Nonlinear Regression Curve Fitting and Regression (Statcrunch) Answers to selected problems

Transcription

1 Nonlinear Regression Curve Fitting and Regression (Statcrunch) Answers to selected problems Act 1&3 1. a) Exponential growth fits well. b) Statcrunch: Ln ( Y ) = ( x ) Exponential function (Log Y transform) fits well. The points on the scatterplot are close to the curve and it has a high r-squared value. c) r 99.6%. 99.6% of the variability in wind power can be explained by the exponential relationship with time (years). d) See above. Regression Criteria a) Both the years and wind power are quantitative measurement data. b) There are no influential outliers. All the points are close to the curve and it has a very high R-squared value. c) The scatterplot has an exponential growth shape and a high R-squared value with the exponential curve. d) The residual plot verses the x values is a little fan shaped, so does not pass the homoscedasticity requirement. e) The histogram of the residuals is bell shaped, so it passes the nearly normal requirement. f) The histogram is also centered near zero, and so passes that requirement as well.

2 . a) Exponential decay will fit the data well. b) statcrunch: Ln ( y ) = Y (x) ; The exponential decay function (Log Y transform) fits pretty well. The points on the scatterplot are somewhat close to the curve and it has a high r-squared value. c) r %. So 83.4% of the variability in the retirement account balance can be explained by the exponential relationship with time (months). d) See above. Regression Criteria a) Both the time in months and retirement account balance are quantitative measurement data with units. b) There may be one outlier. The data point in the first month when there is 78 thousand dollars in the account. This point is pretty far from the exponential curve. It is probably not extremely influential since the curve still has a high R-squared value. c) The scatterplot has an exponential decay shape and a high R-squared value with the exponential curve. The residual plot indicates a parabola pattern indicating that perhaps a quadratic curve might fit this data well also. d) The residual plot verses the x values is a little fan shaped, so does not pass the homoscedasticity requirement. e) The histogram of the residuals is skewed right, so it fails the nearly normal requirement. f) The histogram is still centered near zero, and so passes that requirement.

3 3. a) Exponential growth fits well. b) Statcrunch: Ln ( y ) = ( x ) Exponential function (Log Y transform) fits really well. The points on the scatterplot are close to the curve and it has a high r-squared value. c) r 100%. 100% of the variability in account balance can be explained by the exponential relationship with time (years since 1990). d) See above. Regression Criteria a) Both the years and savings account balance are quantitative measurement data. b) There are no influential outliers. All the points are close to the curve and it has a very high R-squared value. c) The scatterplot has an exponential growth shape and a high R-squared value with the exponential curve. d) The residual plot verses the x values is not fan shaped, so it does pass the homoscedasticity requirement. e) The histogram of the residuals is almost bell shaped, so it passes the nearly normal requirement. f) The histogram is also centered near zero, and so passes that requirement as well.

4 4. a) Exponential decay or a Decreasing Log function will fit the data well. b) Statcrunch: Ln ( y ) = ( x ) The exponential decay function (Log Y transform) fits well. The points on the scatterplot are close to the curve and it has a high r-squared value. c) r %. So 91.3% of the variability in ultrasound response can be explained by the exponential relationship with the metal distance. d) See above. Regression Criteria a) Both the ultrasound response and the metal distance are quantitative variables. b) There are no influential outliers. All the points are relatively close to the curve and it has a high R-squared value. c) The scatterplot has an exponential decay shape and a high R-squared value with the exponential curve. d) The residual plot verses the x values is not fan shaped, so it does pass the homoscedasticity requirement. e) The histogram of the residuals is skewed right, so it does not pass the nearly normal requirement. f) The histogram is centered near zero, and so passes that requirement.

