We conduct a simple regression of size on hhinc. The regression output is reported in Table 1. Interpret the sign and magnitude of the estimated coefficients. Is the sign of the estimated coefficient of hhinc in line with your answer to a)? Table 1 regress size hhinc Source df MS Number of obs 5,407 F(1, 5405) 1613.34 %3D Model 263141566 1 263141566 Prob > F 0.0000 %3D Residual R-squared Adj R-squared %3D Total 1.1447e+09 5,406 211749.457 Root MSE size Coefficient Std. err. P>|t| [95% conf. interval] hhinc .0082545 .0002055 40.17 0.000 0078516 0086574 _cons 800.9835 9.404917 85.17 0.000 782.5461 819.4209

The Quantitative Environmental Learning Project reported widths and lengths (in centimeters) of a sample of 88 Puget Sound butter clams. Output is shown for the regression of length on width.
Pearson correlation of Width and Length = 0.989
The regression equation is
Length = 0.257 + 1.22 Width
Predictor Coef SE Coef T P
Constant 0.25689 0.09293 2.76 0.007
Width 1.22013 0.01940 62.89 XXXXX
S = 0.3253 R-Sq = 97.9% R-Sq(adj) = 97.8%
   Assume the data to be a representative sample taken from the population of all Puget Sound butter clams, with 
b0, b1, and s
 as reported in the output. Which of these do we know exactly for the population?
 β0 
β1 
 σ
 all of the above
 none of the above

The Quantitative Environmental Learning Project reported widths and lengths (in centimeters) of a sample of 88 Puget Sound butter clams. Output is shown for the regression of length on width.
Pearson correlation of Width and Length = 0.989
The regression equation is
Length = 0.257 + 1.22 Width
Predictor Coef SE Coef T P
Constant 0.25689 0.09293 2.76 0.007
Width 1.22013 0.01940 62.89 XXXXX
S = 0.3253 R-Sq = 97.9% R-Sq(adj) = 97.8%   How do we denote typical vertical distance of clam length from the regression line for the population of all butter clams?
 s
 ŷ 
 b0 
β1
 β0
 σ
 r
 b1
 μ y

The Quantitative Environmental Learning Project reported widths and lengths (in centimeters) of a sample of 88 Puget Sound butter clams. Output is shown for the regression of length on width.
Pearson correlation of Width and Length = 0.989
The regression equation is
Length = 0.257 + 1.22 Width
Predictor Coef SE Coef T P
Constant 0.25689 0.09293 2.76 0.007
Width 1.22013 0.01940 62.89 XXXXX
S = 0.3253 R-Sq = 97.9% R-Sq(adj) = 97.8%   Suppose we wanted to use the same data set (butter clam widths and lengths) to set up a confidence interval to estimate how much longer than wide butter clams tend to be. What would be the appropriate procedure?
- -regression
 - paired t 
 -two-sample t
 -chi-square
 -several-sample F

Refer to the data in images.  Test for a significant difference in the variances of theinitial white blood cell count between patients who did andpatients who did not receive a bacterial culture.

In a study of the performance of a new engine design, the weight of 12 cars (in pounds) and the top speed (in mph) were recorded. A regression line was generated and shown to be an appropriate description of the relationship. The results of the regression analysis are below. Depend Variable: Top Speed Variable Constant Coefficient 107.58 s.e. of Coeff t-ratio prob Weight 0.8710 11.12 0.4146 9.67 0.000 2.10 0.062 R squared = 30.6% R squared (adjusted) = 23.7% s = 10.42 with 12-2 = 10 degrees of freedom Part A: Provide the regression equation based off the analysis provided & explain it in context. Part B: List the conditions for inference that need to be verified. Assuming these conditions have been met, does the data provide convincing evidence of a relationship between weight and top speed? Part C: Assuming all conditions for inference have been verified, determine a 95% confidence interval estimate for the slope of the regression line.

OLS Regression is intended to determine which variables cause the variance of a dependent variable, but regression models actually cannot differentiate between cause and effect. A. True B. False

A manager at a local bank analyzed the relationship between monthly salary and three independent variables: Length of service (measured in months), Gender (0 = female, 1 = male), and Job type (0 = clerical, 1 = technical). The following ANOVA summarizes the regression results.ANOVA Source of Variation df Sum of Squares Mean Square F Regression 3 1,004,346.771 334,782.257 5.96 Residual 26 1,461,134.596 56,197.48445   Total 29 2,465,481.367         Coefficients Standard Error t-Stat p-value Intercept 784.92 322.25 2.44 0.02 Service 9.19 3.20 2.87 0.01 Gender 222.78 89.00 2.50 0.02 Job −28.21 89.61 −0.31 0.76 The level of significance is 0.05. Based on the hypothesis tests for the individual regression coefficients, ________.   Multiple Choice   all the regression coefficients are not equal to zero   "Job" is the only significant variable in the model   only months of service and gender are significantly related to monthly salary   "Service"…

The regional manager of a franchise business is interested in understanding how income in a region affects sales. Below is a regression output for sales ($’000) regressed on the average household income of an area ($’000) Linear Fit Sales = 14.5774 + 2.9048*Income Summary of Fit RSquare 0.9683 RSquare Adj 0.9630 Root Mean Square Error 3.1083 Mean of Response 43.6250 Analysis of Variance Source DF Sum of Squares Mean Square F Ratio Model 1 1771.9048 1771.90 183.3946 Error 6 57.9702 9.66 Pro>F C. Total 7 1829.8750 ItI Intercept 14.5774 2.4101 6.05 0.0009* Income 2.9098 0.2145 13.54 < 0.0001* Answer the following questions: (i) What is the average sales across all regions? (ii) Interpret the slope of regression (iii) What is the prediction of the value of sales in a region with an average…

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 Interpret the standard deviation of the regression model. a) We expect most of the sampled diamond prices to fall within $1117.56 of their least squares predicted values. b) We can explain 89.25% of the variation in the sampled diamond prices around their mean using the size of the diamond in a linear model. c) For every 1-carat increase in the size of a diamond, we estimate that the price of the diamond will increase by $1117.56. d) We expect most of the sampled diamond prices to fall within $2235.12 of their least squares predicted values.