A linear regression analysis reveals a strong, negative linear relationship between x and y. Which of the following could possibly be the results from this analysis? (A) ŷ 13.1 27.4x, r = 0.85 (B) == 27.4+ 13.1x, r = -0.95 (D) == (C) ŷ 13.1+ 27.4x, r = 0.95 542 385x, r = -0.15 (E) 0.85 0.25x, r = -0.85
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- The mall has a set of data with employee age (X) and the corresponding number of annual on-the-job-accidents(Y). Analysis on the set finds that the regression equation is Y=100-3X. What is the likely number of accidents for someone aged 30? 97 100 10 none of the aboveNine pairs of data yield a regression equation of y=0.93x + 19.4, with r= 0.967 and an average y value of 64.70. What is best predicted value for y when x =65? Is it 79.85, 25.74, 89.61, 57.82, 70.55, 96.70, 19.40, or 64.70?The flow rate in a device used for air quality measurement depends on the pressure drop x (inches of water) across the device's filter. Suppose that for x values between 5 and 20, these two variables are related according to the simple linear regression model with true regression line y = -0.11 + 0.097x. (a.1) What is the true average flow rate for a pressure drop of 10 in.?(a.2) A drop of 15 in.?(b) What is the true average change in flow rate associated with a 1 inch increase in pressure drop?(c) What is the average change in flow rate when pressure drop decreases by 5 in.?
- A statistics professor wants to use the number of hours a student studies for a statistic final exam (x) to predict the final exam score (y). A regression model was fit based on data collected for a class during the previous semester, with the following results: y =35.0 + 3x Which of the following is the correct interpretation of the regression coefficient (slope)? Select the correct response: When the student does not study for the final exam, the mean final exam score is 35.0. None of the above are an interpretation of the slope For each increase of one hour in studying time, the mean change in the final exam score is predicted to be 35.0 For each increase of one hour in studying time, the mean change in the final exam score is predicted to be 3.0.A business school placement director wants to estimate the mean annual salaries 5 years after students graduate. A random sample of 25 such graduates found a sample mean of 42 740 USD and a sample standard deviation of 4 780 USD. Find a 90% confidence interval for the population mean, assuming that the population distribution is normal.Data from 147 colleges from 1995 to 2005 (Lee,2008) were tested to predict the endowments (in billions) to a college from the average SAT score of students attending the college. The resulting regression equation was Y = -20.46 + 4.06 (X). This regression indicates that: a. for every one-point increase in SAT scores, a college can expect 4.06 billion more in endowments. b. most colleges have very high endowments. c. for every one-point increase in SAT scores, a college can expect 20.46 billion fewer in endowments. d. for every one-dollar increase in endowments, the college can expect a half-point increase in SAT scores.
- The accompanying table shows results from regressions performed on data from a random sample of 21 cars. The response (y) variable is CITY (fuel consumption in mi/gal). The predictor (x) variables are WT (weight in pounds), DISP (engine displacement in liters), and HWY (highway fuel consumption in mi/gal). The equation CITY - 3.17 +0.823HWY was previously determined to be the best for predicting city fuel consumption. A car weighs 2700 lb, it has an engine displacement of 1.6 L, and its highway fuel consumption is 35 mi/gal. What is the best predicted value of the city fuel consumption? Is that predicted value likely to be a good estimate? Is that predicted value likely to be very accurate? Click the icon to view the table of regression equations. The best predicted value of the city fuel consumption is (Type an integer or a decimal. Do not round.). Regression Table I R² Adjusted R2 WT/DISP WT/HWY Predictor (x) Variables P-Value WT/DISP/HWY 0.000 0.942 0.000 0.748 0.000 0.942 0.000…A particular article used a multiple regression model to relate y = yield of hops to x, = average temperature (°C) between date of coming into hop and date of picking and x, = average percentage of sunshine during the same period. The model equation proposed is the following. y = 415.11 – 6.6x1 – 4.50x2 +e (a) Suppose that this equation describes the actual relationship. What mean yield corresponds to a temperature of 20 and a sunshine percentage of 40? (b) What is the mean yield when the average temperature and average percentage of sunshine are 19 and 44, respectively?A particular article used a multiple regression model to relate y = yield of hops to x₁ = mean temperature (°C) between date of coming into hop and date of picking and x₂ = mean percentage of sunshine during the same period. The model equation proposed is the following. y = 415.116.6x₁4.50x2+e (a) Suppose that this equation does indeed describe the true relationship. What mean yield corresponds to a temperature of 20 and a sunshine percentage of 39? (b) What is the mean yield when the mean temperature and percentage of sunshine are 19.1 and 42, respectively? You may need to use the appropriate table in Appendix A to answer this question.
- In a fisheries researchers experiment the correlation between the number of eggs in tge nest and the number of viable (surviving ) eggs for a sample of nests is r=0.67 the equation of the regression line for number of viable eggs y versus number of eggs in the nest x is y =0.72x + 17.07 for a nest with 140 eggs what is the predicted number of viable eggs ?The table below shows the amounts of crude oil (in thousands of barrels per day) produced by a country and the amounts of crude oil (in thousands of barrels per day) imported by a country, for the last seven years. Construct and interpret a 98% prediction interval for the amount of crude oil imported by the this country when the amount of crude oil produced by the country is 5,464 thousand barrels per day. The equation of the regression line is y=-1.106x+15,759.462 Oil_produced,_x Oil_imported,_y5,816 9,3455,741 9,1245,660 9,6325,405 10,0095,155 10,1685,059 10,1055,015 10,055A regression was run to determine if there is a relationship between hours of TV watched per day (x) and number of situps a person can do (y). The results of the regression were: y=ax+b a=-0.666 b=36.053 r2=0.552049 r=-0.743 Use this to predict the number of situps a person who watches 4.5 hours of TV can do (to one decimal place)