5 5. The exponential curve equations were found by doing a LOG(Y) transformation. In order to solve for y and make a prediction we will need to make both sides of the equation the exponent on either base 10 (minitab) or base e (statcrunch). Hence the y values are exponential in nature. 6. The atomic energy data has negative Y values, so we could not do a LOG(Y) transformation as we can only take the LOG of positive values. Act 4 1. a) 0 x 14 ; This represents the years between 1995 and 009. b) I believe wind power is an untapped power source and as most countries are in desperate need of clean energy, wind power will most likely continue to grow exponentially. c) Predict 8458 MW of power in year 7 d) Predict MW of power in year 13 e) Year 0 would be excessive extrapolation.. a) 1 x 5 ; This represents the months between January 010 to February 01. b) In looking at their spending patterns, unless the couple finds another sourse of income, they may run out of money in this account. As long as the account stays open, it will probably follow this pattern. c) Predict the account will have $48.37 left at month 11.5 d) Predict the account will have $40.1 by month 4.5 e) Month 36 would be excessive extrapolation. If we did extrapolate, we would predict the account would have $36.06 f) Month 480 would be excessive extrapolation. Plus I doubt the account will have enough money to stay open. 3. a) 0 x ; This represents the years between 1990 and 01. b) Savings accounts generally do tend to follow a exponential growth model. The more money in the account, the faster they grow. However extrapolation would be dangerous as interest rates are tied to the economy. During a recession, the interest may to drop to zero. c) In year 16, we predict the account to have $ d) In year 1, we predict the account to have $773.51

6 e) Due to the recent recession, I doubt the savings account would continue to grow exponentially. If we did extrapolate and make a guess, we would predict the account to have $ , but this prediction could be very wrong. f) Year 50 would be excessive extrapolation. It is doubtful this account will continue to follow this pattern for 50 years. 4. a) 0.5 x 6 ; This represents the metal distance to be between 0.5 and 6 millimeters. b) It seems that the farther away the metal is the ultrasound response is decreasing. It makes sense to think this response may continue to follow the exponential decay pattern outside the scope of the data. c) We predict the response to be if the metal distance is.83 mm d) We predict the response to be 5.6 if the metal distance is 4.51 mm e) If we do extrapolate and use the formula to predict the ultrasound response at a metal distance of 6.75 mm, we would get a negative ultrasound response. So this shows that extrapolation would be a bad idea for this data and would give huge errors. 5. The equations that were found by doing a LOG(Y) transformation on them are really exponential functions. In order to solve for y and make a prediction we will need to make both sides of the equation the exponent on either the number 10 or e. Hence the y values are exponential in nature. If we are doing a LOG (x) transformation, then the function is actually a logarithmic function. Exponential and Log functions are inverses of each other. If you take exponential data and switch the x and y variables, you would get a logarithmic pattern. 6. Extrapolation is difficult to judge because there is no way to judge error. Standard error only applies to the scope of the x values. If we leave the scope of the x values, anything could happen. People do extrapolate however, because they are curious what will happen next year for example. Context is important as well as thinking about future tendencies. It is important to remember the difference between a little extrapolation (next year) and excessive extrapolation (0 years from now). It is always a bad idea to excessively extrapolate. Act 8 1ab) An opening down parabola fits the data well. Minitab found the quadratic function to be y = x 3.38 x^. The function fits the data well with an r-squared value of 98.7%. 1c) r-squared was 98.7%. 98.7% of the variability in height can be explained by the quadratic relationship to time.

