Question

NEED QUESTION THREE PLEASE!

Appraised Value ($1000 Area (square feet 1380 3120 3520 1130 1030 1720 3920 1490 1860 3430 2000 3660 2500 1220 1390 76 216 238 69 50 119 282 81 132 228 145 251 170 71 29 1. (30 pts) Using the oil well drilling cost data set posted, fit a simple linear regression (SLR) that uses cost as a function a) Write out the equation of the regression line. Interpret the slope and intercept in the context of this problem. Do they make sense? (b) Test against the hypothesis that the slope parameter is zero 4 different ways (ANOVA, t-test for Bi, t-test for p, and a confidence interval for β) (c) What is the R2 for the SLR you have obtained? What does it mean? (d) Plot the standardized residuals against the independent variable. What can you say about the regression using this graph? (HINT: Are there outliers? Does it seem reasonable to claim the data has a linear fit?) 3. (30 pts) Use the housing appraisal data set to fit a SLR of appraised value (in thousands of dollars) as a function of area (in square feet). Answer all the questions from 1 as they would pertain to this data set. WO

Answer 1

Go to spss, input the data

Analyze-Regression-Linear-Dependent(appraused_value)-dependent(area)-Statistics(CI at 95%)-Save(unstandardized)-Continue-ok

Model Summary R Square 961 Adjusted R Square Std. Error of the Estimate 16.90647 980a 958 a. Predictors: (Constant), area b. Dependent Variable appraised_value ANOVAa Sum of Squares 90906.629 df Mean Square Sig Regression Residua Tota 90906.629 318.046 3715.771 94622.400 13 14 285.829 a. Dependent Variable: appraised _value b. Predictors: (Constant), area

Coefficientsa Standardized Unstandardized Coefficients Coefficients 95.0% Confidence Interval for B Std. Error Beta Sig Lower Bound Upper Bound (Constant) 2.776 98017.834 6.564 087 29.588 10.657 004 016 52.612 area 078 068 a. Dependent Variable: appraised _value Residuals Statisticsa Minimum Maximum Mean Predicted Value Residua Std. Predicted Value Std. Residual Std. Deviation 80.58121 16.29148 1.000 964 50.6891275.9323 143.8000 9.7470418.71026 1.640 1.107 1.155 2.942 15 15 a. Dependent Variable: appraised _value
a. The regression line is appraised_value=-29.588+0.078area

For one percent change in area there will be 0.078(slope) change in appraised_value. However with no change in area, the appraised_value will be -29.588(intercept)

Yes, they make sense

b. The tests are performed using different methods .

pvalue< 0.05 hence we reject null hypothesis

-2.776 doesn't fall in 95% CI hence we reject null hypothesis

From anova table Statistics-F is sig. at 0.05

Analyze-correlate-bivariate-ok

Correlations appraised_val ue area appraised_value Pearson Correlation 980 Sig. (2-tailed) 15 980 15 area Pearson Correlation Sig. (2-tailed) 15 15 **. Correlation is significant at the 0.01 level (2-tailed)

Rho is significant at 0.05

c. R sq . obtained is 0.961 indicating the fit is good

d. Graphs-chart builder-scatter/dot(simple)-standardized residual in y axis and area in x-ok

it is not reasonable to say it has a linear fit from the graph due to many outliers.

2.00000 1.00000 00000O 1.00000 -2.00000 3.00000 1000.00 1500.00 2000.00 2500.00 3000.00 3500.00 4000.00 area