Analysis of Variance Table Response: Price
Df Sum Sq Mean Sq F value Pr(>F)
Living.Area 1 1.3501e+12 1.3501e+12 362.0394 < 2e-16 ***
Bedrooms 1 2.3642e+10 2.3642e+10 6.3394 0.01241 *
Fireplaces 1 7.6232e+07 7.6232e+07 0.0204 0.88642
Residuals 259 9.6588e+11 3.7293e+09
--- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 >
Using= 0.05, perform an F test of overall linear relationship. State the hypotheses, the value of F-test statistic, p-value, and your conclusion.
The test is based on the statistic F=[SS(Regression)/3]/[SS(Residual)/259]
Then under the null, F~F3,259
From the output,
SS(Regression)=1.3501e+12+2.3642e+10+7.6232e+07=1.373818e+12
SS(Residual)=9.6588e+11
Observed F=(1.373818e+12/3)/(9.6588e+11/259) =122.7961
F.05,3,259=2.639455
p value=P(F3,259>122.7961)= 0.00000 approximately
Thus the predictor variables under consideration are significantly worthwhile in predicting Price.
Analysis of Variance Table Response: Price Df Sum Sq Mean Sq F value Pr(>F) Living.Area 1 1.3501e+12 1.3501e+12 362.0394 < 2e-16 *** Bedrooms 1 2.3642e+10 2.3642e+10 6.3394 0.01241 * Fireplaces...
Using R output provided
1). Perform hypothesis testing for B(beta)1=2 using
A(alpha)=0.05
> summary(ls) Call: Residuals: Min 1Q Median 3Q Max 0.20283 -0.14691 -0.02255 0.06655 0.44541 Coefficients: (Intercept) 0.365100.099043.686 0.003586 ** Signif. codes: 0 '***' 0.001 '0.01 '*'0.05 '.' 0.1''1 Estimate Std. Error t value Pr>Itl) 0.96683 0.18292 5.286 0.000258** Residual standard error: 0.1932 on 11 degrees of freedom Multiple R-squared 0.7175, Adjusted R-squared: 0.6918 F-statistic: 27.94 on 1 and 11 DF, p-value: 0.0002581 anovaCLs) Analysis of Variance Table Response:...
A company manager is interested in analyzing the relationship between years of working experience and the salary of their employees. He has collected the data from 30 employees of their years of experience and the salary. Below provided is a partial regression output from R. Use the provided information to answer below questions Coefficients: (Intercept) YearsExperience Estimate Std. Error t value Pr(>ltl) 25792.2 2273.1 9450.0 --- Signif. codes: O '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1'' 1 Analysis of...
2. 2. After we fit the model, the R commander output is provided below. Coefficients: (Intercept) -5.128e+03 1.103e+02 46.49 2e-16** Estimate std. Brror t value Pr(lt|) TEMP PERT TEM: FERT 1.45se-01 9.692e-03 -15.01 1.06e-12 3.110e+01 1.344e+00 23.13 2e-16* 1.397e+02 3.140e+00 44.51 < 2e-16** TEMPSQ FERTSO -1.334e-01 6.853e-03 19.46 6.46e-15 -1.144e+00 2.741e-02 41.74 <2e-16 signif. codes: 00.001 0.01 0.05 011 Residual standard error: 1.679 on 21 degrees of freedom Multiple R-squared: 0.993, F-statistic: 596.3 on 5 and 21 DF, p-value: 2.2e-16...
please provide the output in R
aov (hw labels) >summary (fit) > fit Df Sum Sq Mean Sq F value Pr(>F) 2 493.3 labels 246.7 5.763 0.0176 Residuals 12 513.6 42.8 Signif. codes: 0 0.001 0.01 0.05 0.1 1 12. For this problem, you will use R to conduct an ANOVA F-test. The R code from #11 should be useful for this problem. In #11, the dean only collected samples of size five from each engineering majors. Suppose he collects...
6. Consider the additive two way analysis of variance model where Σα:la.-012b1ßi = 0 and the Eij are independent normal randonn vari- ables with zero means and variance ơ2.Let i-1 (a) Show that are unbiased estimators of the respective parameters. (b) (Devore, 1987) In an experiment to assess the effect of the angle of pull on the force required to separate electrical connectors, four different angles (factor A) were used and each of 5 connectors (factor B) were pulled once...
(a) Using the above t-test data to determine whether or not there
is a linear relationship between the two variables.
(b) Using the above ANOVA F-test data to determine whether or not
there is a linear relationship between the two variables.
(c) How do the results in (a) compare to those in (b)?
We were unable to transcribe this imageAnalysis of Variance Table Response: DatSGPA Dat $ACT 1 3. 588 3. 5878 9. 2402 0.002917 Df Sum Sq Mean sq...
1.-Interpret the following regression model Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -7.819e+05 7.468e+04 -10.470 < 2e-16 *** Lot.Size -5.359e-01 1.163e-01 -4.610 4.67e-06 *** Square.Feet 1.108e+02 1.109e+01 9.986 < 2e-16 *** Num.Baths 2.985e+04 9.650e+03 3.094 0.00204 ** API.2011 1.226e+03 9.034e+01 13.568 < 2e-16 *** dis_coast -7.706e+00 2.550e+00 -3.022 0.00259 ** dis_fwy 1.617e+01 1.232e+01 1.312 0.18995 dis_down 5.364e+00 3.299e+00 1.626 0.10429 I(dis_fwy * dis_down) -4.414e-04 5.143e-04 -0.858 0.39098 Pool 1.044e+05 2.010e+04 5.194 2.59e-07 *** --- Signif. codes: 0 ‘***’ 0.001...
The R code will help to answer
the question.
8. DeGroot&Shervish (2002) consider an experiment to study the combined effects of taking a stimulant and a tranquilizer. In this experiment three types of stimulant and four types of tranquilizer are administered to a group of rabbits. Each rabbit received one of the stimulants, then 20 minutes later, one of the tranquilizers. One hour later their response time (in microseconds) to a stimulus was measured. The results were: Tranquilizer Stimulant 1...
2.-Interpret the following regression model Call: lm(formula = Sale.Price ~ Lot.Size + Square.Feet + Num.Baths + API.2011 + dis_coast + I(dis_fwy * dis_down * dis_coast) + Pool, data = Training) Residuals: Min 1Q Median 3Q Max -920838 -84637 -19943 68311 745239 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -7.375e+05 7.138e+04 -10.332 < 2e-16 *** Lot.Size -5.217e-01 1.139e-01 -4.581 5.34e-06 *** Square.Feet 1.124e+02 1.086e+01 10.349 < 2e-16 *** Num.Baths 3.063e+04 9.635e+03 3.179 0.00153 ** API.2011 1.246e+03 8.650e+01 14.405 < 2e-16...
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