7 1d) standard deviation of the residual errors = 8.74 feet. The points in the scatterplot are an average of 8.74 from the quadratic curve. If we use the quadratic function to predict the height of the rock, we could have an average error of 8.74 feet too high or too low. 1e) x coord of vertex = The rock will reach a maximum height after seconds. The maximum height was about 61.5 feet. Act 10 1a) 0 < x < 3.6 seconds. 1b) Extrapolation might be bad, as the rock will reach a height of zero rather quickly. 1c) After 1.8 seconds we predict the height will be 35.5 feet. 1d) After 3. seconds we predict the height will be 10.7 feet 1e) If we extrapolate, after 4 seconds we predict the height of the rock to be at 13.9 feet. 1f) We can not plug in 0 seconds. We would get a negative height. The rock stops when its height reaches zero. a) 1 < x < 1 months b) The parabolic pattern will probably repeat each year. We will not be able to extrapolate with this function as the solar energy will increase again during the summer. c) kwh of power at mid-march d) kwh of power at mid-october ef) We can not extrapolate as the quadratic model will be always decreasing for months 13 and above. The answer will have a lot of error. 3a) 3 < x < 50 hours 3b) Yes. It will probably continue to follow the parabolic pattern outside the scope of the data. The more hours the employees work will drive costs up even higher. 3c) We predict costs of $5,930 if the employees work 40 hours per week. 3d) We predict costs of $7411 if the employees work 35 hours per week. 3e) The data will probably follow the parabolic pattern. If the employees work 30 hours they will not meet demand and costs will increase. If we extrapolate, we predict costs of $31979 if the employees work 30 hours per week. 3f) We should not excessively extrapolate.

8 Unit 5 Review Sheet Answers 1a) The scatterplot shows an exponential growth pattern, hence a LOG(Y) transformation seems appropriate. The equation of the exponential function is LN (y) = (x). Solving for y gives : y = e^( (x)) A quadratic model may also fit in the scope of the data. 1b) The exponential function does fit the data well. 1c) The data is quantitative and ordered pair with the number of eagle pairs given in each year. The scatterplot does not show any influential outliers. The points in the scatterplot are close to the exponential curve. With a high r-squared value and low standard error. However the residual plot shows an S pattern indicating that perhaps a cubic function may fit the data better. The residual plot does not show a fan shape, so it does meet the homoscedasticity requirement. The histogram of the residuals shows a bell shape with center not exactly at zero, but at least close to zero. 1d) Using excel, the standard deviation of the residual errors = So the points in the scatterplot are an average of from the exponential curve. Also if we use the exponential equation to predict the number of eagle pairs, we could have an average error of 188 too high or too low. 1e) r-squared was 98.7%. So 98.7% of the variability in eagle pairs can be explained by the exponential relationship with time (years). Confounding variables may include the cutting down of forests, as well as pesticides and poachers. 1f) The x values range between year 13 and year 50. This represents the years between 1963 and 000. We predict there were about 3793 eagle pairs in 1993 (year 43). 1g) We should not use this data to predict the number of eagles in 030, as this would be excessive extrapolation. 1h) In order to solve for y during the LOG(Y) transform, we will have to make both sides the exponent on 10. Hence y is exponential in nature. a) The quadratic function is y x 6.58x. The quadratic function fits the eagle data well. The points in the scatterplot are close to the curve. It also has a high r-squared value and a relatively low standard error. b) The data is quantitative and ordered pair with the number of eagle pairs given in each year. The scatterplot does not show any influential outliers. The points in the scatterplot are close to the exponential curve, with a high r-squared value and low standard error. However the residual plot shows an S pattern indicating that perhaps a cubic function may fit the data better than the quadratic. The residual plot does not show a fan shape, so it does meet the

