The defect test results of the regression model are reported as follows: Modified Wald test for groupwise heteroskedasticity in fixed effect regression model HO: sigma (i)^2 = sigma^2 for all i chi2 (2094) = 2.1e+05 Prob>chi2 = 0.0000 What defects does the model have? How to fix it?
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- Olympic Pole Vault The graph in Figure 7 indicates that in recent years the winning Olympic men’s pole vault height has fallen below the value predicted by the regression line in Example 2. This might have occurred because when the pole vault was a new event there was much room for improvement in vaulters’ performances, whereas now even the best training can produce only incremental advances. Let’s see whether concentrating on more recent results gives a better predictor of future records. (a) Use the data in Table 2 (page 176) to complete the table of winning pole vault heights shown in the margin. (Note that we are using x=0 to correspond to the year 1972, where this restricted data set begins.) (b) Find the regression line for the data in part ‚(a). (c) Plot the data and the regression line on the same axes. Does the regression line seem to provide a good model for the data? (d) What does the regression line predict as the winning pole vault height for the 2012 Olympics? Compare this predicted value to the actual 2012 winning height of 5.97 m, as described on page 177. Has this new regression line provided a better prediction than the line in Example 2?The number of pounds of steam used per month by a chemical plant is thought to be related to the average ambient temperature (in F) for that month. The past year’s usage and temperatures are in the following table: Assuming that a simple linear regression model is appropriate, fit the regression model relating stem usage (y) to the average temperature (x). What is the estimate of Sigma2? What is the estimate of expected stem usage when the average temperature is 55 F? What change in mean stem usage is expected when the monthly average temperature changes by 1 F? Suppose that the monthly average temperature is 47 F. Calculate the fitted value of y and the corresponding residual. Test for significance of regression using α=0.01 (Use ANOVA). Calculate the r2 of the model. Find a 99% CI for B1 .The following result perspective in RapidMiner shows a multiple linear regression model. Based on the diagram, the model for our dependent variable Y is Predicted Y= (Insulation *0.420)+(Temperature *0.071)+(Avg_Age*0.065)+(Home_Size *0.311)+7.589 Attribute Insulation Temperature Avg Age Home Size (Intercept) O True O False Coefficient 3.323 -0.869 1.968 3.173 134.511 Std. Error 0.420 0.071 0.065 0.311 7.589 Std. Coefficient 0.164 -0.262 0.527 0.131 ? Tolerance 0.431 0.405 0.491 0.914 ? t-Stat 7.906 -12.222 30.217 10.210 17.725
- The least-squares regression equation is y=784.6x+12,431 where y is the median income and x is the percentage of 25 years and older with at least a bachelor's degree in the region. The scatter diagram indicates a linear relation between the two variables with a correlation coefficient of 0.7962. Predict the median income of a region in which 25% of adults 25 years and older have at least a bachelor's degree.One is interested in the ceteris paribus relationship between the dependent variable y, and the explanatory variable ₁1. For this, one collects data on two control variables Xi2 and Xiz and runs two LS regressions. Regression 1: yi = B₁x₁1 + Ui Regression 2: y = B₁xil+ B₂x₁2 + B3x13 + Ei Let ₁ denote the LS estimate of ₁ of Regression 1 and let ₁ denote the LS estimate of ₁ in Regression 2. Assume that the true model is given by Regression 2 and that all variables are centered. a) Would you expect a difference between ₁ and ₁ when is highly correlated with 2 and 3 and the partial effects of xi2 and xi3 on yi are also high? b) Would you expect a difference between ₁ and 3₁ when ₁ has almost no correlation with and Xi3 but X2 and 3 are highly correlated? Xi2 c) Which of the two estimators is more efficient if x₁1 is highly correlated with 2 and 13, and Xiz have small partial effects on yi? but i2 d) Which of the two estimators is more efficient if x₁1 is almost uncorrelated with 2 and…Define the ADL and GLS Estimators of Regression.
- A study was conducted to see if a person's income will affect their well-being. We want to create a model with Y = well-being index, and X = income in $1,000. The regression model is: y = beta0 + beta1*x %D y = beta0 + beta1*x + error %3D y_hat = beta0_hat + beta1_hat*x + error O y_hat = beta0_hat + beta1_hat*xInterpreting technology: The following display from the TI-84 Plus calculator presents the least-squares regression line for predicting the price of a certain stock y from the prime interest rate in percent x LinReg y=+abx a=2.29525776 b=0.38970413 r^2=0.4319044662 r=0.65719439 Regression line equation: 2.295225776+.38970413x What is the correlation between the interest rate and the yield of the stock?Compare the simple linear model in Question 2 and the exponential model in Question 3, do you observe any improvements after conducting the exponential model relative to the linear model? Support your conclusion with details (such as R-Square, homogeneity of variance and normality test) this is Q2 and my answer to it: Fit the simple linear regression model TIME = β0 + β1ENZ + ε. Write down the estimated regression function and examine the residual plot and normality test. Describe what you observed and make brief comments. Hint: you need to check the ANOVA table (that is the F-Statistic and its p-value on the last line of the summary(yourmodel) output)), parameter estimates tables, R-Square, residual plot and normality test. Coefficients: (Intercept) enz -108.716 3.967 TIME= -108.716 + 3.967 (ENZ) + ε By checking normality using the Shapiro-Wilk test, the p-value is 0.0000 less than 0.05. The null is rejected, and the data is not normally distributed.…
- :"RegressionIt is argued that less time spent on social media will result in improved course marks among ECO242 students. To test whether this is the case you collect data from 20 students on their final marks (Y) and number of facebook posts during the semester (X). You make the following calculations:∑XY=9057,∑X2=2470 ∑X = 190 ∑Y=1164Next, you run the following regression:marks=β^1+β^2facebookpostswhere 'marks' is the final mark for the course in percentage points, and 'facebookposts' is the number of facebook posts completed during the course. Question 1 .The value for the slope parameter is? Question 2 The value for the intercept parameter is? Question 3 If the standard error for the intercept parameter estimate is 1.43701, construct a 95% confidence interval for the parameter. Pr( ?. ≤β1≤ ?. . )=95% Question 4 .If the standard error for the slope parameter estimate is 0.129308, construct a 95% confidence interval for the parameter. Pr( ? ≤β2≤ ?. )=95% Questioion 5.Assuming…BUS – 173 (Applied Statistics) # Determining if the following statement are True and False. If False, write the correct statement. Also Solve the MCQ 01. The possible range of coefficient of determination (r^2) is 0 to 1 (a) True (b) False 02. The possible range of coefficient of correlation (r) is -1 to +1 (a) True (b) False 03. In the simple linear regression model, b represents the (a) least square estimate of A (b) least square estimate of B (c) Random error term (d) Not givenInterpret the least squares regression line of this data set. Meteorologists in a seaside town wanted to understand how their annual rainfall is affected by the temperature of coastal waters. For the past few years, they monitored the average temperature of coastal waters (in Celsius), x, as well as the annual rainfall (in millimetres), y. Rainfall statistics • The mean of the x-values is 11.503. • The mean of the y-values is 366.637. • The sample standard deviation of the x-values is 4.900. • The sample standard deviation of the y-values is 44.387. • The correlation coefficient of the data set is 0.896. The correct least squares regression line for the data set is: y = 8.116x + 273.273 Use it to complete the following sentence: The least squares regression line predicts an additional annual rainfall if the average temperature of coastal waters increases by one degree millimetres of Celsius.
