Simple linear regression results:Dependent Variable: DIAMETERIndependent Variable: RATODIAMETER= -1314.0475 + 199.06132 RATOSample size: 13R (correlation coefficient) = 0.97602564R-sq = 0.95262604%3DEstimate of error standard deviation: 45.080042Parameter estimates:Parameter EstimateStd. Err.Alternative DFT-StatP-valueIntercept-1314.0475 110.8098211 -11.858583 <0.0001Slope199.06132 13.3844071114.872629 <0.0001Analysis of variance table for regression model:Source DFSSMSF-statP-valueModel1 449514.92 449514.92 221.19509 <0.0001Error11 22354.312 2032.2102Total12 471869.23 11.37 Repair and replacement costs of water pipes. Refer to theIHS Journal of Hydraulic Engineering (September 2012)5 H20PIPE study of water pipes, Exercise 11.21 (p. 630). Refer, again,to the Minitab simple linear regression printout (p. 631)relating y = the ratio of repair to replacement cost ofcommercial pipe to x = the diameter (in millimeters) ofthe pipe.a. Locate the value of s on the printout.b. Give a practical interpretation of s.

a. The notation s represents the value of the standard error of the model and in the given model the…

Answered: 11.37 Repair and replacement costs of…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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mean of x = 2.882, standard deviation = 1.634 Mean of y = 2.588, standard deviation = 0.246 The correlation coefficient is 0.159 slope coefficient for regression line = .0239 Find the intercept for the regression line.
mean of x = 2.882, standard deviation = 1.634 Mean of y = 2.588, standard deviation = 0.246 The correlation coefficient is 0.159 slope coefficient for regression line = .0239 y intercept = 2.52 write the regression model
To estimate the value of a dependent variable based on the values of one or more independent variables, the process used is called: Correlation Regression Slope Intercept Residual
Compute the predicted value of IPT for the mean value of PSS (static facial expressions (higher PSS)) The regression formula is: The terms of the equation and their values are:
The accompanying Minitab regression output is based on data that appeared in the article "Application of Design of Experiments for Modeling Surface Roughness in Ultrasonic Vibration Turning."+ The response variable is surface roughness (um), and the independent variables are vibration amplitude (um), depth of cut (mm), feed rate (mm/rev), and cutting speed (m/min), respectively. The regression equation is Ra = -0.972 - 0.0312a + 0.557d + 18.3f + 0.00282v Predictor Coef SE Coef Constant -0.9723 0.3923 -2.48 0.015 -0.03117 0.01864 -1.67 0.099 d 0.5568 0.3185 1.75 0.084 18.2602 0.7536 24.23 0.000 0.002822 0.003977 0.71 0.480 S = 0.822059 Source R-Sq = 88.6 R-Sq (adj) = 88.04 DS MS 0.000 Regression Residual Error 4 401.02 100.25 148.35 76 51.36 0.68 Total 80 452.38 (a) How many observations were there in the data set? observations (b) Interpret the coefficient of multiple determination. O 8.0% of the observed variation in feed rate can be explained by the model relationship with vibration…
What is the relationship between diamond price and carat size? 307 diamonds were sampled and a straight-line relationship was hypothesized between y = diamond price (in dollars) and x = size of the diamond (in carats). The simple linear regression for the analysis is shown below: Least Squares Linear Regression of PRICE Predictor Variables Coefficient Std Error т P Constant -2298.36 158.531 -14.50 0.0000 Size 11598.9 230.111 50.41 0.0000 Resid. Mean Square (MSE) R-Squared Adjusted R-Squared 0.8925 1248950 0.8922 Standard Deviation 1117.56 Interpret the coefficient of determination for the regression model. A) We expect most of the sampled diamond prices to fall within $2235.12 of their least squares predicted values. B) For every 1-carat increase in the size of a diamond, we estimate that the price of the diamond will increase by $1117.56. C) There is sufficient evidence to indicate that the size of the diamond is a useful predictor of the price of a diamond when testing at alpha =…
A study on how the time of exercise affects heart rate had the following outputSimple Linear regression results:Dependent Variable: Heart RateIndependent Variable: Exercise timeSample size: 81R (correlation coefficient) = 0.4188R-sq = 0.1754Estimate of error standard deviation: 0.32416Parameter estimates: Parameter Estimate Std. Err. DF T-Stat P-Value Intercept 1.80938 1.9352 79 0.935 0.3524 Slope 0.4752395 0.11594 79 4.099 <0.0001 According to the output, if I exercise for time=163, what should be my heart rate?Use 4 decimal placesAfter exercising everybody has different heart rates, which means there is a lot of variability in heart rates. How much of that variability is explained by exercise time?Use 4 decimal places
Following regression output is for the consumption of water( millions of litres) in a city predicted by temperature (degree Celsius). Summary Output SSxx= 1003.875 SSxy= 16600.75 SSyy= 307135.5 average X= 24.375 average Y= 484.75 Adjusted R squared 0.8761 observations 8 please find: regressions statistics ANOVA intercept/ temperature ( coefficients, standard error, t stat, p value) regression line equation coefficient of correlation point estimate forecast 95% confidence interval estimate (UCL and LCL)
Which of the variables is the indepenent variable and dependent variable for the following question. fit a simple linear regression model to predict latitudes using average monthly range lat= latitudes range= the average monthly range between mean montly maximum and minimum temperatures for a selected set of US cities.
The following regression has been estimated: Dependent Variable: disability 1 Independent Variables: (1) Constant, (2) Person's age (ecage26), (3) male1 "male1" is equal to 1 for males and 0 for females; disability1 is equal to 1 for disabled and 0 for not disabled regress disabilityl ecage26 malel Source | Total | Model | 1137.25995 2 Residual | 9104.04738 47702 disabilityl I ecage26 | malel SS cons df 10241.3073 47704 Coef. Std. Err. .0084218 -.0084567 -.0891858 MS 568.629974 .19085253 .214684457 .0001093 .0040058 .0060068 t 77.04 -2.11 -14.85 47705 Number of obs = F( 2, 47702) = 2979.42 Prob > F R-squared Adj R-squared = Root MSE P>|t| 0.000 0.035 0.000 [95% CI] = = 0.0000 0.1110 0.1110 .43687 .0082076 -.0163082 -.1009592 .0086361 -.0006052 -.0774125 a) Report the estimated values (the actual numbers) of the coefficient estimates (including the constant) and explain what they measure. b) What are some economic reasons why the estimated coefficients might take on the signs (and…
A study was done to compare tree height with trunk thickness. The following output was generated from the regression model.Simple linear regression results:Dependent Variable: Tree HeightIndependent Variable: Trunk SizeHeight = 39.041668 + 8.668677 TrunkSample size: 25R (correlation coefficient) = 0.4418R-sq = 0.1952Estimate of error standard deviation: 10.09977Parameter estimates: Parameter Estimate Std. Err. Intercept 39.041668 15.1736 Trunk 8.668677 3.670058 Assuming the conditions are met test if trunk size is a good predictor of tree height
▼ Set background Clear frame 6.2 2) The program used to create this scatterplot found the line-of-best-fit and reported the r-squared value as r^2 = 0.805 for the relationship between arm-span and height for several individuals. What is the correlation coefficient? Is it positive or negative? Explain how you know. 1850- 1800- 1750- 1700- 00 1650- 1600- 1550- arm-span (cm) height (cm) 160.0 R Sq Linear-0805 : =S Open c 2 IN

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