9 homoscedasticity requirement. The histogram of the residuals shows a bell shape with center at zero. c) the standard deviation of the residual errors = 139.9, so if we use this quadratic function to predict the number of eagle pairs, we could have an average error of 140 eagle pairs. Also points in the scatterplot are about from the quadratic curve. d) r-squared was 99.5%. So 99.5% of the variability in eagle pairs can be explained by the quadratic relationship with time (years). e) The x values range between year 13 and year 50. This represents the years between 1963 and 000. We predict there were about 5715 eagle pairs in 1998 (year 48). This prediction could be off by an average of 140 eagle pairs too many or too few. f) We should not use this data to predict the number of eagles in 05 as this would be excessive extrapolation. gh) The quadratic curve predicts that the fewest eagle pairs happened in year This corresponds to the year In this year, the eagles came closest to extinction. It is estimated that the number of eagle pairs during 1969 fell to only 9. i) When comparing the quadratic and exponential functions, we have to consider many requirements. Both functions had high r-squared values, though the quadratic functions r- squared value was slightly higher and the standard deviation of the residual errors was slightly lower than the exponential function. Both functions met the requirements of regression, though the quadratic function s histogram was more bell shaped and better centered at zero. Overall the quadratic seems a better fit for the eagle data. 3a) The scatterplot shows an exponential or logarithmic decay pattern, hence a decreasing log function LOG(x) transformation seems appropriate. The equation of the log function is Y = LN (x). A quadratic model may also fit in the scope of the data. 3b) The natural log function does fit the data well. It has a reasonably high r-squared value (86.9%), though the standard error is a little higher than expected (1036). 3c) The data is quantitative and ordered pair with the number of drunk driving fatalities given in each year. The scatterplot does show some points a little bit far from the curve but they are not overly influential and seem to follow the log pattern. The points in the scatterplot are close to the log curve, with a high r-squared value. However the residual plot shows an S pattern indicating that perhaps a cubic function may fit the data better. The residual plot does not show a fan shape, so it does meet the homoscedasticity requirement. The histogram of the residuals is skewed right and not centered at zero. 3d) the standard deviation of the residual errors = This indicates that points in the scatterplot were from the log curve on average. If we use the Log curve to predict the

10 number of drunk driving fatalities, we could have an average error of 1036 too many or too few. 3e) r-squared was 86.9%. So 86.9% of the variability in fatalities by drunk drivers can be explained by the logarithmic relationship with time (years). Confounding variables may include the level of intoxication, the weight of the individual, the traffic pattern, or the number of cars on the road. 3f) The x values range between year and year 5. This represents the years between 198 and 005. The log function estimates that the number of drunk driving fatalities during 003 was about This prediction could be off by as many as 1036 fatalities. 3g) We should not use this data to predict the number of drunk driving fatalities in 040, as this would be excessive extrapolation. 3h) The y value is equal to a number times LN (x) plus another number. That makes the y logarithmic in nature. 4a) The quadratic function is y x 0.74x. The quadratic function fits the drunk driving data well. The points in the scatterplot are relatively close to the curve. It also has a high r-squared value and a relatively low standard error. 4b) The data is quantitative and ordered pair with the number of drunk driving fatalities given in each year. The scatterplot does not show any influential outliers. The points in the scatterplot are close to the quadratic curve, with a high r-squared value and a relatively low standard error. However the residual plot shows an S pattern indicating that perhaps a cubic function may fit the data better than the quadratic. The residual plot does not show a fan shape, so it does meet the homoscedasticity requirement. The histogram of the residuals skewed right and not centered at zero. 4c) the standard deviation of the residual errors = If we use this quadratic function to predict the number of drunk driving fatalities, we could have an average error of 85 deaths. Also points in the scatterplot are about from the quadratic curve. 4d) r-squared was 9.1%. So 9.1% of the variability in drunk driving fatalities can be explained by the quadratic relationship with time (years). 4e) The x values range between year and year 5. This represents the years between 198 and 005. The quadratic function estimates that the number of drunk driving fatalities during 003 was about This prediction could be off by as many as 84 fatalities on average. 4f) We should not use this data to predict the number of fatalities in 030 as this would be excessive extrapolation. 4gh) The quadratic curve predicts that the fewest drunk driving fatalities happened in year This corresponds to the year 00. In this year, the function predicts the number of

11 drunk driving fatalities reached a minimum. It is estimated that the minimum number of fatalities during 00 fell to deaths. The quadratic function predicts that the number of deaths will increase after 00, while the log curve predicts they will continue to decrease. This may be something to consider when looking at the future use of these functions. 4i) When comparing the quadratic and log functions, we have to consider many requirements. Both functions had high r-squared values, though the quadratic functions r-squared value was slightly higher. The standard deviation of the residual errors for the quadratic function was significantly lower than the log function. Both functions met most of the requirements of regression, though both histograms were not bell shaped and not centered at zero. Overall the quadratic seems a better fit for the data based on the lower standard error.