Phone Radiation Listed below are the measured radiation absorption rates (in W/kg) corresponding to these cell phones: iPhone 5S, BlackBerry Z30, Sanyo Vero, Optimus V, Droid Razr, Nokia N97, Samsung Vibrant, Sony Z750a, Kyocera Kona, LG G2, and Virgin Mobile Supreme. The data are from the Federal Communications Commission (FCC). The media often report about the dangers of cell phone radiation as a cause of cancer. The FCC has a standard that a cell phone absorption rate must be 1.6 W/kg or less. If you are planning to purchase a cell phone, are any of the measures of center the most important statistic? Is there another statistic that is most relevant? If so, which one?
- Cell Phone Radiation Listed below are the measured radiation absorption rates (in W/kg) corresponding to these cell phones: iPhone 5S, BlackBerry Z30, Sanyo Vero, Optimus V, Droid Razr, Nokia N97, Samsung Vibrant, Sony Z750a, Kyocera Kona, LG G2, and Virgin Mobile Supreme. The data are from the Federal Communications Commission.The following scatterplot shows the mean annual carbon dioxide (CO,) in parts (CO2) per million (ppm) measured at the top of a mountain and the mean annual air temperature over both land and sea across the globe, in degrees Celsius (C). Complete parts a through h on the right. f) View the accompanying scatterplot of the residuals vs. CO2. Does the scatterplot of the residuals vs. CO, show evidence of the violation of any assumptions behind the regression? 16.800 A. Yes, the outlier condition is violated. 16.725 O B. Yes, the linearity and equal variance assumptions are violated. 16.650 C. Yes, the equal variance assumption is violated. 16.575 O D. No, all assumptioris are okay. 16.500 O E. Yes, all the assumptions are violated. 325.0 337.5 350.0 362.5 CO2 (ppm) OF Yes, the linearity assumption is violated. his vear, What mean temperature doesA paper gives data on x = change in Body Mass Index (BMI, in kilograms/meter) and y = change in a measure of depression for patients suffering from depression who participated in a pulmonary rehabilitation program. The table below contains a subset of the data given in the paper and are approximate values read from a scatterplot in the paper. BMI Change (kg/m²) -0.5 0.7 0.5 0.1 0.8 1 1.5 1.2 1 0.4 0.4 Depression Score Change -1 4 4 8 13 14 16 18 12 14 The accompanying computer output is from Minitab. Fitted Line Plot Depression score change = 6.598 + 5.327 BMI change 20- 5.10254 R-Sq R-Sq (adj) 20.06% 27.32% 15- 10- 5- 0- -0.5 0.0 0.5 1.0 1.5 BMI change R-sq 5.10254 27.32% Coefficients Term Coef SE Coef T-Value P-Value VIF Constant 6.598 2.19 3.01 0.0132 BMI change 5.327 2.75 1.94 0.0812 1.00 Regression Equation Depression score change = 6.598 + 5.327 BMI change (a) What percentage of observed variation in depression score change can be explained by the simple linear regression model?…
