The output shown was obtained from MINITAB. Predictor Constant x S 2.167 R-Sq 91.3%

Chapter 14, Problem 25

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The output shown was obtained from MINITAB. Predictor Constant x S 2.167 R-Sq 91.3% R-Sq(adj) 90.6% Coef 12.396 1.3962 StDev 1.381 0.1245 T 8.97 11.21 P 0.000 0.000 The regression equation is y 12.4 1.40 x (a) The least-squares regression equation is yn = 1.3962x + 12.396. What is the predicted value of y at x = 10? (b) What is the mean of y at x = 10 ? (c) The standard error, se, is 2.167. What is an estimate of the standard deviation of y at x = 10? (d) If the requirements for inference on the least-squares regression model are satisfied, what is the distribution of y at x = 10? 25. Zestimate Sale Price362 370309 315365.5 371.9215 218184 186.5252.5 260247.5 250.8244 251Source: zillow.com(a) Draw a scatter diagram of the data, treating the Zestimateas the explanatory variable and sale price as the responsevariable.(b) Determine the least-squares regression line. Test whetherthere is a relation between the Zestimate and sale price at thea = 0.05 level of significance.click Excel Options. Select Add-Ins. In the Add-Insbox, select Analysis ToolPak. Click OK.2. Enter the explanatory variable in column A andthe response variable in column B.3. Select the Data menu and then select DataAnalysis . . . .4. Select the Regression option.5. With the cursor in the Y-range cell, highlight therange of cells that contains the response variable.With the cursor in the X-range cell, highlightthe range of cells that contains the explanatoryvariable. Click OK.StatCrunchHypothesis Test on the Slope1. Enter the explanatory variable in column var1 andthe response variable in column var2.2. Select Stat>Regression>Simple Linear. Choosevar1 for the X-variable, choose var2 for theY-variable. Click Next.3. Select the Hypothesis Tests radio button. Choosethe appropriate values in the null hypothesis forboth the intercept and slope. Choose the directionof the alternative hypothesis. Click Calculate.Confidence Interval for the Slope1. Enter the explanatory variable in column var1 andthe response variable in column var2.2. Select Stat>Regression>Simple Linear. Choosevar1 for the X-variable, choose var2 for theY-variable. Click Next.3. Select the Confidence Intervals radio button.Choose the confidence level. Click Calculate.Technology Step-By-StepTesting the Least-Squares Regression ModelTI-83/84 PlusHypothesis Test on the Slope 1. Enter the explanatory variable in L1 and theresponse variable in L2. 2. Press STAT, highlight TESTS, and selectF:LinRegTTest.... 3. Be sure that Xlist is L1 and Ylist is L2. Makesure that Freq: is set to 1. Select the directionof the alternative hypothesis. Place the cursor onCalculate and press ENTER.Confidence Interval for the Slope 1. Enter the explanatory variable in L1 and theresponse variable in L2. 2. Press STAT, highlight TESTS, and selectG: LinRegTint .... 3. Be sure that Xlist is L1 and Ylist is L2. Make surethe Freq: is set to 1. Select the confidence level.Highlight Calculate. Press ENTER.MINITAB 1. With the explanatory variable in C1 and theresponse variable in C2, select the Stat menuand highlight Regression. Highlight and selectRegression . . . . 2. Select the explanatory variable (MINITAB callsthem predictors) and response variable andclick OK.Excel1. Make sure the Data Analysis Tool Pack is activatedby selecting the Microsoft Office Button, and then(c) A home with a Zestimate of $370,000 recently sold for$150,000. Determine the least-squares regression linewith this home included. Test whether there is a relationbetween the Zestimate and sale price at the a = 0.05 level ofsignificance. Do you think this observation is influential?